10.1 Meiosis
connections between the chromatids can be (a)
shown clearly. (b)
(c)
Ater crossing over has occurred the chromatids (d)
condense by supercoiling. The tight pairing
between the homologous chromosomes ends, but
they are still held together at each point where
crossing over has occurred. This is because the
two chromatids o each chromosome remain
closely aligned, but chromatids in dierent
chromosomes are now linked to each other. The
result is an X-shaped, knot-like structure called
a chiasma.
Chiasmata hold homologous chromosomes
together or a while, but then slide to the end o
the bivalent, allowing the chromosomes to move
to opposite poles o the cell.
At one stage in prophase I all of the chromatids of two homologous chromosomes become tightly paired up
together. This is called synapsis.
four chromatids in total, long and thin at this stage centromeres
The DNA molecule of one of the chromatids is cut. A second cut is made at exactly the same point in the
DNA of a non-sister chromatid.
DNA is cut at the same point
in two non-sister chromatids
The DNA of each chromatid is joined up to the DNA of the non-sister chromatid. This has the eect of
swapping sections of DNA between the chromatids.
In the later stages of prophase I the tight pairing of the homologous chromosomes
ends, but the sister chromatids remain tightly connected. When each cross-over
has occurred there is an X-shaped structure called a chiasma.
chiasma
Meiosis I
Homologous chromosomes separate in meiosis.
The frst meiotic division is unique while the second round resembles
mitosis. There are a number o ways in which meiosis I diers rom
mitosis and meiosis II:
443
10 Genetics and evolution (aHl)
pole of cell i) sister chromatids remain associated with each other
AB
ab ii) the homologous chromosomes behave in a coordinated ashion
Ab in prophase
aB
aB iii) homologous chromosomes exchange DNA leading to genetic
Ab recombination
ab
AB iv) meiosis I is a reduction division in that it reduces the chromosome
number by hal.
Figure 4 Random orientation
The processes that result in the creation o genetic variety o gametes
are initiated in meiosis I. The segregation o homologous chromosomes
occurs during anaphase I resulting in two haploid cells, each with only
one copy o each homologous pair.
Independent assortment
Independent assortment ofgenes is due to the random
orientation ofpairs ofhomologous chromosomes in meiosis I.
When Mendels work was rediscovered at the start o the 2 0th century,
the mechanism that causes independent assortment o unlinked genes
was soon identifed. O bservations o meiosis in a grasshopper, Brachystola
magna, had shown that homologous chromosomes pair up during
meiosis and then separate, moving to opposite poles. The pole to which
each chromosome in a pair moves depends on which way the pair is
acing. This is random. Also, the direction in which one pair is acing
does not aect the direction in which any o the other pairs are acing.
This is called independent orientation.
I an organism is heterozygous or a gene, then in its cells one chromosome
in a pair will carry one allele o the gene and the other chromosome will
carry the other allele. In meiosis, the orientation o the pair o chromosomes
will determine which allele moves to which pole. Each allele has a 50 per
cent chance o moving to a particular pole. Similarly or another gene,
located on another chromosome, or which the cell is heterozygous, there
is a 50 per cent chance o an allele moving to a particular pole. Because
there is random orientation o chromosome pairs, the chance o two alleles
coming together to the same pole is 25 per cent (see fgure 4) .
Figure 4 shows why an individual that has the genotype AaBb can produce
our dierent types o gamete: AB, Ab, aB and ab. It also shows why there
is an equal probability o each being produced.
Meiosis II
Sister chromatids separate in meiosis II.
Ater meiosis I, the daughter cells enter meiosis II without passing
through interphase. Meiosis II is similar to mitosis in that the replicated
chromosome is separated into chromatids. Sister chromatids are
separated but they are likely to be non-identical sister chromatids due to
the occurrence o crossing over.
444
10.2 inheritance
10.2 i
Understnding applictions
Unlinked genes segregate independently as a Completion and analysis o Punnett squares or
result o meiosis. dihybrid traits.
Gene loci are said to be linked i on the same Morgans discovery o non-Mendelian ratios in
chromosome. Drosophila.
Variation can be discrete or continuous. Polygenic traits such as human height may
also be infuenced by environmental actors.
The phenotypes o polygenic characteristics
tend to show continuous variation.
Chi-squared tests are used to determine
whether the dierence between an observed
and expected requency distribution is
statistically signicant.
Nture of science Skills
Looking or patterns, trends and discrepancies: Calculation o the predicted genotypic and
Mendel used observations o the natural world phenotypic ratio o ospring o dihybrid
to nd and explain patterns and trends. Since crosses involving unlinked autosomal genes.
then, scientists have looked or discrepancies
and asked questions based on urther Identication o recombinants in crosses
observations to show exceptions to the rules. involving two linked genes.
For example, Morgan discovered non-Mendelian
ratios in his experiments with Drosophila. Use o a chi-squared test on data rom dihybrid
crosses.
Segregtion nd independent ssortment
Unlinked genes segregate independently as a result
o meiosis.
Segregation is the separation of the two alleles of every gene that occurs
during meiosis. Independent assortment is the observation that the alleles of
one gene segregate independently of the alleles of other genes.
Genes found on different chromosomes are unlinked and do segregate
independently as a result of meiosis. However, genes which are on the same
chromosome are linked and therefore do not segregate independently. The
exception is for linked genes that are far apart on the chromosome. Crossing
over between genes occurs more frequently the further the separation of
genes and can make it appear that the genes are unlinked.
The examples discussed below are based on the assumption that different
alleles segregate independently.
445
10 Genetics and evolution (aHl)
Punnett squares for dihybrid traits
Completion and analysis of Punnett squares for dihybrid traits.
In a dihybrid cross, the inheritance o two genes To create a Punnett square:
is investigated together. Mendel perormed
dihybrid crosses. As an example, he crossed Step 1 : determine the genotypes o the parents.
pure-breeding peas that had round yellow seeds
with pure-breeding peas that had wrinkled Step 2: identiy the dierent varieties o
green seeds. gametes the parents can produce. Note by
Mendels principle o segregation, one copy o
All the F (rst-generation) hybrids had round each gene is present in the gamete. A common
1 mistake is or students to include two copies or
no copies.
yellow seeds. This is not surprising, as these
characters are due to dominant alleles. When Step 3: Set up a Punnett grid or your
Mendel allowed the F plants to sel-pollinate, he cross, with as many rows as there are
unique male gametes (sperm) and as many
1 columns as there are unique emale gametes
( eggs) .
ound that our dierent phenotypes appeared in
the F2 generation: Step 4: Fill in the ospring genotypes inside
the table by matching the egg allele at the
round yellow one o the original parental top o the column with the sperm allele rom
phenotypes the row.
round green a new phenotype Step 5: Determine the genotype ratio or the
predicted ospring.
wrinkled yellow another new phenotype
Step 6: Determine the phenotype ratio or the
wrinkled green the other original parental predicted ospring.
phenotype.
The Punnett grid (gure 1 ) shows how a ratio
I the genotype o the F1 hybrids is S sYy, the o F2 phenotypes is predicted, on the basis o
gametes produced by these hybrids could independent assortment. Create a tally chart to
veriy that the predicted phenotypic ratio is:
contain either S or s with either Y or y. The our
9 yellow round: 3 green round: 3 yellow
possible gametes are S Y, S y, sY and sy. I the wrinkled: 1 green wrinkled
inheritance o these two genes is independent, SY Sy sY sy
then the chance o a gamete containing S or s SSYY SSYy SsYY SsYy
SY
will not aect its chance o containing either Y
SSYy SSyy SsYy Ssyy
or y. The chance o a gamete containing each Sy
allele is _12_, so the comb ined chance o co ntaining SsYY SsYy ssYY ssYy
two spe cic a = _41_. This the ory that sY
lleles is _1_ _1_
SsYy Ssyy ssYy ssyy
2 2 sy
the alleles o two genes pass into gametes without Figure 1 Punnett grid for a dihybrid cross
infuencing each other is called independent
assortment.
A Punnett square is a diagram that is used to
help orm predictions about the outcome o a
particular breeding event where independent
assortment o alleles is occurring. It is used to
directly determine the probability o a particular
genotype but can also be used to determine the
probability o a particular phenotype. It is a table
that is used to systematically combine every
possible combination o maternal allele and
paternal allele.
446
10.2 inheritance
Making prdicitions using Punntt squars
Calculation o the predicted genotypic and phenotypic ratio o ofspring o dihybrid
crosses involving unlinked autosomal genes.
Use the following questions to develop your 2 In peas the allele or smooth seed (S) is
skill in dihybrid cross calculations. dominant over the allele or wrinkled seed (s) .
1 A armer has rabbits with two particular The allele or yellow seed ( Y) is dominant over
traits, each controlled by a separate gene. the allele or green seed (y) .
Coat colour brown is completely dominant A pure breeding tall plant with smooth
to white. Tailed is completely dominant to seeds was crossed with a pure breeding short
tail-less. A brown, tailed male rabbit that is plant with wrinkled seeds. All the F1 plants
heterozygous at both loci is crossed with a were tall with smooth seeds. Two o these
white, tail-less emale rabbit. A large number F1 plants were crossed and our dierent
o ospring is produced with only two phenotypes were obtained in the 320
phenotypes: brown and tailed, white and tail- plants produced.
less, and the two types are in equal numbers. How many tall plants with wrinkled seeds
(i) State both parents genotypes and the would you expect to nd?
gametes that are produced by each during 3 In Drosophila the allele or normal wings ( W)
the process o meiosis. is dominant over the allele or vestigial
wings (w) and the allele or normal body (G)
Male genotype: .......................................
Female genotype: ....................................... is dominant over the allele or ebony
Male gametes: ....................................... body ( g) . I two Drosophila with the
Female gametes: ....................................... genotypes Wwgg and wwGg are crossed
(ii) Predict the genotypic and phenotypic ratios together, what ratio o phenotypes is expected
o the F2 generation. Show your working. in the ospring?
excptions to Mndls ruls
Looking or patterns, trends and discrepancies: Mendel used observations o the
natural world to nd and explain patterns and trends. Since then, scientists have
looked or discrepancies and asked questions based on urther observations to
show exceptions to the rules. For example, Morgan discovered non-Mendelian
ratios in his experiments with Drosophila.
Thomas Hunt Morgan discovered non-Mendelian chromosomal theory o inheritance. He believed
ratios in his experiments with the ruit fy, that the variation that he saw in organisms was
Drosophila melanogaster. Morgan wasnt the rst better explained by environmental infuence.
scientist to use the ruit fy as a research organism, However, his own subsequent observations
but his success popularized its use. Many students o the pattern o inheritance o white eyes
o biology will have some experience o breeding led him to reconsider his own perspective. At
ruit fies either actually or virtually. the same time as his results reinorced aspects
o Mendels conclusions, Morgans studies
At the start o his investigations, Morgan was identied exceptions to Mendels principle o
critical o Mendels theory o inheritance and independent assortment.
wasnt convinced by aspects o the emerging
447
10 Genetics and evolution (aHl)
activity Impications of Morgans discovery of sex
inkage
How can the presence o the three
white-eyed fies among 1,200 in Morgans discovery o non-Mendelian ratios in Drosophila.
Morgan's experiment be explained
in the F generation? Ater breeding thousands o Drosophila in his ruit fy room at
C olumbia University, Morgan noticed a single ruit fy with white eyes
1 instead o the normal red colour. He mated this white-eyed specimen
to an ordinary red-eyed fy. Although the rst generation involving
Figure 2 The fy on the right, with the over 1 ,200 ospring was all red-eyed except or three fies, white-eyed
red compound eye, is the common, fies appeared in much larger numbers in the second generation. In
or wild type. The fy on the let is the second generation, approximately three red-eyed fies appeared or
a mutant type known as White every white-eyed fy as predicted by Mendels principle o dominance
Miniature Forked. It has white eyes, and recessiveness. What surprised Morgan was that all o the white-
shorter wings than the normal fy, and eyed fies in the second generation were male. Mendels principle o
the bristles on its ace and body are dominance and recessiveness would predict a three to one ratio o red
distorted and orked. D. melanogaster to white in both males and emales, but all o the emales had red eyes.
has been used or many years in
genetic studies because it is easy to Morgan began to reverse his earlier position to entertain the possibility
raise in large numbers, reproduces that association o eye colour and sex in ruit fies had a physical
rapidly, and many o its mutations basis in the chromosomes. O ne o Drosophilas our chromosome pairs
are easy to spot under a low-powered was thought by other researchers to be used or sex determination.
light microscope Morgans own idea was that gender was determined by quantity o
chromatin. Males possess the XY chromosome pair while fies with
the XX chromosome are emale. S ince the Y chromosome is smaller,
Morgan was technically correct. However, i the actor or eye colour
was located exclusively on the X chromosome, Morgan could use
Mendelian rules or inheritance o dominant and recessive traits to
explain his observations. Further, the chromosomal theory could
explain why sex and eye colour did not assort independently.
