NOTA BIOLOGY SEM1 2021_2022

EXAMPLE 1

In a population of cows at genetic equilibrium, 640 cows have long horns (dominant trait), while remaining 360 cows have
short horns (recessive trait). Answer the following questions.

a) What is the frequency of dominant and recessive alleles? Total number of
samples:1000 → 3 dp

Frequency of homozygous recessive, q2 = 360

(640 + 360) 1. Always start the calculation by finding

q2 = 0.36 the value of q first.

Frequency of recessive alleles, q = √ 0.36 2. Compulsory to write what do the
= 0.6
symbol represent at the beginning of
Given that, p + q = 1
p = 1–q calculation. For instants:
▪ Frequency of homozygous
Frequency of dominant alleles, p = 1 – 0.6
= 0.4 recessive, q2
▪ Frequency of recessive allele, q
▪ Frequency of dominant allele, p
3. Don’t forget to always write the

equation p + q = 1

b) Calculate the frequency of homozygous dominant, heterozygous and homozygous recessive.

Frequency of homozygous dominant, p2 = (0.4)2
= 0.16

Frequency of heterozygous, 2pq = 2(0.4)(0.6) Always show the calculation. Key in the
Frequency of homozygous recessive, q2 = 0.48
values into the formula first, and write the
= (0.6)2 answer. Just don’t straight away write the
= 0.36
answer.

c) How many heterozygous genotype cows with long horns?

Frequency of heterozygous, 2pq = 2(0.4)(0.6)
= 0.48

Number of heterozygous genotypes cow = 2pq x total number of cows
= 0.48 x 1000
= 480 cows

d) Determine the frequency of cows with long horn phenotypes.

Frequency of cows with long horn phenotypes = p2 + 2pq
= (0.4)2 + 2(0.4)(0.6)
= 0.16 + 0.48
= 0.64

e) If all the short horn cows were killed due to mad-cow diseases (bovine spongiform encephalopathy , BSE), while the
rest were randomly mating to produce next generation.
i. Calculate the frequency of dominant and recessive allele of the new generation.
ii. Percentage of cows with short horn.

Number of cows were left = 1000 - 360 Remember, each organism carries
= 640 cows a pair of alleles for particular gene.
Example :

New total alleles = 640 x 2 For 10 homozygous dominant
1280 alleles individuals (AA), meaning that we
have 20 alleles A.
Number of long horn cows with homozygous dominant genotypes = 160 cows
For 10 heterozygous individual has
a pair of alleles (Aa), meaning that
we have 10 allele A and 10 allele
a.

Number of long horn cows with heterozygous genotypes = 480 cows

Number of short horn cows with homozygous recessive genotypes = 360 cows (killed)

Frequency of dominant alleles = Number of dominant alleles in the population

(p) Total number of alleles in the gene pool

= (160 x 2) + 480

1280

= 800 Total number of You cannot use the Hardy-
1280 samples:1280 → 3 dp Weinberg equation here!

= 0.625 All surviving organism will be
randomly mating with each other to
Frequency of recessive alleles = Number of recessive alleles in the population produce new generation.
(q) Total number of alleles in the gene pool
= 480 So, you must use the basic
1280 formula to calculate the allele
= 0.375 frequencies in the new generation

Percentage of cows with short horn? .

Frequency of homozygous recessive, q2 = (0.375)2
= 0.141

Percentage = q2 x 100
= 0.141 x 100
= 14.1%

EXAMPLE 2

Rabbit’s ears can be either short or floppy, where short ears are dominant over floppy ears. There are 653 rabbits in a
population where 549 of them are rabbits with short ears.
a) Find the frequency of the dominant and recessive alleles
b) Find the frequency of individuals with dominant, heterozygous, and recessive genotypes.
c) The next generation of rabbits has 560 individuals with short ears and 840 individuals with floppy ears. Is the

population in Hardy-Weinberg Equilibrium? Solve for p and q.

Answer (a) :

Number of floppy ear rabbits (homozygous recessive genotype) = 653 - 549
= 104

Frequency of homozygous recessive, q2 = 104 Total number of
653 samples:653 → 2 dp

q2 = 0.16

Frequency of recessive allele, q = √ 0.16
= 0.4

Given that, p + q = 1
p = 1–q

Frequency of dominant allele, p = 1 – 0.4
= 0.6
Answer (b) :
Frequency of homozygous dominant, p2 = (0.6)2
= 0.36

Frequency of heterozygous, 2pq = 2(0.6)(0.4)
Frequency of homozygous recessive, q2 = 0.48

= (0.4)2
= 0.16

Answer (c):

Recall that previous generation had allele frequencies of = 0.6 and = 0.4

The next generation of rabbits has 560 individuals with short ears and 840
individuals with floppy ears. Is the population in Hardy-Weinberg Equilibrium?

Frequency of homozygous recessive, q2 = 840 Total number of
samples:1400 → 3 dp
(560 +840)

q2 = 0.6

Frequency of recessive allele, q = √ 0.6
= 0.775

Given that, p + q = 1
p = 1–q

Frequency of dominant allele, p = 1 – 0.775
= 0.225

No, the population is not in a state of Hardy-Weinberg Equilibrium because
the allele frequencies are not the same as the next generation.

EXAMPLE 3

One in 1000 people in the US get cystic fibrosis, which is caused by a homozygous recessive condition. If the
population is in equilibrium with respect to this locus,
a) What should be the frequency of carriers in the US?
b) Out of 8000 people, how many should be carriers?
c) What is the ratio of carriers to people with cystic fibrosis?