Figure 3 Coloured scanning electron linked genes
micrograph o a ruit fy (Drosophila
melanogaster) our wing mutant. Gene loci are said to be linked i they are on the
Two o the mutant's our wings are same chromosome.
visible (blue) on one side o its body
(brown) . The multi-aceted right eye Further investigations led Morgan to discover more mutant traits in
o the fy (red) is also visible. The wild- Drosophila about two dozen between 1 91 1 and 1 91 4. O ne o Morgans
type fy has two wings students showed that a yellow-bodied mutant characteristic was
inherited in the same way as white eyes. Further, these two mutations
were not inherited independently. They were able to establish the notion
o gene linkage through their experiments.
Morgan, and other geneticists working in the early part o the 20th century,
went on to discover a group o genes that were all located on the
X chromosome o Drosophila. B y careul crossing experiments they were
able to show that these genes were arranged in a linear sequence along
the X chromosome. Groups o genes were then assigned to the other
chromosomes in Drosophila, again arranged in a specic sequence. The
same pattern has been ound in other species each particular gene is
ound in a specic position on one chromosome type. This is called the
448
10.2 inheritance
locus of a gene. If two chromosomes have the same sequence of genes
they are homologous. Homologous chromosomes are not usually
identical to each other because, for at least some of the genes on them,
the alleles will be different.
Since Morgans time it has been discovered that all the genes on a
chromosome are part of one D NA molecule. In Drosophila there are
eight chromosomes in diploid nuclei. In males one of these is an X and
another is a Y chromosome. In females two of them are X chromosomes.
The other six chromosomes are common to males and females they are
called autosomes.
D iploid nuclei have two of each type of autosome, so in Drosophila there
are three types of autosome. Geneticists working early in the 20th century
found four groups of linked genes in Drosophila, corresponding to the three
types of autosome and to the X chromosome.
There are two types of linkage autosomal gene linkage, when the
genes are on the same autosome, and sex linkage, when the genes are
located on the X chromosome.
I-X II III IV
0.0 yellow body 1.3 star eyes 2.0 eyeless
1.5 white eyes 6.1 curly wings 3.0 shaven
13.7 crossveinless 13.0 dumpy wings bristles
21.0 singed bristles 26.0 sepia eyes
27.5 tan body
38.1 miniature wings
43.0 sable body 41.0 dichaete wings
48.5 black body 47.0 radius incompletus
51.5 scalloped wings 54.5 purple eyes 47.5 aristapedia
56.7 forked bristles 55.2 apterous
58.2 stubble bristles
57.0 bar eyes 58.5 spineless bristles
67.0 vestigial wings 70.7 ebony body
72.0 lobe eyes
75.5 curved wings
104.5 brown eyes
Figure 4 Map showing linkage groups o the our ruit fy chromosomes. Chromosome 1 is the 60
X chromosome
percent of population 50
types of variaion
40
Variation can be discrete or continuous.
30
The differences between individual organisms are referred to as
variation. Where individuals fall into a number of distinct categories, 20
the variation is discrete or discontinuous. Blood types are an example
of discrete variation. While there are several blood types, there are no 10
in-between categories. Figure 5 shows the frequency of each of the blood
type phenotypes in a population sample from Iceland. 0 AB AB
0 blood type
Figure 5 Blood type distribution in Iceland
449
10 Genetics and evolution (aHl)
ApBp ApBw AwBp AwBwfrequency continuous variation
ApApBpBp ApApBwBp AwApBpBp AwApBwBp
ApBp The phenotypes o polygenic characteristics tend to show
ApApBpBw ApApBwBw AwApBpBw AwApBwBw continuous variation.
ApBw
ApAwBpBp ApAwBwBp AwAwBpBp AwAwBwBp There are examples o inheritance in which two or more genes
AwBp aect the same character. The genes have an additive eect. Mendel
ApAwBpBw ApAwBwBw AwAwBpBw AwAwBwBw discovered an example o this in beans, where a cross between a
AwBw purple-fowered plant and a white-fowered plant gave purple-
fowered plants in the F generation, but when these were sel-
Figure 6 Results of a cross involving
polygenic inheritance 1
1:2:1 pollinated, the expected 3:1 ratio did not occur and instead a range
1:4:6:4:1 o fower colours was seen. This can be explained i there are two
1:6:15:20:15:6:1 unlinked genes, with co-dominant alleles (see gure 6) . Sel-
1:8:28:56:70:56:28:8:1 pollination o the F1 should give ve dierent shades o fower colour
in a ratio o 1 :4:6:4:1 . I the number o unlinked genes, with co-
Figure 7 Pascal's triangle dominant alleles, were larger then there would be more phenotypic
variants. The number and requency o variants can be predicted using
250 alternate rows o Pascals triangle (gure 7) . A requency distribution
200 is shown (gure 8) or a character aected by ve genes with co-
150 dominant alleles. As the number o genes increases, the distribution
100 becomes increasingly close to the normal distribution. Many
50 characters in humans and other organisms are close to the normal
distribution, or example the mass o bean seeds, height in humans
Figure 8 Variation due to polygenic inheritance and intelligence in humans. The closeness to a normal distribution
suggests that more than one gene is involved. This situation is called
Figure 9 polygenic inheritance.
environmntal infun
Polygenic traits such as human height may also be
infuenced by environmental actors.
When the variation due to polygenic inheritance is examined
careully, it is usually ound to be continuous there is a complete
range o variation, rather than the distinct classes that might be
predicted rom Mendelian inheritance. This is because the dierences
in phenotype between the classes are subtle and the eects o
the environment blur these dierences so much that they are
undetectable.
Skin colour in humans is an example o continuous variation. It is
partly due to the environment sunlight stimulates the production o
the black pigment melanin in the skin. It is also due to the infuence
o several genes, so is an example o polygenic inheritance.
Figure 9 shows an image o identical twins who were both
competitive athletes with long-term dierences in their dietary
and exercise regimes. A number o traits which show continuous
variation are visible in the photo including height. Note the
dierences in their height in the second image, despite having
identical genomes.
450
10.2 inheritance
Figure 1 0 shows genetically identical mice that dier in terms o the
nutrition received by their mothers.
The pups o mice on the standard diet generally had golden ur. B ut a
high proportion o those born to mice on the enriched diet had dark
brown ur.
Figure 10 Genetically identical mice showing variation in size and coat colour as a
consequence o environmental diferences in utero
Identifying recombinants
Identifcation o recombinants in crosses involving two linked genes.
William Bateson, Edith Saunders and Reginald Although the observed percentages do not
t the 9:3:3:1 ratio, results like this were not
Punnett discovered the rst exception to the unexpected. Genes were thought to be part
o chromosomes by a number o scientists at
law o independent assortment in 1 903. When the time, and as there are ar more genes than
chromosomes, some genes must be ound
they crossed sweet pea plants with purple together on the same chromosome. Alleles o
these genes would thereore not ollow the law o
fowers and long pollen grains with plants independent assortment and would pass together
into a gamete.
with red fowers and round pollen grains, all
This is seen in the results or the sweet pea cross
the F1 hybrids had purple fowers and long there were more o the purple long and red round
pollen grains. When these F1 plants were sel- plants than expected. This is because these were
pollinated, our phenotypes were observed the original parental combinations o alleles. This
pattern o inheritance is called gene linkage. Since
in the F2 generation, but not in the amiliar 1 903, many more examples have been ound,
9:3:3:1 ratio. The cross was repeated with larger always with a higher requency o the parental
combinations than predicted rom Mendelian ratios.
numbers and the F contained the numbers o
2 A genetic diagram explaining the cross in
sweet peas is shown in gure 1 1 , using lines to
plants shown in table 1 . symbolize the chromosomes on which the linked
genes are located.
Pyp obsvd obsvd 9:3:3:1
fquy % %
purple long
purple round 4,831 69.5 56.25
red long 390 5.6 18.75
red round 393 5.6 18.75
Table 1 1,338 19.3 6.25
451
10 Genetics and evolution (aHl)
purple red was described as a part of meiosis in sub-topics 3.3
and 1 0.1 . Figure 1 2 shows how crossing over
long round gives new allele combinations. The formation of a
pp chromosome or DNA with a new combination of
parents P P alleles is recombination. An individual that has
L L this recombinant chromosome and therefore has a
different combination of characters from either of
gametes P p the original parents is called a recombinant.
L
purple
long
F1 P p
L
44% 6% Pp
L
44% 6%
P pP p DNA replication
LL P Pp p
LL
P P PP pP PP p
L L LL L L L pairing of homologous
chromosomes and crossing over
p p Pp pp Pp p
L L LL L L L
P P PP pP PP p Pp
LL L
p p Pp pp Pp p 1st division
LL P P of meiosis p p
Figure 11 Cross involving gene linkage LL
2nd division
The linkage between pairs of genes in a p
linkage group is not usually complete and new of meiosis
combinations of alleles are sometimes formed.
This happens as a result of crossing over, which P Pp
LL
Figure 12 Formation of recombinants
Data-based questions: Gene linkage in Zea mays
Corncobs are often used for showing inheritance white starchy 279
patterns. All the grains on a cob have the same white waxy 420
female parent, and with careful pollination they
can also have the same male parent. A variety Using this data, deduce whether the genes
with coloured and starchy grains was crossed for coloured/white and starchy/waxy
with a variety with white and waxy grains. The F1 are linked. [2]
grains were all coloured and starchy. The F1 plants A variety with coloured and shrunken grains
grown from these grains were crossed (F F ). was crossed with a variety with white and
1 1
non-shrunken grains. The F1 grains were all
1 Calculate the expected ratio of F2 plants, coloured and non-shrunken. The F plants
assuming that the genes for coloured/white
1
and starchy/waxy grains are unlinked. Use
grown from these grains were test crossed
a genetic diagram to show how you reached
using pollen from a homozygous recessive
your answer. [3]
variety with white shrunken grains.
2 The actual frequencies were: 3 Calculate the expected ratio of F2 plants,
assuming that the genes are unlinked,
coloured starchy 1,774
using a diagram to show how you
coloured waxy 263 [2]
reached your answer.
452
10.2 inheritance
4 The actual requencies were: Using this data, deduce whether the genes
coloured non-shrunken 638 or coloured/white and non-shrunken/
shrunken are linked. [2]
coloured shrunken 21,379 5 Deduce whether the genes or starchy/waxy
and non-shrunken/shrunken are linked. [1 ]
white non-shrunken 21 ,096
white shrunken 672
chi-squared tests are used to determine TOK
whether the dierene between an observed and
expeted requeny distribution is statistially W mg psusv
signifant ssl pso
b psd s fv, o,
Use o a chi-squared test on data rom dihybrid crosses. os, odmd s
mpulv?
In 1 901 , Bateson reported one o the frst post-Mendelian studies o a
cross involving two traits. White leghorn chickens with large single In popular discourse, there is
combs, were crossed to Indian game owl with dark eathers and small a distrust oknowledge claims
pea combs. All o the F1 were white with pea combs, and the ratio o supported by statistics. One
F phenotypes involving 1 90 ospring was: 1 1 1 white pea, 37 white aphorism popularized by Mark
Twain is that there are three kinds o
2 lies: lies, damned lies and statistics.
The misuse ostatistics can be
single, 34 dark pea and 8 dark single. The expected ratio is 9:3:3:1 . The inadvertent or it can be intentional.
observed ratio was dierent. Were the dierences between observed Some examples include:
and expected due to sampling error or were the dierences statistically
signifcant, suggesting that the traits do not assort independently? This conclusions can be based
can be tested using the chi-squared test. on statistical analysis o
samples that are selected
There are two possible hypotheses: with bias and are thereore
not representative o the
H : the traits assort independently population
0
rejection o the alternate
H1 : the traits do not assort independently hypothesis can be mistakenly
interpreted as proo o the null
We can test these hypotheses using a statistical procedure the chi- hypothesis
squared test.
a sample size that is
Method for chi-squared test too small is likely to be
poorly representative o
1 Draw up a contingency table o observed requencies, which are the the population even i it is
numbers o individuals o each phenotype. selected without bias
2 Calculate the expected requencies, assuming independent experimenters may discount
assortment, or each o the our phenotypes. Each expected requency data that they believe does
is calculated rom values on the contingency table using the expected not conorm to theory.
probability rom the Punnett grid multiplied by the actual total.
The efects osuch issues can be
W p W sgl Dk p Dk sgl tol minimized through a diligent and
190 honest approach meaning that it
Observed 111 37 34 8 190 is the user rather than the tool that
Expected needs to be the subject oscrutiny.
( __9___ ) 190 ( )__3___ 190 ( __3___ ) 190 ( )__1___ 190
16 16 16 16
= 106.9 = 35.6 = 35.6 = 11.9
3 Determine the number o degrees o reedom, which is one less than
the total number o classes (4 1 ) = 3 degrees o reedom.
453
10 Genetics and evolution (aHl)
4 Find the critical region or chi-squared rom a table o chi-squared
values, using the degrees o reedom that you have calculated and a
signifcance level (p) o 0.05 (5%) . The critical region is any value o
chi-squared larger than the value in the table.
Critical values of the 2 distribution
p
df 0.995 0.975 0.9 0.5 0.1 0.05 0.025 0.01 0.005 df
1 0.000 0.000 0.016 0.455 2.706 3.841 5.024 6.635 7.879 1
2 0.010 0.051 0.211 1.386 4.605 5.991 7.378 9.210 10.597 2
3 0.072 0.216 0.584 2.366 6.251 7.815 9.348 11.345 12.838 3
At the 0.05 level o signifcance, the critical value is 7.81 5.
5 Calculate chi-squared using this equation:
2 = (_obs - _exp) 2
exp
= _( 1 1 1 1_06.9) 2 + _( 3 7 3_5 . 6) 2 + _( 3 4 3_5 .6) 2 + _( 8 1 1 .9) 2
1 06.9 35.6 35.6 11.9
= 1.56
6 Compare the calculated value o chi-squared with the critical region.
I the calculated value is in the critical region, there is evidence
at the 5% level or an association between the two traits; i.e., the
traits are linked. We can rej ect the hypothesis H0.
I the calculated value is not in the critical region, because it
is equal to or below the value obtained rom the table o chi-
squared values, H0 is not rejected. There is no evidence at the 5%
level or an association between the two traits.
The probability value is outside o the critical region (0.9 >p>0.5) so we
reject the alternate hypothesis and accept the null hypothesis.
Data-based questions: Using the chi-squared test
Warren and Hutt ( 1 93 6) test-crossed a double 1 Construct a contingency table o [4]
heterozygote or two pairs o alleles in hens: observed values.
one or the presence (Cr) or absence (cr) o
a crest and one or white (I) or non-white (i) 2 Calculate the expected values, assuming
plumage.
independent assortment. [4]
For their F2 cross, there was a total o 754
ospring. 3 Determine the number o degrees o [2]
reedom.
337 were white, crested;
4 Find the critical region or chi-squared at a
337 were non-white, non-crested;
signifcance level o 5%. [2]
34 were non-white crested; and
5 Calculate chi-squared. [4]
46 were white, non-crested.
6 State the two alternative hypotheses, H0
and H1 and evaluate them using the calculated
value or chi-squared. [4]
454
10.3 Gene Pools anD sPeciation
10.3 G p d p
Understnding applictions
A gene pool consists o all the genes and their Identiying examples o directional, stabilizing
diferent alleles, present in an interbreeding and disruptive.
population.
Speciation in the genus Allium by polyploidy.
Evolution requires that allele requencies
change with time in populations. Skills
Reproductive isolation o populations can be Comparison o allele requencies o
temporal, behavioural or geographic. geographically isolated populations.
Speciation due to divergence o isolated Nture of science
populations can be gradual.
Looking or patterns, trends and discrepancies:
Speciation can occur abruptly. patterns o chromosome number in some genera
can be explained by speciation due to polyploidy.
Gene pools
A gene pool consists o all the genes and their diferent
alleles, present in an interbreeding population.
The most commonly accepted denition o a species is the biological
species concept. This denes a species as a group o potentially
interbreeding populations, with a common gene pool that is
reproductively isolated rom other species. Some populations o the same
species are geographically isolated so it is possible or multiple gene pools
to exist or the same species.
Individuals that reproduce contribute to the gene pool o the next
generation. Genetic equilibrium exists when all members o a population
have an equal chance o contributing to the uture gene pool.
allele frequency nd evolution
Evolution requires that allele requencies change with
time in populations.
Evolution is dened as the cumulative change in the heritable characteristics
o a population over time. Evolution can occur due to a number o reasons
such as mutations introducing new alleles, selection pressures avouring
the reproduction o some varieties over others and barriers to gene fow
emerging between dierent populations. I a population is small, random
events can also have a signicant eect on allele requency.
455
10 Genetics and evolution (aHl)
activity Patterns of natural selection
In the cross depicted in gure 1, Identiying examples o directional, stabilizing and
the requency ofower colour disruptive selection.
phenotypes in Japanese our
oclocks is shown over three Fitness o a genotype or phenotype is the likelihood that it will be
generations. The genotype CRCR ound in the next generation. Selection pressures are environmental
yields red fowers, the genotype actors that act selectively on certain phenotypes resulting in natural
CW CW yields white fowers selection. There are three patterns o natural selection: stabilizing
and because the alleles are selection, disruptive selection, and directional selection.
co-dominant, the genotype CRCW
yields pink fowers: In stabilizing selection, selection pressures act to remove extreme
varieties. For example, average birth weights o human babies are
in the rst generation, 50% o avoured over low birth weight or high birth weight. A clutch is the
the population is red and 50% number o eggs a emale lays in a particular reproductive event. Small
is white clutch sizes may mean that none o the ospring survive into the
next generation. Very large clutch sizes may mean higher mortality
in the second generation, as the parent cannot provide adequate nutrition and resources and
100% o the fowers are pink may impact their own survival to the next season. This means that a
medium clutch size is avoured.
in the third generation, there
are 50% pink, 25% white and In disruptive natural selection, selection pressures act to remove
25% red. intermediate varieties, avouring the extremes. One example is in the
red crossbill Loxia curvirostra. The asymmetric lower part o the bill
Show that the allele requency is o red crossbills is an adaptation to extract seeds rom conier cones.
50% CR and 50% CW in each othe An ancestor with a straight bill could have experienced disruptive
three generations. While phenotype selection, given that a lower part o the bill crossed to either side
requencies can change between enables a more efcient exploitation o conier cones. Both let over
generations, it is possible that allele right and right over let individuals exist within the same population
requency is not changing. This allowing them to access seeds rom cones hanging in dierent positions.
population is not evolving because
allele requencies are not changing. In directional selection, the population changes as one extreme o a
range o variation is better adapted.
eggs
Dt-bsed questions: Stabilizing selection
CRCR
A population o bighorn sheep ( Ovis canadensis) on Ram Mountain
sperm CRCR CRCW in Alberta, Canada, has been monitored since the 1 970s. Hunters
can buy a licence to shoot male bighorn sheep on the mountain. The
F1 generation CRCW CWCW large horns o this species are very attractive to hunters, who display
all CRCW them as hunting trophies.
CWCW Most horn growth takes place between the second and the ourth
year o lie in male bighorn sheep. They use their horns or fghting
F2 generation other males during the breeding season to try to deend groups o
1:2:1 emales and then mate with them. Figure 2 shows the mean horn
length o our- year- old males on Ram Mountain, between 1 975 and
Figure 1 A change in phenotypic 2002.
frequency between generations
does not necessarily indicate that a) Outline the trend in horn length over the study period.
evolution is occurring
b) Explain the concept o directional selection reerring to this
example.
456
mean horn length /cm 10.3 Gene Pools anD sPeciation
c) Discuss the trade-o between short and long horns as an
adaptation in this case.
80
70
60
50
40
0
1970 1975 1980 1985 1990 1995 2000 2005
year
Figure 2
Source: Reprinted with permission from Macmillan Publishers Ltd: David W. Coltman, Undesirable
evolutionary consequences of trophy hunting, Nature, vol. 426, issue 6967, pp. 655658
D-bd qu a) Identiy the mode value or mass at birth.
Researchers carried out a study on 3,760 children b) Identiy the optimum mass at birth or
born in a London hospital over a period o 1 2 survival.
years. Data was collected on the childrens mass
at birth and their mortality rate. The purpose o c) Outline the relationship between mass at
the study was to determine how natural selection birth and mortality.
acts on mass at birth. The chart in fgure 3 shows
the requency o babies o each mass at birth. The d) Explain how this example illustrates the
line superimposed on the bar chart indicates the pattern o natural selection called stabilizing
percentage mortality rate (the children that did selection.
not survive or more than 4 weeks) .
800 100
frequency of mass at birth600
mortality/% (log scale)
400 10
200
0
0.00 0.50 1.00 1.50 2.00 2.50 3.00 3.50 4.00 4.50 5.00
mass at birth/kg
Figure 3
Source: W H Dowderswell, (1984) Evolution, A Modern Synthesis, page 101
457
10 Genetics and evolution (aHl)
Data-based questions a. ghting
In coho salmon ( Oncorhynchus kisutch), some b. sneaking.
males reach maturity as much as 50% earlier and
as small as 30% o the body size o other males in c) Identiy a size o male sh that never gets
the population. Success in spawning (breeding) within 1 00 cm (1 m) by ollowing either
depends on the male releasing sperm in close s tr a te g y.
proximity to the egg-laying emale. Small and
large males employ dierent strategies to gain d) Explain how this example illustrates the
access to emales. The small-sized males called pattern o natural selection known as
jacks are specialized at sneaking. The large-sized disruptive selection.
males are specialized at ghting and coercing
emales to spawn. In contrast, intermediate-sized proximity to female(cm) 40 2 10 10 5
males are at a competitive disadvantage to both 83 12 ghting
jacks and large males as they are more targeted 40
or ghts which they lose and are more likely
to be prevented rom sneaking. The graph in 120 sneaking 28
gure 4 shows the average proximity to emales 3
achieved by the two strategies.
3 8 3
a) Determine the mean proximity to emales 6
achieved by 3539 cm males by: 200
a. sneaking 2529 3539 4549 5559 6569
male body size(cm)
b. ghting.
Nature, Vol. 313, No. 5997, pp. 4748, 3 January 1985
b) Determine the size range that gets nearest to
the emales by: Figure 4 Efect o body size and courting strategy on proximity
to emales
there are diferen caegories o reproducive
isolaion
Reproductive isolation of populations can be temporal,
behavioural or geographic.
Speciation is the ormation o a new species by the splitting o an
existing population. Various barriers can isolate the gene pool o one
population rom that o another population. Speciation may occur when
this happens. I the isolation occurs because o geographic separation o
populations, then the speciation is termed allopatric speciation.
The cichlids (sh) are one o the largest amilies o vertebrates. Most
species o cichlids occur in three East Arican lakes, Lake Victoria,
Lake Tanganyika and Lake Malawi. Annual fuctuations in water
levels lead to isolation o populations that are then subject to dierent
selection pressures. When the rainy season comes, the populations are
recombined but can then be reproductively isolated. This can result in
the ormation o new species.
Sometimes isolation o gene pools occurs within the same geographic
area. I speciation occurs, then the process is termed sympatric
speciation. For example, isolation can be behavioural. When closely
458
10.3 Gene Pools anD sPeciation
D-bd qu: Lacewing songs (a) 4 5 10 15 20 25 30
Songs are part o the process o mate selection 2
in members o dierent species within the genus 0
Chrysoperla ( lacewings) . Males and emales o the -2
same species have precisely the same song and -4
during the pre-mating period take turns making
the songs. The oscillograph or two species o 0
lacewings are shown in gure 5.
1 Compare the songs o the two species o (b) 4
lacewings. [3] 2
2 Explain why dierences in mating songs 0
might lead to speciation. [3] -2
3 The ranges o the two species currently -4
overlap. Suggest how dierences in song
could have developed: 1 2 3 4 5 6 7 8 9 10 11 12
a) by allopatric speciation [4] Figure 5 Pre-mating songs of lacewings: (a) C. lucasina
b) by sympatric speciation. and (b) C. mediterranea. C. lucasina ranges across most
of Europe and eastward into western Asia, as well as
across the northern quarter of Africa. C. mediterranea
ranges across southern to central Europe and across the
north African Mediterranean
related individuals dier in their courtship behaviour, they are oten
only successul in attracting members o their own population.
There can be temporal isolation o gene pools in the same area.
Populations may mate or fower at dierent seasons or dierent times o
day. For example, three tropical orchid species o the genus Dendrobium
each fower or a single day. Flowering occurs in response to sudden
drops in temperature in all three species. However, the lapse between
the stimulus and fowering is 8 days in one species, 9 in another, and
1 0 to 1 1 in the third. Isolation o the gene pools occurs because, at the
time the fowers o one species are open, those o the other species have
already withered or have not yet matured.
diferent populations have iferent allele requencies
Comparison of allele frequencies of geographically isolated populations.
Online databases such as the Allele Frequency PanI is a gene in cod sh that codes or an
D atabase ( AlFreD ) hosted by Yale University integral membrane protein called pantophysin.
contains the requencies o a variety o Two alleles o the gene, PanIA and PanIB, code
human populations. Most human populations or versions o pantophysin that dier by
are no longer in geographic isolation because our amino acids in one region o the protein.
o the ease o travel and the signicant culture to Samples o cod sh were collected rom
culture contact that exists due to globalization. 23 locations in the north Atlantic and were
Nonetheless, patterns o variation do exist, tested to nd the proportions o PanIA and
especially when comparing remote island PanIB alleles in each population. The results are
populations with mainland populations. shown in pie charts, numbered 1 23, on the
459
10 Genetics and evolution (aHl)
map in fgure 6. The proportions o alleles in
a population are called the allele requencies.
The requency o an allele can vary rom 0.0
to 1 .0. The light grey sectors o the pie charts
show the allele requency o PanIA and the
black sectors show the allele requency
o PanIB.
1 State the two populations with the highest
PanIB allele requencies. [2]
2 Deduce the allele requencies o a population
in which hal o the cod fsh had the genotype
PanIA PanIA, and hal had the genotype PanIA
PanIB. [2]
3 Suggest two populations which are likely
geographically isolated. [2]
4 Suggest two possible reasons why the PanIB
allele is more common in population 1 4 than Figure 6
population 21 . [2] Source: R A J Case, et al., (2005) , Marine Ecology Progress
Series, 201, pages 267278
TOK Gradualism in speciationfrequency
What role does expectation Speciation due to divergence o isolated populationstime
have in determining the can be gradual.
response of scientists to
unexpected discoveries? There are two theories about the pace o evolutionary change. Gradualism,
as depicted in fgure 7, is the idea that species slowly change through a
The coherence test o truth series o intermediate orms. The axis label structure might reer to such
flters knowledge claims things as beak length in birds or cranial capacity in hominids.
through existing theories
that are well established. structure
I the new knowledge
claim does not ft, it is Figure 7 In the gradualist framework, new species emerge from a long sequence of
more likely to be greeted intermediate forms
with skepticism. While
polyploidy does occur in Gradualism was, or a long time, the dominant ramework in
fsh and amphibians, it has palaeontology. However, it was conronted by gaps in the ossil record,
always been unexpected i.e. an absence o intermediate orms. Gradualism predicted that
in mammals. The sex evolution occurred by a long sequence o continuous intermediate
determination system in orms. The absence o these intermediate orms was explained as
mammals is very sensitive imperections in the ossil record.
to extra sex chromosomes.
Since the existence o a
tetraploid mammal was frst
claimed, the response has
been skepticism. Though
there is still no reasonable
answer to the question o the
T. barrerae origin.
460
10.3 Gene Pools anD sPeciation
Punctuated equilibrium gra d u a l i s m
Speciation can occur abruptly. morphology
Punctuated equilibrium holds that long periods of relative stability in a time
species are punctuated by periods of rapid evolution. According to the punctuated equilibrium
theory of punctuated equilibrium, gaps in the fossil record might not be
gaps at all, as there was no long sequence of intermediate forms. Events Figure 8
such as geographic isolation (allopatric speciation) and the opening of
new niches within a shared geographic range can lead to rapid speciation.
Rapid change is much more common in organisms with short generation
times like prokaryotes and insects.
Figure 8 compares the two models. The top model shows the gradualist
model with slow change over geological time. The punctuated
equilibrium model on the bottom consists of relatively rapid changes
over a short period of time followed by periods of stability.
Polyploidy can lead to speciation
Looking for patterns, trends and discrepancies: patterns of chromosome number
in some genera can be explained by speciation due to polyploidy.
A polyploid organism is one that has more than ambiguous: probes detected only two copies of each
two sets of homologous chromosomes. Polyploidy autosome pair but it has also been observed that
can result from hybridization events between there are several genes that exist in four copies.
different species. There are also polyploids whose
chromosomes originate from the same ancestral
species. This can occur when chromosomes
duplicate in preparation for meiosis but then
meiosis doesnt occur. The result is a diploid gamete
that when fused with a haploid gamete produces
a fertile offspring. In other words, the polyploid
has now become reproductively isolated from the
original population. The polyploid plant can self-
pollinate or it can mate with other polyploid plants.
Polyploidy can lead to sympatric speciation.
Polyploidy occurs most commonly in plants, though Figure 9 Tympanoctomys barrerae
it does also occur in less complex animals. The red Figure 10 Octomys mimax
viscacha (Tympanoctomys barrerae), a rodent from
Argentina, has the highest chromosome number of
any mammal and it has been hypothesized that this
is the result of polyploidy. Its chromosome number
is 1 02 and its cells are roughly twice normal size. Its
closest living relative is Octomys mimax, the Andean
viscacha-rat of the same family, whose 2n = 56.
Researchers propose that an Octomys-like ancestor
produced tetraploid offspring (i.e. 4n = 1 1 2) that
were reproductively isolated from their parent
species, eventually shedding some of the additional
chromosomes gained at this doubling. Recent
scholarship has tested this hypothesis but results are
461
10 Genetics and evolution (aHl)
Polyploidy has occurred frequently in Allium
Speciation in the genus Allium by polyploidy.
Estimates of the number of species of angiosperms Many species of Allium reproduce asexually and
that have experienced a polyploidy event range polyploidy may confer an advantage over diploidy
between 50 to 70%. under certain selection pressures.
The Allium genus includes onions, leeks, garlic Wild onion (Allium canadense) is a native of North
and chives, and as such has played an important America. The diploid number for the plant is 1 4.
role in the food of multiple cultures. Determining However, there are variants such as A. c. ecristatum
the number of species in the genus presents a ( 2 n = 2 8) and A. c. lavendulae ( 2 n = 2 8) .
challenge to taxonomists as polyploidy events
are common within the genus. These result in a Allium angulosum and Allium oleraceum are
number of reproductively isolated but otherwise two species that occur in Lithuania. One is a
similar populations. diploid plant with 1 6 chromosomes and one is a
tetraploid plant with 32 chromosomes.
Figure 11 Metaphase chromosomes of Allium angulosum, Figure 12 Metaphase chromosomes of Allium oleraceum,
2n=16 2n=32
462
Questions
Questions (i) Deduce the chromosome number
1 Identiy the stages o meiosis shown in o nuclei in their lea cells. Give two
fgures 1 3 and 1 4.
reasons or your answer. [3]
Figure 13
( ii) S uggest a disadvantage to S. arcticum
and S. olafi o having more D NA than
other bog mosses. [1 ]
d) It is unusual or plants and animals to
have an odd number o chromosomes in
their nuclei. Explain how mosses can have
odd numbers o chromosomes in their
lea cells. [2]
Sphagnum M f nmbr f
pc D n a/p g chrmm
S. aongstroemii 0.47 19
S. arcticum 0.95
S. balticum 0.45 19
S. fmbriatum 0.48 19
S. olafi 0.92
S. teres 0.42 19
S. tundrae 0.44 19
S. warnstorfi 0.48 19
Table 1
Figure 14 3 The mechanisms o speciation in erns have
been studied in temperate and tropical habitats.
2 The DNA content o cells can be estimated One group o three species rom the genus
using a stain that binds specifcally to DNA. A Polypodium lives in rocky areas in temperate
narrow beam o light is then passed through orests in North America. Members o this
a stained nucleus and the amount o light group have similar morphology (orm and
absorbed by the stain is measured, to give an structure) . Another group o our species
estimate o the quantity o DNA. The results in rom the genus Pleopeltis live at dierent
table 1 are or lea cells in a species o bog moss altitudes in tropical mountains in Mexico and
(Sphagnum) rom the Svalbard islands. Central America. Members o this group are
morphologically distinct.
a) Compare the DNA content o the bog
Data rom the dierent species within each
mosses. [2] group was compared in order to study the
mechanisms o speciation.
b) Suggest a reason or six o the species o
Genetic identity was determined by comparing
bog moss on the Svalbard islands all the similarities o certain proteins and genes
in each species. Values between 0 and 1 were
having the same number o chromosomes assigned to pairs o species to indicate the degree
o similarity in genetic identity. A value o 1 would
in their nuclei. [2] mean that all the genetic actors studied were
identical between the species being compared.
c) S. arcticum and S. olafi probably arose as
new species when meiosis ailed to occur in
one their ancestors.
463
10 Genetics anD evolution
a) Compare the geographic distributions o c) Suggest how the process o speciation
the two groups. [1 ] could have occurred in Polypodium. [1 ]
b) ( i) Identiy, giving a reason, which group, d) Explain which o the two groups has most
Polypodium or Pleopeltis, is most probably been genetically isolated or
genetically diverse. [1 ] the longest period o time. [2]
(ii) Identiy the two species that are most
similar genetically. [1 ] 4 In Zea mays, the allele or coloured seed (C) is
dominant over the allele or colourless seed (c) .
The allele or starchy endosperm (W) is dominant
over the allele or waxy endosperm (w) . Pure
breeding plants with coloured seeds and starchy
endosperm were crossed with pure breeding
plants with colourless seeds and waxy endosperm.
Po. sibiricum a) State the genotype and the phenotype o
the F1 individuals produced as a result o
this cross.
genotype .................................................
Po. amorphum 0.435 phenotype ............................................. [2]
0.608
0.338 b) The F1 plants were crossed with plants
that had the genotype c c w w. C alculate
Po. appalachianum
the expected ratio o phenotypes in the
F2 generation, assuming that there is
independent assortment.
Pl. polyepis Expected ratio ...................................... [3]
Pl. crassinervata
The observed percentages o phenotypes in
the F2 generation are shown below.
coloured starchy 37%
colourless starchy 1 4%
coloured waxy 16%
colourless waxy 33%
Pl. conzattii The observed results dier signifcantly
Pl. mexicana rom the results expected on the basis o
independent assortment.
Pl. polyepis c) State the name o a statistical test that could
be used to show that the observed and the
0.925 0.836 expected results are signifcantly dierent.
Pl. conzattii Pl. mexicana [1 ]
d) Explain the reasons or the observed results
0.792 0.870 o the cross diering signifcantly rom the
Pl. crassinervata expected results. [2]
Figure 15 The approximate distribution in North America o
the three species o Polypodium (Po.) and a summary o
genetic identity
Source: C Haufer, E Hooper and J Therrien, (2000) , Plant Species
Biology, 15, pages 223236
464
11 AnImAL phYsIOLOGY (AhL)
CIEtroLduLctioB I O L O G Y
Immunity is based on recognition of self and waste products and some animals also balance
destruction of foreign material. The roles of the water and solute concentrations. Sexual
musculoskeletal system are movement, support reproduction involves the development and
and protection. All animals excrete nitrogenous fusion of haploid gametes.
11.1 Antibody production and vaccination
Udertadig Alicatio
Every organism has unique molecules on the Antigens on the surace o red blood cells
surace o their cells. stimulate antibody production in a person with
a dierent blood group.
B lymphocytes are activated by T lymphocytes
in mammals. Smallpox was the frst inectious disease
o humans to have been eradicated by
Plasma cells secrete antibodies. vaccination.
Activated B cells multiply to orm a clone o Monoclonal antibodies to hCG are used in
plasma cells and memory cells. pregnancy test kits.
Antibodies aid the destruction o pathogens. skill
Immunity depends upon the persistence o Analysis o epidemiological data related to
memory cells. vaccination programmes.
Vaccines contain antigens that trigger immunity nature of ciece
but do not cause the disease.
Consider ethical implications o research:
Pathogens can be species-specifc although Jenner tested his vaccine or smallpox on
others can cross species barriers. a child.
White cells release histamine in response to
allergens.
Histamines cause allergic symptoms.
Fusion o a tumour cell with an antibody-
producing plasma cell creates a hybridoma cell.
Monoclonal antibodies are produced by
hybridoma cells.
465
11 ANIMAL PHYSIOLOGY ( AHL)
Antigens in blood transfusion
Every organism has unique molecules on the surace o
their cells.
Any oreign molecule that can trigger an immune response is reerred
to as an antigen. The most common antigens are proteins and very large
polysaccharides. Such molecules are ound on the surace o cancer cells,
parasites and bacteria, on pollen grains and on the envelopes o viruses.
As an example, gure 1 shows a representation o an infuenza virus.
Hemagglutinin and neuraminidase are two antigens ound on the
surace o the virus. Hemagglutinin allows the virus to stick to host cells.
Neuraminidase helps with the release o newly-ormed virus particles.
The surace o our own cells contains proteins and polypeptides.
Immune systems unction based on recognizing the distinction between
oreign antigens and sel. Figure 2 shows a mixture o pollen grains
rom several species. The antigens on the surace o these grains are
responsible or triggering immune responses that are called allergies or
hay ever in common language.
hemagglutinin
Figure 2 Pollen grains lipid membrane
other protein
genetic material (RNA)
neuraminidase
Figure 1 Infuenza virus
Antigens in blood transfusion
Antigens on the surace o red blood cells stimulate antibody production in a
person with a diferent blood group.
Blood groups are based on the presence or antigen A results. Blood type AB involves the
absence o certain types o antigens on the surace presence o both types o antigens.
o red blood cells. Knowledge o this is important
in the medical procedure called transusion where OA
a patient is given blood rom a donor. The AB O
blood group and the Rhesus (Rh) blood group are
the two most important antigen systems in blood
transusions as mismatches between donor and
recipient can lead to an immune response.
In gure 3, the dierences between the three A, B B AB
and O phenotypes are displayed. All three alleles
involve a basic antigen sequence called antigen H. Key N acetyl-galactosamine fucose
In blood type A and B, this antigen H is modied red blood cell N acetyl-glucosamine galactose
by the addition o an additional molecule. I the
additional molecule is galactose, antigen B results. Figure 3
I the additional molecule is N-acetylgalactosamine,
466
11.1 Antibody production An d vAccin Ation
I a recipient is given a transusion involving the
wrong type o blood, the result is an immune
response called agglutination ollowed by
hemolysis where red blood cells are destroyed and
blood may coagulate in the vessels (fgure 4) .
red blood cells with antibodies from agglutination hemolysis
surface antigens from recipient ( cl u m p i n g)
an incom atable donor
Figure 4
Blood typing involves mixing samples o blood Figure 5
with antibodies. Figure 5 shows the result o a
blood group test showing reactions between blood
types (rows) and antibody serums (columns) . The
frst column shows the bloods appearance prior
to the tests. There are our human blood types: A,
B , AB and O . Type A blood has type A antigens
( surace proteins) on its blood cells. Type B blood
has type B antigens. Mixing type A blood with
anti-A+B serum causes an agglutination reaction,
producing dense red dots that are dierent rom
the control in the frst column. Type B blood
undergoes the same reaction with anti-B serum
and anti A+B serum. AB blood agglutinates in all
three anti- serums. Type O blood has neither the A
or B antigen, so it does not react to the serums.
The specifc immune response
B lymphocytes are activated by T lymphocytes in mammals.
The principle o challenge and response has been used to explain
how the immune system produces the large amounts o the specifc
antibodies that are needed to fght an inection, and avoid producing
any o the hundreds o thousands o other types o antibodies that
could be produced. Antigens on the surace o pathogens that have
invaded the body are the challenge. The response involves the
ollowing stages.
Pathogens are ingested by macrophages, and antigens rom them are
displayed in the plasma membrane o the macrophages. Lymphocytes
called helper T cells each have an antibody-like receptor protein in
their plasma membranes, which can bind to antigens displayed by
macrophages. O the many types o helper T cell, only a ew have
receptor proteins that ft the antigen. These helper T cells bind and are
activated by the macrophage.
467
11 ANIMAL PHYSIOLOGY ( AHL)
1 Macrophage ingests pathogen The activated helper T cells then bind to lymphocytes called B cells.
and displays antigens from it Again, only B cells that have a receptor protein to which the antigen
binds are selected and undergo the binding process. The helper T cell
2 Helper T cell specic to the activates the selected B cells, both by means o the binding and by
antigen is activated by the release o a signalling protein.
macrophage
The role of plasma cells
3 B cell specic to the antigen is activated Plasma cells secrete antibodies.
by proteins from the helper T cell
Plasma cells are mature B lymphocytes (white blood cells) that produce
5 B cell also divides to 4 B cell divides repeatedly and secrete large number o antibodies during an immune response.
produce memory cells to produce antibody- Figure 7 shows a plasma cell. The cells cytoplasm (orange) contains an
secreting plasma cells unusually extensive network o rough endoplasmic reticulum (rER) .
rER manuactures, modifes and transports proteins, in this case, the
antibodies. The cell produces a lot o the same type o protein meaning
that the range o genes expressed is lower than a typical cell. This
explains the staining pattern o the nucleus where dark staining indicates
unexpressed genes.
Clonal selection and memory cell formation
Activated B cells multiply to form a clone of plasma cells
and memory cells.
6 Antibodies produced by the clone of The activated B cells divide many times by mitosis, generating a clone o
plasma cells are specic to antigens plasma cells that all produce the same antibody type. The generation o
on the pathogen and help to destroy it. large numbers o plasma cells that produce one specifc antibody type is
known as clonal selection.
Figure 6 The stages in antibody production
The antibodies are secreted and help to destroy the pathogen in ways
described below. These antibodies only persist in the body or a ew
weeks or months and the plasma cells that produce them are also
gradually lost ater the inection has been overcome and the antigens
associated with it are no longer present.
Although most o the clone o B cells become active plasma cells, a
smaller number become memory cells, which remain long ater the
inection. These memory cells remain inactive unless the same pathogen
inects the body again, in which case they become active and respond
very rapidly. Immunity to an inectious disease involves either having
antibodies against the pathogen, or memory cells that allow rapid
production o the antibody.
Figure 7 A plasma cell The role of antibodies
Antibodies aid the destruction of pathogens.
Antibodies aid in the destruction o pathogens in a number o ways.
Opsonization: They make a pathogen more recognizable to
phagocytes so they are more readily enguled. Once bound, they can
link the pathogen to phagocytes.
468
11.1 Antibody production An d vAccin Ation
Neutralization of viruses and bacteria: Antibodies can prevent toK
viruses rom docking to host cells so that they cannot enter the cells.
Wha a game he ells
Neutralization of toxins: Some antibodies can bind to the toxins s a he essee f
produced by pathogens, preventing them rom aecting susceptible cells. smallx skles?
Activation of complement: The complement system is a collection Once wild smallpox had
o proteins which ultimately lead to the peroration o the membranes been eradicated there
o pathogens. Antibodies bound to the surace o a pathogen activate remained the challenge o
a complement cascade which leads to the ormation o a membrane what to do with samples o
attack complex that orms a pore in the membrane o the pathogen smallpox still in the hands
allowing water and ions to enter into the cell ultimately causing the o researchers and the
cell to lyse. military. Despite calls or
the remaining stockpiles to
Agglutination: Antibodies can cause sticking together or be eradicated by the WHO,
agglutination o pathogens so they are prevented rom entering cells both the US and Russia have
and are easier or phagocytes to ingest. The large agglutinated mass delayed complying with this
can be fltered by the lymphatic system and then phagocytized. The directive.
agglutination process can be dangerous i it occurs as a result o an
incorrect blood transusion. Game theory is a branch o
mathematics that makes
Figure 8 summarizes some o the modes o action o antibodies. predictions about human
behaviour when negotiations
function of antibodies are being undertaken. In
terms o payo, i one side
agglutination activation of complement reneges and the other
complement proceeds on the basis o
reduces number of trust, the gain to the deal
pathogenic units to breaker is maximized. In
be engulfed this case, they are no longer
threatened by the adversary
bacterium lysis but retain the ability to
threaten. I both parties
bacteria neutralization renege, the risk remains
that the virus will be used
opsonization blocks adhesion of bacteria and docking as a weapon in both the frst
p h a go cy te attack and in retaliation.
of viruses to cells blocks Maximum net gain or all
coating would involve both parties
antigen activity complying with the directive
with antibody enhances phagocytosis but this involves trust and
virus of toxins risk taking.
Figure 8
bacterium
toxin
Immunity
Immunity depends upon the persistence o memory cells.
Immunity to a disease is due either to the presence o antibodies that
recognize antigens associated with the disease, or to memory cells
that allow production o these antibodies. Immunity develops when
the immune system is challenged by a specifc antigen and produces
antibodies and memory cells in response. Figure 9 distinguishes a
primary immune response (launched the frst time the pathogen inects
469
11 ANIMAL PHYSIOLOGY ( AHL)
the body) and the secondary immune response which is launched the
second time the pathogen inects the body. Memory cells ensure that
the second time an antigen is encountered, the body is ready to respond
rapidly by producing more antibodies at a aster rate.
secondary response
concentration of antibody primary response
0 10 20 30 40 50 60
t i m e /d a y s
rst encounter second encounter
with antigen with antigen
Figure 9 The secondary immune response
Figure 10 Vaccines lead to immunity
Vaccines contain antigens that trigger immunity but do
not cause the disease.
A vaccine is introduced into the body, usually by inj ection. The vaccine
may contain a live attenuated (weakened) version o the pathogen, or
some derivative o it that contains antigens rom the pathogen. This
stimulates a primary immune response. I the actual microorganism
enters the body as a result o inection, it will be destroyed by the
antibodies in a secondary immune response.
Figure 1 0 shows a phagocyte engulfng a Mycobacterium bovis bacterium
(orange) . This is the strain o the bacterium used in the vaccination or
tuberculosis (TB) . The bacteria are live but attenuated (weakened) and
not as pathogenic as their relative Mycobacterium tuberculosis. The vaccine
primes the immune system to produce antibodies that act on both
species o bacteria, without causing the disease, so that it responds more
rapidly i inected with Mycobacterium tuberculosis ( TB ) bacteria.
Ethical considerations of Jenners vaccine experiments
Consider ethical implications of research: Jenner tested his vaccine for smallpox
on a child.
Edward Jenner was an 1 8th century scientist she would never develop smallpox. He inected
who noted that a milkmaid claimed that an eight-year-old boy with cowpox. Ater a
because she had caught the disease cowpox brie illness, the boy recovered. Jenner then
470
11.1 Antibody production An d vAccin Ation
purposely inected the boy with smallpox to Jenners experiments were perormed well beore the
conrm that he had the ability to resist the ormulation o any statements o ethical principles
disease. or the protection o human research subjects. The
Nuremberg Trials condemned medical experiments
He was the rst person to use human beings on children. These trials that ollowed the Second
as research subjects in testing a vaccine. He did not World War resulted in the Nuremberg Code or
do any preliminary laboratory research nor any the protection o research subjects, and later the
preliminary animal studies beore experimenting World Health Organizations International Ethical
with human beings, his subject was a small child Guidelines or Biomedical Research Involving Human
well below the age o consent, and he deliberately Subjects (1 993) . Jenners experiments would not be
inected him with an extremely virulent, oten approved by a modern ethical review committee.
atal, disease-causing agent.
The eradication of smallpox
Smallpox was the frst inectious disease o humans to have been eradicated
by vaccination.
The eorts to eradicate smallpox are an example the disease could be maintained and
o the contributions that intergovernmental re-emerge. This is the reason a yellow
organizations can make to address issues o global ever eradication eort ailed in the
concern. The rst such eort was launched in 1 950 early 1 900s.
by the Pan American Health Organization. The Symptoms o inection emerge quite quickly
World Health Assembly passed a resolution in 1 959 and are readily visible allowing teams to
to undertake a global initiative to eradicate smallpox. ring vaccinate all o the people who might
It met with mixed success until a well-unded have come in contact with the aficted
Smallpox Eradication Unit was established in 1 967. person. In contrast, eorts to eradicate
The last known case o wild smallpox was in 1 977 polio have been hampered because inected
in Somalia, though there were two accidental persons do not always present readily
inections ater this. The campaign was successul recognized symptoms.
or several reasons: Immunity to smallpox is long-lasting unlike
Only humans can catch and transmit such conditions as malaria where reinection is
smallpox. There is no animal reservoir where more common.
Vaccines and epidemiology
Analysis o epidemiological data related to vaccination programmes.
Epidemiology is the study o the distribution, the World Health O rganization ( WHO ) , UNIC EF
patterns and causes o disease in a population. The and the Rotary Foundation. S imilarly, UNIC EF is
spread o disease is monitored in order to predict leading a worldwide initiative to prevent tetanus
and minimize the harm caused by outbreaks as through vaccination.
well as to determine the actors contributing to the
outbreak. Epidemiologists would be involved in A small number o polio cases are the result o
planning and evaluating vaccination programmes. a ailure in vaccination programmes. Figure 1 1
shows the incidence o wild rather than vaccine-
An eort to achieve the global eradication o polio induced polio cases in India over a seven-year
was begun in 1 988, as a combined eort between period. Epidemiologists would investigate to
471
11 ANIMAL PHYSIOLOGY ( AHL)
determine the causes of the two peaks in numbers. 1800 1600
Figure 1 2 shows the geographic distribution 1600
of polio cases over a 1 3-year period in India.
Epidemiologists would use information about number of cases 1400
geographic distribution to determine origins of
outbreaks so they could focus resources on those 1200
areas. They could track incidence to determine
the effectiveness of reduction campaigns. It is 1000 873
heartening to know that by 201 2, India had been 800
declared polio-free.
600 676
The concern is that polio-free countries can still see
some polio cases if infected individuals cross borders. 400 265 268 255 134
200 66
0
2000 2001 2002 2003 2004 2005 2006 2007
year
Figure 11
1998 1999 2000 2001
1,934 cases 1,126 cases 265 cases 268 cases
2002 2003 2004 2005
1,600 cases 255 cases 134 cases 66 cases
2006 2007 2008 2009
676 cases 874 cases
2010 559 cases
2011
Figure 12
472
11.1 Antibody production An d vAccin Ation
daa-ase qess: Polio incidence in 2012
Figure 1 3 provides data about polio incidence 4 Identiy one country where the situation appears
in the three countries where wild polio was still 5 to have improved between 201 1 and 201 2. (2)
endemic as o mid-201 2.
Given that in 1 988 there were an estimated
1 Defne the term endemic (1) 350,000 cases o polio globally, discuss the
success o the polio eradication programme. (5)
2 Identiy the three countries where polio
was still endemic as o mid-201 2. (1) 6 Suggest some o the challenges an epidemiologist
3 Identiy the strain o polio virus which is might ace in gathering reliable data. (5)
the most prevalent. (1) 7 Research to fnd the status o polio eradication
in these countries.
wild poliovirus (WPV) cases
Afghanistan
year-to-date 2012 year-to-date 2011 total in 2011 date of most
WPV1 WPV2 W1W3 total WPV1 WPV2 W1W3 total 80 recent case
13 0 0 13 11 0 0 11 30 June 2012
Pakistan
year-to-date 2012 year-to-date 2011 total in 2011 date of most
WPV1 WPV3 W1W3 total WPV1 WPV3 W1W3 total 198 recent case
20 2 1 23 58 1 0 59 22 June 2012
Nigeria
year-to-date 2012 year-to-date 2011 total in 2011 date of most
WPV1 WPV3 W1W3 total WPV1 WPV3 W1W3 total 62 recent case
42 13 0 55 14 6 0 20 22 June 2012
Global
total cases YTD 2012 YTD 2011 total in 2011
globally 96 274 650
in endemic countries 91 91 341
in non-endemic countries 5 183 309
Figure 13
Zoonosis are a growing global health concern
Pathogens can be species-specifc although others can
cross species barriers.
Pathogens are oten highly specialized with a narrow range o hosts. There
are viruses that are specifc to birds, pigs and bacteria or example. There are
bacterial pathogens that only cause disease in humans. Humans are the only
known organism susceptible to such pathogens as syphilis, polio and measles,
but we are resistant to canine distemper virus, or example. The bacterium
Mycobacterium tuberculosis does not cause disease in rogs because rogs rarely
reach the 37 C temperature necessary to support the prolieration o the
bacterium. Rats injected with the diphtheria toxin do not become ill because
their cells lack the receptor that would bring the toxin into the cell.
473
11 ANIMAL PHYSIOLOGY ( AHL)
Figure 14 A thermal scanning camera is being A zoonosis is a pathogen which can cross a species barrier. This is an
used to monitor the skin temperature o emerging global health concern. Bubonic plague, Rocky Mountain spotted
passengers arriving at Nizhny Novgorod airport, ever, Lyme disease, bird fu and West Nile virus are all zoonotic diseases. The
in Russia. Raised skin temperature can be an major actor contributing to the increased appearance o zoonotic diseases
indicator o ever rom illnesses. Such cameras is the growth o contact between animals and humans by such means as
have been used widely to screen or possible humans living in close contact with livestock or disruption o habitats.
carriers o various possible zoonotic epidemic
infuenzas such as bird fu and swine fu For example, in the late 1 990s in Malaysia, intensive pig arming in the
habitat o bats inected with the Nipah virus eventually saw the virus
move rom the bats to the pigs to the humans and resulted in over
1 00 human deaths.
The immune system produces histamines
White cells release histamine in response to allergens.
Mast cells are immune cells ound in connective tissue that secrete
histamine in response to inection. Histamine is also released by
basophils which circulate in the blood. Histamine causes the dilation o
the small blood vessels in the inected area causing the vessels to become
leaky. This increases the fow o fuid containing immune components to
the inected area and it allows some o the immune components to leave
the blood vessel resulting in both specic and non-specic responses.
Efects o histamines
Histamines cause allergic symptoms.
Histamine is a contributor to a number o symptoms o allergic reactions.
Cells in a variety o tissues have membrane-bound histamine receptors.
Histamine plays a role in bringing on the symptoms o allergy in the nose
(itching, fuid build-up, sneezing, mucus secretion and infammation) .
Histamine also plays a role in the ormation o allergic rashes and in the
dangerous swelling known as anaphylaxis.
To lessen the eects o allergic responses, anti-histamines can be taken.
Figure 15 The rash across the body o this The process or creating hybridoma cells
male patient is due to the release o excessive
histamines in response to taking Amoxicillin Fusion of a tumour cell with an antibody-producing
(penicillin) antibiotic plasma cell creates a hybridoma cell.
Monoclonal antibodies are highly specic, puried antibodies that are
produced by a clone o cells, derived rom a single cell. They recognize
only one antigen.
plasma cells
isolate
immunize mouse spleen B
cells
antigen and dye
used to screen to nd
desired hybridoma
hybridomas
cell culture myeloma cells
474 Figure 16
11.1 Antibody production An d vAccin Ation
To produce the clone o cells that will manuacture a monoclonal
antibody, the antigen recognized by the antibody is inj ected into a
mouse, or other mammal. In response to this challenge, the mouses
immune system makes plasma B cells that are capable o producing
the desired antibody. Plasma cells are removed rom the spleen o the
mouse. They will be o many dierent types with only some producing
the desired antibody.
The B cells are used with cancer cells called myeloma cells. The
cells ormed by usion o plasma B cells and myeloma cells are called
hybridoma cells.
production of monoclonal antibodies Figure 17hCG dye
hCG dye
Monoclonal antibodies are produced by hybridoma cells. A hCG
hCGhCG hCG dye
Because the ull diversity o B cells are used with the myeloma cells, hCG dye
many dierent hybridomas are produced and they are individually
tested to fnd one that produces the required antibody. dye dye
dye
Once identifed, the desired hybridoma cell is allowed to divide and B
orm a clone. These cells can be cultured in a ermenter where they dye
will secrete huge amounts o monoclonal antibody. Figure 1 7 shows a dye dye
2000-litre ermenter used in the commercial production o monoclonal dye
antibodies. The hybridoma cell is multiplied in the ermenter to produce
large numbers o genetically identical copies, each secreting the antibody dye
produced by the original lymphocyte.
C
Monoclonal antibodies are used both or treatment and diagnosis o dye
diseases. Examples include the test or malaria that can be used to dye
identiy whether either humans or mosquitoes are inected with the dye
malarial parasite, the test or the HIV pathogen or the creation o dye
antibodies or injection into rabies victims.
D dye dye
pregnancy tests emloy monoclonal dye
antibodies dye
Monoclonal antibodies to hCG are used in pregnancy Figure 18
test kits.
A
Monoclonal antibodies are used in a broad range o diagnostic tests,
including tests or HIV antibodies and or an enzyme released during 1 Explain how a blue band appears
heart attacks. Pregnancy test kits are available that use monoclonal
antibodies to detect hCG (human chorionic gonadotrophin) . hCG is at point C if the woman is
uniquely produced during pregnancy by the developing embryo and
later the placenta. The urine o a pregnant woman contains detectable pregnant. [3]
levels o hCG.
2 Explain why a blue band does
Figure 1 8 shows how the pregnancy test strip works. At point C,
there are antibodies to hCG immobilized in the strip. At point B not appear at point C if the
there are ree antibodies to hCG attached to a dye. At point D there
are immobilized antibodies that bind to the dye-bearing antibodies. woman is not pregnant. [3]
Urine applied to the end o a test strip washes antibodies down
the strip. 3 Explain the reasons for the use
of immobilized monoclonal
antibodies at point D, even
though they do not indicate
whether a woman is pregnant
or not. [3]
475
11 ANIMAL PHYSIOLOGY ( AHL)
11.2 Movement
Udertadig Applicatio
Bones and exoskeletons provide anchorage or Antagonistic pairs o muscles in an insect leg.
muscles and act as levers.
skill
Movement o the body requires muscles to
work in antagonistic pairs. Annotation o a diagram o the human elbow.
Drawing labelled diagrams o the structure o a
Synovial joints allow certain movements but
not others. sarcomere.
Analysis o electron micrographs to fnd the
Skeletal muscle fbres are multinucleate and
contain specialized endoplasmic reticulum. state o contraction o muscle fbres.
Muscle fbres contain many myofbrils. nature of ciece
Each myofbril is made up o contractile Fluorescence was used to study the cyclic
sarcomeres. interactions in muscle contraction.
The contraction o the skeletal muscle is
achieved by the sliding o actin and myosin
flaments.
Calcium ions and the proteins tropomyosin and
troponin control muscle contractions.
ATP hydrolysis and cross-bridge ormation are
necessary or the flaments to slide.
Figure 1 Boe ad exokeleto achor mucle
476 Bones and exoskeletons provide anchorage or muscles
and act as levers.
Exoskeletons are external skeletons that surround and protect most o
the body surace o animals such as crustaceans and insects. Figure 1
shows a scanning electron micrograph o a spider next to exoskeletons
that have been moulted.
Bones and exoskeletons acilitate movement by providing an anchorage
or muscles and by acting as levers. Levers change the size and direction
o orces. In a lever, there is an eort orce, a pivot point called a ulcrum
and a resultant orce. The relative positions o these three determine the
class o lever.
In fgure 2, the diagram shows that when a person nods their head
backward, the spine acts as a frst- class lever, with the ulcrum ( F) being
ound between the eort orce ( E) provided by the splenius capitis muscle
and the resultant orce (R) causing the chin to be extended.
The grasshopper leg acts as a third-class lever as the ulcrum is at the body
end and the eort orce is between the ulcrum and the resultant orce.
11.2 MoveMent
Muscles are attached to the insides o exoskeletons but to the outside biceps
o bones. contracted
F E triceps
E relaxed
R E
R R elbow extended scapula
E F F biceps humerus
R relaxed
E ER triceps
F R F radius contracted
F (c) Third-class lever
(a) First-class lever ulna
(b) Second-class lever
Figure 2
skeletal mucle are antagonitic Figure 3 The biceps and triceps are
antagonistic muscles
Movement of the body requires muscles to work in
antagonistic pairs.
Skeletal muscles occur in pairs that are antagonistic. This means that
when one contracts, the other relaxes. Antagonistic muscles produce
opposite movements at a j oint. For example, in the elbow, the triceps
extends the orearm while the biceps fex the orearm.
daa-bas qusis: Flight muscles
In one research project, pigeons (Columba livia) 1 Deduce the number o downstrokes o
the wing during the whole fight.
were trained to take o, fy 35 metres and land [1 ]
on a perch. During the fight the activity o 2 Compare the activity o the
sternobrachialis muscle during the
two muscles, the sternobrachialis (SB) and the three phases o the fight.
thoracobrachialis (TB) , was monitored using [3]
electromyography. The traces are shown in gure 4.
The spikes show electrical activity in contracting 3 Deduce rom the data in the
electromyograph how the
muscles. Contraction o the sternobrachialis causes thoracobrachialis is used.
a downward movement o the wing. [1 ]
take o fast ight landing 4 Another muscle, the supracoracoideus, is
antagonistic to the sternobrachialis. State
SB the movement produced by a contraction
o the supracoracoideus. [1 ]
5 Predict the pattern o the electromyograph
TB trace or the supracoracoideus muscle
during the 35-metre fight. [2]
400 ms
Figure 4 Electrical activity in the sternobrachialis (SB) and the thoracobrachialis (TB) muscles during fight o a pigeon
477
11 ANIMAL PHYSIOLOGY ( AHL)
An insect leg has antagonistic muscles
Antagonistic pairs of muscles in an insect leg.
The grasshopper, like all insects, has three pairs o appendages. The
hindlimb o a grasshopper is specialized or jumping. It is a jointed
appendage with three main parts. Below the joint is reerred to as
the tibia and at the base o the tibia is another joint below which is
ound the tarsus. Above the j oint is reerred to as the emur. Relatively
massive muscles are ound on the emur.
When the grasshopper prepares to jump, the fexor muscles will
contract bringing the tibia and tarsus into a position where they
resemble the letter Z and the emur and tibia are brought closer
together. This is reerred to as fexing. The extensor muscles relax
during this phase. The extensor muscles will then contract extending
the tibia and producing a powerul propelling orce.
extensor extensor tibia
muscle relaxes muscle contracts extends
tibia exor muscle
exes relaxes
exor
muscle
contracts
Figure 6 Composite high-speed photograph Figure 5
of a grasshopper (Order Orthoptera)
jumping from the head of a nail
The human elbow is an example of a synovial joint
Annotation of a diagram of the human elbow.
humerus bone to which the The point where bones meet is called a joint. Most
biceps and triceps are attached joints allow the bones to move in relation to each
other this is called articulation. Most articulated
triceps extends biceps exes joint-capsule seals joints have a similar structure, including cartilage,
the joint and helps to synovial fuid and joint capsule.
the joint the joint prevent dislocation
Cartilage is tough, smooth tissue that covers
synovial uid the regions o bone in the joint. It prevents
lubricates the joint contact between regions o bone that might
and prevents friction otherwise rub together and so helps to prevent
riction. It also absorbs shocks that might cause
ulna bone to which radius bone to which bones to racture.
the triceps is attached the biceps is attached Synovial fuid lls a cavity in the joint
between the cartilages on the ends o the
cartilage covers the bones. It lubricates the joint and so helps to
prevent the riction that would occur i the
bones and prevents friction cartilages were dry and touching.
Figure 7 The elbow joint The joint capsule is a tough ligamentous covering
to the joint. It seals the joint and holds in the
synovial fuid and it helps to prevent dislocation.
478
11.2 MoveMent
Diferent joints allow diferent ranges omovement
Synovial joints allow certain movements but not others.
The structure of a joint, including the joint capsule and the
ligaments, determines the movements that are possible. The knee
joint can act as a hinge joint, which allows only two movements:
flexion (bending) and extension (straightening) . It can also act as a
pivot joint when flexed. The knee has a greater range of movement
when it is flexed than when it is extended. The hip joint, between
the pelvis and the femur, is a ball and socket j oint. It has a greater
range of movement than the knee joint in that it can flex and
extend, rotate, and move sideways and back. This latter type of
movement is called abduction and adduction.
exion outward rotation
abduction
adduction inward rotation
hyperextension extension outward rotation
inward rotation
Figure 8 Range of motion at the shoulder
479
exion
abduction
extension adduction
Figure 9 Range of motion at the hip
11 ANIMAL PHYSIOLOGY ( AHL)
structure o ucle fbre
Skeletal muscle fbres are multinucleate and contain
specialized endoplasmic reticulum.
The muscles that are used to move the body are attached to bones, so
they are called skeletal muscles. When their structure is viewed using
a microscope, stripes are visible. They are thereore also called striated
muscle. The two other types o muscle are smooth and cardiac.
Striated muscle is composed o bundles o muscle cells known as muscle
fbres. Although a single plasma membrane called the sarcolemma
surrounds each muscle fbre, there are many nuclei present and muscle
fbres are much longer than typical cells. These eatures are due to the
act that embryonic muscle cells use together to orm muscle fbres.
Figure 1 0 shows a muscle fbre.
sarcolemma
nucleus
myobril
sarcoplasmic
reticulum
one sarcomere Figure 10
light band Z-line dark band A modifed version o the endoplasmic reticulum, called the sarcoplasmic
reticulum, extends throughout the muscle fbre. It wraps around every
Figure 11 The ultrastructure o the myofbril, conveying the signal to contract to all parts o the muscle fbre at
muscle fbre once. The sarcoplasmic reticulum stores calcium. Between the myofbrils are
large numbers o mitochondria, which provide ATP needed or contractions.
480
myofbril
Muscle fbres contain many myofbrils.
Within each muscle fbre there are many parallel, elongated structures
called myofbrils. These have alternating light and dark bands, which
give striated muscle its stripes. In the centre o each light band is a disc-
shaped structure, reerred to as the Z-line.
11.2 MoveMent
structure o myofbril
Each myofbril is made up o contractile sarcomeres.
The micrograph in fgure 1 3 shows a longitudinal section through a
myofbril. A number o repeating units that alternate between light and
dark bands are visible. Through the centre o each light area is a line
called the Z-line. The part o a myofbril between one Z-line and the
next is called a sarcomere. It is the unctional unit o the myofbril.
The pattern o light and dark bands in sarcomeres is due to a precise
and regular arrangement o two types o protein flament thin actin
flaments and thick myosin flaments. Actin flaments are attached
to a Z-line at one end. Myosin flaments are interdigitated with actin
flaments at both ends and occupy the centre o the sarcomere. Each
myosin flament is surrounded by six actin flaments and orms cross-
bridges with them during muscle contraction.
The arcomere Figure 12 A transverse section through a
skeletal muscle fbre showing numerous
Drawing labelled diagrams othe structure o a sarcomere. myofbrils. A nucleus is shown in the bottom let
light dark band light
band band
thick myosin
laments
thin actin
laments
Z-line sarcomere Z-line
Figure 14 The structure o a sarcomere
When constructing diagrams o a sarcomere, ensure to demonstrate Figure 13
understanding that it is between two Z-lines. Myosin flaments should
be shown with heads. Actin flaments should be shown connected to
Z-lines. Light bands should be labelled around the Z-line. The extent
o the dark band should also be indicated.
daa-bas qusis: Transverse sections of striated muscle
The drawings in fgure 1 5 show myofbrils in 1 Explain the dierence between a transverse
transverse section.
and a longitudinal section o muscle. [2]
2 Deduce what part o the myofbril is
represented by the drawings as small dots. [2]
3 Compare the pattern o dots in the three
diagrams. [3]
4 Explain the dierences between the [3]
diagrams in the pattern o dots.
Figure 15
481
11 ANIMAL PHYSIOLOGY ( AHL)
cross-bridge mechanis of skeletal uscle contraction
detaches
The contraction o the skeletal muscle is achieved by the
sliding o actin and myosin flaments.
During muscle contraction, the myosin flaments pull the actin flaments
inwards towards the centre o the sarcomere. This shortens each sarcomere
and thereore the overall length o the muscle fbre (see fgure 1 6) .
The contraction o skeletal muscle occurs by the sliding o actin and
myosin flaments. Myosin flaments cause this sliding. They have heads
that can bind to special sites on actin flaments, creating cross-bridges,
through which they can exert a orce, using energy rom ATP. The heads
are regularly spaced along myosin flaments and the binding sites are
regularly spaced along the actin flaments, so many cross-bridges can
orm at once (see fgure 1 7) .
(a) relaxed muscle
binding site actin Z-line light band light band Z-line
myosin
myosin head dark band
myosin
lament
formation of cross- actin
bridge in presence
of calcium ions
light band shortens, dark band remains
indicating actin the same length
slides along myosin
movement of actin sarcomere contracts
cross-bridge (b) contracted muscle
moves actin along
Figure 16 Diagram of relaxed and contracted sarcomeres
shape of myosin head
changes
Figure 17
Deterining the state of skeletal uscle contraction
Analysis o electron micrographs to fnd the state o contraction o muscle fbres.
Relaxed relaxed sarcomere In a relaxed sarcomere, the Z-lines are arther
muscle contracted sarcomere apart, the light bands are wider and overall
the sarcomere is longer. In the centre o the
Contracted sarcomere, there is another line called the
muscle M-line. In a relaxed sarcomere, there is a more
visible light band on either side o the M-line.
Figure 18 Electron micrograph of relaxed and contracted
sarcomeres
482
11.2 MoveMent
The control o skeletal muscle ATP causes the breaking o the cross-bridges by
contraction attaching to the myosin heads, causing them to
detach rom the binding sites on actin.
Calcium ions and the proteins
tropomyosin and troponin control Hydrolysis o the ATP, to AD P and phosphate,
muscle contractions. provides energy or the myosin heads to
swivel outwards away rom the centre o the
In relaxed muscle, a regulatory protein called sarcomere this is sometimes called the cocking
tropomyosin blocks the binding sites on actin. When o the myosin head.
a motor neuron sends a signal to a muscle fbre to
make it contract, the sarcoplasmic reticulum releases New cross-bridges are ormed by the binding
calcium ions. These calcium ions bind to a protein o myosin heads to actin at binding sites
called troponin which causes tropomyosin to move, adjacent to the ones previously occupied
exposing actins binding sites. Myosin heads then (each head binds to a site one position urther
bind and swivel towards the centre o the sarcomere, rom the centre o the sarcomere) .
moving the actin flament a small distance.
Energy stored in the myosin head when it was
The role o ATp in the sliding o cocked causes it to swivel inwards towards
flaments the centre o the sarcomere, moving the
actin flament a small distance. This sequence
ATP hydrolysis and cross-bridge o stages continues until the motor neuron
ormation are necessary or the stops sending signals to the muscle fbre.
flaments to slide. Calcium ions are then pumped back into the
sarcoplasmic reticulum, so the regulatory
For signifcant contraction o the muscle, protein moves and covers the binding sites on
the myosin heads must carry out this action actin. The muscle fbre thereore relaxes.
repeatedly. This occurs by a sequence o stages:
1 myosin laments have heads which 2 ATP binds to the myosin heads
form cross-bridges when they are and causes them to break the
attached to binding sites on actin cross-bridges by detaching
laments. from the binding sites.
movement ADP + P ATP
5 the ADP and phosphate are 3 ATP is hydrolysed to ADP and
released and the heads push the phosphate, causing the myosin
actin lament inwards towards heads to change their angle.
the centre of the sarcomere-
this is called the power stroke. ADP + P the heads are said to be cocked
in their new position as they are
ADP + P storing potential energy from ATP.
4 the heads attach to binding sites on
actin that are further from the centre of
the sarcomere than the previous sites.
Figure 19
483
11 ANIMAL PHYSIOLOGY ( AHL)
The use o fuorescence to study contraction
Fluorescence has been used to study the cyclic interactions in muscle contraction.
Fluoresence is the emission o electromagnetic In another experiment, researchers cut apart
radiation, oten visible light, by a substance ater it Nitella axillaris cells. These cells are unique in
has been illuminated by electromagnetic radiation that they have a network o actin laments
o a dierent wavelength. The fuorescence underlying their membranes. Researchers
can oten be detected in a light microscope and attached fuorescent dye to myosin molecules in
captured on lm or later analysis. an eort to show that myosin can walk along
actin laments.
Some o the classic experiments in the history o
muscle research have depended on fuorescence. The uorescent dye
coelenterate Aequorea victoria (gure 20) produces a bead attached to myosin
calcium-sensitive bioluminescent protein, aequorin.
Scientists studied the contraction o giant single myosin ATP
muscle bres o the acorn barnacle Balanus nubilus by actin lament from
injecting samples o the muscle with aequorin. When
muscles were stimulated to contract in the study, ADP Nitella axillaris
initially there was strong bioluminescence coinciding
with the release o Ca2+ rom the sarcoplasmic actin
reticulum. The light intensity began to decrease
immediately ater the cessation o the stimulus. With this technique, the researchers were able to
demonstrate the ATP-dependence o myosin- actin
interaction.
The graph in gure 21 shows the velocity
o myosin molecules as a unction o ATP
concentration.
5
lament velocity, m/s 4
3
2
1
0 100 150 200 400 1000
0 50 ATP, M
Figure 21
Figure 20 Aequorea victoria
484
11.3 the Kidne y And osMoregulAtion
11.3 t k a ma
Udertadig Applicatio
Animals are either osmoregulators or Consequences o dehydration and
osmoconormers. o ve rh y d ra ti o n .
The Malpighian tubule system in insects and the Treatment o kidney ailure by hemodialysis or
kidney carry out osmoregulation and removal o kidney transplant.
nitrogenous wastes.
Blood cells, glucose, proteins and drugs are
The composition o blood in the renal artery is detected in urinary tests.
dierent rom that in the renal vein.
skill
The ultrastructure o the glomerulus and
Bowmans capsule acilitate ultrafltration. Drawing and labelling a diagram o the human
ki d n e y .
The proximal convoluted tubule selectively
reabsorbs useul substances by active transport. Annotation o diagrams o the nephron.
The loop o Henl maintains hypertonic nature o ciece
conditions in the medulla.
Curiosity about particular phenomena:
The length o the loop o Henl is positively investigations were carried out to determine
correlated with the need or water conservation how desert animals prevent water loss in their
in animals. wastes.
ADH controls reabsorption o water in the
collecting duct.
The type o nitrogenous waste in animals is
correlated with evolutionary history and habitat.
Diferet repoe to chage i omolarity
i the eviromet
Animals are either osmoregulators or osmoconormers.
Osmolarity reers to the solute concentration o a solution. Many
animals are known as osmoregulators because they maintain a
constant internal solute concentration, even when living in marine
environments with very dierent osmolarities. All terrestrial animals,
reshwater animals and some marine organisms like bony fsh are
osmoregulators. Typically these organisms maintain their solute
concentration at about one third o the concentration o seawater and
about 1 0 times that o resh water.
Osmoconormers are animals whose internal solute concentration tends
to be the same as the concentration o solutes in the environment.
485
11 ANIMAL PHYSIOLOGY ( AHL)
data-base questions lowers the reezing point. 2 delta is equivalent to
about 1 00% ocean seawater, 0.2 delta is equivalent
The striped shore crab Pachygrapsus crassipes to about 1 0% ocean seawater, and 3 .4 delta is
(gure 1 ) is ound on rocky shores over the equivalent to about 1 70% seawater.
west coast o North and Central America as
well as in Korea and Japan. P. crassipes is oten 1 Determine the solute concentration o (1)
exposed to dilute salinities in tide pools and crab blood at which the concentration o
reshwater rivulets, but it only rarely encounters surrounding water is 1 delta.
salt concentrations much higher than that o the
ocean. Samples o crabs were placed in water 2 D etermine the range over which P. crassipes
concentrations o varying osmolarity and samples
o blood were taken to determine osmolarity is able to keep its blood solute concentration
o the blood. In this experiment, the unit o
osmolarity is measured in units based on reezing airly stable. (1)
point depression. When solutes are added to water
they disrupt hydrogen bonding. Freezing requires 3 Predict what the graph would look like i
additional hydrogen bonding so adding solute P. crassipes was not able to osmoregulate. ( 1 )
4 D iscuss whether P. crassipes is an (3)
osmoconormer or an osmoregulator.
3.0
line of isosmoticity
Pachygrapsus delta 2.0
1.0 ocean 3.0
seawater
Figure 1 The striped shore crab is exposed to varying salt 0
concentrations in its habitat 0 1.0 2.0
water delta
Figure 2
The malpighian tubule syste
The Malpighian tubule system in insects and the
kidney carry out osmoregulation and removal of
nitrogenous wastes.
Arthropods have a circulating fuid, known as hemolymph, that combines
the characteristics o tissue fuid and blood. Osmoregulation is a orm o
homeostasis whereby the concentration o hemolymph, or blood in the case
o animals with closed circulatory systems, is kept within a certain range.
When animals break down amino acids, the nitrogenous waste product
is toxic and needs to be excreted. In insects, the waste product is usually
in the orm o uric acid and in mammals it is in the orm o urea.
Insects have tubes that branch o rom their intestinal tract. These are
known as Malpighian tubules. Cells lining the tubules actively transport
ions and uric acid rom the hemolymph into the lumen o the tubules.
This draws water by osmosis rom the hemolymph through the walls o
the tubules into the lumen. The tubules empty their contents into the
486
11.3 the Kidne y And osMoregulAtion
gut. In the hindgut most of the water and salts are reabsorbed while the
nitrogenous waste is excreted with the feces.
hindgut 4 dehydrated uric acid paste
is released with other waste
2 the tubules empty uric acid
into the gut
midgut semisolid wastes
M a l p i gh i a n M a l p i gh i a n Na+ H20
midgut tubule tubules K+
hindgut 3 some ions are actively reabsorbed
in the hindgut and some water follows
Figure 3
H20
Na+ K+ uric acid
1 uric acid, Na+ and K+ are H20
transported into the tubules
and water follows by osmosis
Drawing the human kidney cortex
medulla
Drawing and labelling a diagram o the renal artery
human kidney. renal vein pelvis of
kidney
When drawing a diagram of the kidney, the shape
should be roughly oval with a concave side to
which the renal artery and vein are attached.
Drawings should clearly indicate the cortex shown
at the edge of the kidney. It should be shown
with a thickness of about _1_ the entire width. The
5
medulla should be shown inside the cortex, with
pyramids. The renal pelvis should be shown on
the concave side of the kidney. The pelvis should ureter (carries urine from the kidney)
drain into the ureter. The renal artery should have Figure 4 Structure of the kidney
a smaller diameter than the renal vein.
Comparing the composition of blood in the renal
artery and the renal vein
The composition o blood in the renal artery is diferent
rom that in the renal vein.
Kidneys function in both osmoregulation and excretion. The kidneys are
responsible for removing substances from the blood that are not needed
or are harmful. As a result, the composition of blood in the renal artery,
487
11 ANIMAL PHYSIOLOGY ( AHL)
through which blood enters the kidney, is dierent rom that in the
renal vein, through which blood leaves.
Substances that are present in higher amounts in the renal artery than
the renal vein include:
Toxins and other substances that are ingested and absorbed but are
not ully metabolized by the body, or example betain pigments in
beets and also drugs.
Excretory waste products including nitrogenous waste products,
mainly urea.
Other things removed rom the blood by the kidney that are not
excretory products include:
Excess water, produced by cell respiration or absorbed rom ood in
the gut.
Excess salt, absorbed rom ood in the gut.
These are not excretory products because they are not produced by body
cells. Removal o excess water and salt is part o osmoregulation. While
blood in the renal artery might contain a variable water or salt content,
blood in the renal vein will have a more constant concentration because
osmoregulation has occurred.
The kidneys lter o about one th o the volume o plasma rom the blood
fowing through them. This ltrate contains all o the substances in plasma
apart rom large protein molecules. The kidneys then actively reabsorb
the specic substances in the ltrate that the body needs. The result o this
process is that unwanted substances pass out o the body in the urine. These
substances are present in the renal artery but not the renal vein.
data-ase questins: Blood supply to the kidney
Table 1 shows the fow rate o blood to the kidney 2 C alculate the volume o oxygen delivered
and other organs, the rate o oxygen delivery and to the organs per litre o blood. [2]
oxygen consumption. All o the values are given 3 In the brain, 34 per cent o the oxygen
per 1 00 g o tissue or organ. The rates are or a that is delivered is consumed. Calculate
person in a warm environment. the same percentage or the other organs. [4]
1 Compare the rate o blood fow to the 4 Discuss the reasons or the dierence
kidney with fow to the other organs. [2] between the kidney and the other organs
bl fw oxygen oxygen in the volume o blood fowing to the organ,
rate elivery cnsumptin
and the percentage o oxygen in the blood
(ml min1 (ml min1 (ml min1
that is consumed. [4]
100 g1) 100 g1) 100 g1) 5 Some parts o the kidney have a high
Brain 54.0 10.8 3.70 percentage rate o oxygen consumption,
Skin 13.0 2.6 0.38 or example the outer part o the medulla.
Skeletal 2.7 0.5 0.18 This is because active processes requiring
muscle energy are being carried out. Suggest one
(resting) 87.0 17.4 11.0 process in the kidney that requires energy. [1 ]
Heart 420.0 84.0 6.80
muscle 6 Predict, with a reason, one change in
Kidney
blood fow that would occur i the person
Table 1
were moved to a cold environment. [2]
488
11.3 the Kidne y And osMoregulAtion
A nal set o dierences between the composition o blood in the renal toK
artery and the renal vein is due to the metabolic activity o the kidney
itsel. Blood leaving the kidney through the renal vein is deoxygenated A a a a b vp
relative to the renal artery because kidney metabolism requires oxygen. jf f ama a?
It also has a higher partial pressure o carbon dioxide because this is a Figure 5 shows some o the techniques
waste product o metabolism. Even though glucose is normally ltered that have been used to investigate
and then entirely reabsorbed, some glucose is used by the metabolism o kidney unction. The animals used
the kidney and thereore the concentration is slightly lower in the renal include rats, mice, cats, dogs and pigs.
vein compared to the renal artery. 1 What are the reasons or carrying out
Plasma proteins are not ltered by the kidney so should be present in the kidney research?
same concentration in both blood vessels. Presence in the urine indicates 2 What criteria should be used to
abnormal unction. This is looked or during clinical examination o a
urine sample. decide i a research technique is
ethically acceptable or not?
The ultrastructure of the glomerulus 3 Apply your criteria to the three
techniques outlined in fgure 5 to
The ultrastructure o the glomerulus and Bowmans determine whether they are ethically
capsule acilitate ultrafltration. acceptable.
4 Who should make the decisions about
B lood in capillaries is at high pressure in many o the tissues o the body, the ethics o scientifc research?
and the pressure orces some o the plasma out through the capillary
wall, to orm tissue fuid. Living animal is anaesthetized and its kidney
is exposed by surgery. Fluid is sampled rom
In the glomerulus o the kidney, the pressure in the capillaries is particularly nephrons using micropipettes. Animal is then
high and the capillary wall is particularly permeable, so the volume o fuid sacrifced so that the position othe sample point
orced out is about 1 00 times greater than in other tissues. The fuid orced in the kidney can be located.
out is called glomerular ltrate. The composition o blood plasma and ltrate
is shown in table 2. The data in the table shows that most solutes are ltered 6
out reely rom the blood plasma, but almost all proteins are retained in the 5
capillaries o the glomerulus. This is separation o particles diering in size 4
by a ew nanometres and so is called ultrafltration. All particles with a 3
relative molecular mass below 65,000 atomic mass units can pass through. 2
The permeability to larger molecules depends on their shape and charge. 1
Almost all proteins are retained in the blood, along with all the blood cells.
Animal is killed and kidneys are removed and
c (p m3 f b pama) rozen. Samples otissue are cut rom regions
okidney that can be identifed. Temperature at
Solutes plasma fltrate which thawing occurs is ound, to give a measure
Na+ ions (mol) osolute concentration.
Cl- ions (mol) 151 144
glucose (mol) nephron
urea (mol) 110 114
proteins (mg) external uid
55 Animal is killed and kidneys are dissected to
obtain samples onephron. Fluids are perused
55 through nephron tissue, using experimental
external uids to investigate the action othe wall
740 3.5 othe nephron.
Table 2 Figure 5
The structure o a section o the lter unit is shown in gure 6 and gure 7. 489
Figure 6 is a coloured transmission electron micrograph (TEM) o a section
through a kidney glomerulus showing its basement membrane (brown line
running rom top right to bottom let) . The basement membrane separates
the capillaries (the white space at the let is the lumen o a capillary) . Note
the gaps in the wall o the capillary which are reerred to as enestrations.
The smaller projections rom the membrane are podocyte oot
processes, which attach the podocytes (specialized epithelial cells) to the
11 ANIMAL PHYSIOLOGY ( AHL)
Figure 6 membrane. The podocytes unction as a barrier through which waste
products are ltered rom the blood.
podocytes strangely shaped cells
with nger-like projections which wrap There are three parts to the ultraltration system.
around capillaries in the glomerulus
and provide support 1 Fenestrations between the cells in the wall o the capillaries. These
are about 1 00 nm in diameter. They allow fuid to escape, but not
fenestrated basement membrane blood cells.
wall of the lter 2 The basement membrane that covers and supports the wall o
the capillaries. It is made o negatively-charged glycoproteins, which
capillary orm a mesh. It prevents plasma proteins rom being ltered out, due
to their size and negative charges.
3 Podocytes orming the inner wall o the Bowmans capsule.
These cells have extensions that wrap around the capillaries o the
glomerulus and many short side branches called oot processes.
Very narrow gaps between the oot processes help prevent small
molecules rom being ltered out o blood in the glomerulus.
I particles pass through all three parts they become part o the
glomerular ltrate.
Figure 8 shows the relationship between the glomerulus and the
Bowmans capsule.
aerent arteriole
podocytes
basement membrane
eerent arteriole fenestrated
wall of capillary
proximal
convoluted tubule
blood red nucleus of lumen of
plasma blood cell capillary wall cell Bowmans capsule
Figure 7 Structure o the flter unit Figure 8
o the kidney
data-base questions: Ultrafltration o charged and uncharged dextrans
Dextrans are polymers o sucrose. Dierent unit o rat glomeruli. Animal experiments
sizes o dextran polymer can be synthesized, like this can help us to understand how the
allowing their use to investigate the eect o kidney works and can be done without causing
particle size on ultraltration. Neutral dextran suering to the animals.
is uncharged, dextran sulphate has many
negative charges, and DEAE is dextran with 1 State the relationship between the size [1 ]
many positive charges. o particles and the permeability to
them o the lter unit o the
Figure 9 shows the relationship between glomerulus.
particle size and the permeability o the lter
490
11.3 the Kidne y And osMoregulAtion
2 a) Compare the permeability o the lter relative ltration rate 1.0
0.9
unit to the three types o dextran. [3] 0.8
0.7
b) Explain these dierences in [3] 0.6 DEAE
p e rm e a b ility. 0.5 neutral
0.4 dextran
3 One o the main plasma proteins is 0.3 dextran
albumin, which is negatively charged 0.2 sulphate
and has a particle size o approximately 0.1
4.4 nm. Using the data in the graph, 2.4 2.8 3.2 3.6 4.0 4.4
explain the diagnosis that is made i 0 particle size / nm
albumin is detected in a rats urine. 2.0
[3]
Figure 9 Relationship between particle
size o dextrans and fltration rate
The role of the proximal convoluted tubule microvilli mitochondria
The proximal convoluted tubule selectively reabsorbs invaginations of lumen
useul substances by active transport. outer membrane basement membrane containing
ltrate
The glomerular ltrate fows into the proximal convoluted tubule. Figure 10 Transverse section o the
The volume o glomerular ltrate produced per day is huge about proximal convoluted tubule
1 80 dm3. This is several times the total volume o fuid in the body
and it contains nearly 1 .5 kg o salt and 5.5 kg o glucose. As the
volume o urine produced per day is only about 1 .5 dm3 and it
contains no glucose and ar less than 1 .5 kg o salt, almost all o the
ltrate must be reabsorbed into the blood. Most o this reabsorption
happens in the rst part o the nephron the proximal convoluted
tubule. Figure 1 0 shows this structure in transverse section. The
methods used to reabsorb substances in the proximal convoluted
tubule are described in table 3. By the end o the proximal tubule all
glucose and amino acids and 8 0 per cent o the water, sodium and
other mineral ions have been absorbed.
sm : are moved by active transport rom ltrate to space outside the Av
tubule. They then pass to the peritubular capillaries. Pump proteins are located
in outer membrane o tubule cells. The drawing below shows the
structure oa cell rom the wall o
c : are attracted rom ltrate to space outside the tubule because o the proximal convoluted tubule.
charge gradient set up by active transport o sodium ions. Explain how the structure othe
proximal convoluted tubule cell, as
g: is co-transported out o ltrate and into fuid outside the tubule, by shown in the diagram, is adapted to
co-transporter proteins in outer membrane o tubule cells. Sodium ions move carry out selective reabsorption.
down concentration gradient rom outside tubule into tubule cells. This provides
energy or glucose to move at the same time to fuid outside the tubule. The 10 m
same process is used to reabsorb amino acids.
Wa: pumping solutes out o ltrate and into the fuid outside the tubule
creates a solute concentration gradient, causing water to be reabsorbed rom
ltrate by osmosis.
Table 3
491
11 ANIMAL PHYSIOLOGY ( AHL)
Te nepron Loop of Henl a tube shaped like a hairpin,
consisting o a descending limb that carries the
Annotation of diagrams of the nephron. ltrate deep into the medulla o the kidney,
and an ascending limb that brings it back out to
The basic unctional unit o the kidney is the the cortex.
nephron. This is a tube with a wall consisting o
one layer o cells. This wall is the last layer o cells Distal convoluted tubule another highly
that substances cross to leave the body it is an twisted section, but with ewer, shorter
epithelium. There are several dierent parts o microvilli and ewer mitochondria.
the nephron, which have dierent unctions and
structures (see gure 1 1 ) : Collecting duct a wider tube that carries the
ltrate back through the cortex and medulla to
Bowmans capsule proximal convoluted tubule the renal pelvis.
distal convoluted tubule
Blood vessels associated with the nephron
a eren t venule are blood vessels. Blood fows though them in
arteriole peritubular the ollowing sequence:
capillaries
eerent Afferent arteriole brings blood rom the
arteriole glomerulus renal artery.
collecting duct Glomerulus a tight, knot-like, high-
pressure capillary bed that is the site o
vasa recta blood ltration.
ascending limb Efferent arteriole a narrow vessel that
of loop of Henl restricts blood fow, helping to generate
descending limb high pressure in the glomerulus.
of loop of Henl
Peritubular capillaries a low-pressure
Figure 11 The nephron and associated blood vessels. The capillary bed that runs around the convoluted
human kidney contains about a million nephrons tubules, absorbing fuid rom them.
B owmans cap sule a cup- shaped structure Vasa recta unbranched capillaries that
with a highly porous inner wall, which collects are similar in shape to the loops o Henl,
the fuid ltered rom the blood. with a descending limb that carries blood
deep into the medulla and an ascending
Proximal convoluted tubule a highly limb bringing it back to the cortex.
twisted section o the nephron, with cells in the
wall having many mitochondria and microvilli Venules carry blood to the renal vein.
projecting into the lumen o the tube.
Te role of te loop of henl
The loop of Henl maintains hypertonic conditions
in the medulla.
The overall eect o the loop o Henl is to create a gradient o solute
concentration in the medulla. The energy to create the gradient is
expended by wall cells in the ascending limb. Here sodium ions are
pumped out o the ltrate to the fuid between the cells in the medulla
called the interstitial fuid. The wall o the ascending limb is unusual in
that it is impermeable to water, so water is retained in the ltrate, even
492
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