Answer (a) Total number of
Frequency of homozygous recessive, q2 = 1 samples:1000 → 3 dp
1000
q2 = 0.001

Frequency of recessive allele, q = √ 0.001
= 0.032

Given that, p + q = 1
p = 1–q

Frequency of dominant allele, p = 1 – 0.032

= 0.968

Frequency of carrier, 2pq = 2(0.968)(0.032)
= 0.062

Answer (b)
Number of carriers = 2pq x total number of people
= 0.062 x 8000
= 496 carriers.

Answer (c)
62 carriers : 1 cystic fibrosis

EXAMPLE 4

In a population of 500 wildflowers, purple flower phenotype is dominant over white flower phenotype. If the alleles
involved are A and a, and there are 320 AA plants, 160 Aa plants and 20 aa plants.
a) Calculate the genotype frequencies from the observed genotype numbers.
b) Apply the Hardy-Weinberg principle to calculate the expected genotype frequencies from the allele frequencies

in the population.
c) Determine whether the population is in Hardy-Weinberg equilibrium or not? give a reason.

Given = 320
purple flower plant with AA genotype = 160
purple flower plant with Aa genotype = 20
white flower plant with aa genotype

a. Calculation of observed genotypes frequencies: Total number of
samples:500 → 2 dp
Frequency of aa genotype plants = 20
500

= 0.04

Frequency of Aa genotype plants = 160
500

= 0.32

Frequency of AA genotype plants = 320
500

= 0.64

b. Calculation of expected genotypes frequencies:

Frequency of homozygous recessive, q2 = 20
500

q2 = 0.04

Frequency of recessive allele, q = √ 0.04
= 0.2

Given that, p + q = 1

p = 1-q
= 1 – 0.2

= 0.8

Frequency of heterozygous, 2pq = 2(0.8)(0.2)
= 0.32

Frequency of homozygous dominant, p2 = (0.8)2
= 0.64

c. The population is in Hardy-Weinberg equilibrium, because the observed genotype frequencies is same
as the expected frequencies.

If the population is in Hardy-Weinberg equilibrium, the observed genotype frequencies
will be (roughly) the same as the expected frequencies. A Chi-Square test is used to
determine if the observed and expected genotype frequencies are significantly different
from each other or not.

6.0 EXPRESSION OF

BIOLOGICAL INFORMATION

6.0 EXPRESSION OF BIOLOGICAL
INFORMATION

6.1 DNA and 6.2 DNA Replication 6.3 Protein 6.4 Gene Regulation
Genetic synthesis: & Expression - lac
transcription and
Information translation operon

Concept of Central Semi-conservative Transcription and The concept of
Dogma replication of DNA translation operon & gene

The enzymes and Codon and regulation
proteins involved Genetic codes
in DNA Replication The components
Transcription and of operon
The mechanism of the stages
DNA Replication The components
and the enzymes involved (initiation, of lac operon and
elongation and their function in E.
involved
termination) in the coli
formation of mRNA
The mechanism of
strand 5’ to 3’ the operon in the

Translation and absence and
the stages presence of
involved
lactose

Initiation Elongation (codon Termination
recognition,
peptide bond
formation and
translocation)

COURSE LEARNING OUTCOMES

6.1 DNA and genetic information

a) State the concept of Central Dogma (CLO1)

6.2 DNA Replication
a) Explain semi-conservative replication of DNA (CLO 3)
b) Explain the enzymes and proteins involved in DNA Replication (CLO 3)
c) Explain the mechanism of DNA Replication and the enzymes involved (CLO 3)

6.3 Protein synthesis: transcription and translation.
a) Explain briefly transcription and translation (CLO 1)
b) Introduce codon and its relationship with sequence of amino acid using genetic code
table (CLO 3)
c) Explain transcription and the stages involved (initiation, elongation and termination) in the
formation of mRNA strand 5’ to 3’ (CLO 3)
d) Explain translation and the stages involved in translation: (CLO 3)
i. Initiation, Elongation (codon recognition, peptide bond formation and
translocation)
ii. Termination

6.4 Gene Regulation & Expression - lac operon.
a) Explain the concept of operon & gene regulation. (CLO 3)
b) State the components of operon (CLO 1)
c) Explain the components of lac operon and their function in E. coli (CLO 3)
d) Explain the mechanism of the operon in the absence and presence of lactose (CLO 3)

6.1 DNA and genetic information

COURSE LEARNING OUTCOMES

a) State the concept of Central Dogma (CLO1)

i. The dogma is a framework for understanding the flow of genetic information between DNA,
RNA & protein, in living organisms.

ii. The DNA replicates its information in a process that involves many enzymes; known as DNA
replication.

iii. DNA information can be copied into mRNA during transcription.
iv. In eukaryotic cells, the mRNA proceeds with post-transcriptional processing/ RNA processing

before migrates from nucleus to the cytoplasm.

v. mRNA as a template carries coded information from DNA (nucleus) to cytoplasm & ribosomes

use it for protein synthesis. This process is called translation.

Figure1: Central Dogma

(https://en.wikipedia.org/wiki/Central_dogma_of_molecular_biology)

6.2 DNA Replication

COURSE LEARNING OUTCOMES

a) Explain semi-conservative replication of DNA (CLO 3)

1. Definition: The process by which a DNA molecule is copied/ DNA synthesis (Campbell 11th ed.)
2. Three alternative models for DNA Replication:

i. Semi-conservative model – the two strands of the parental molecules separate, and each
functions as a template for synthesis of a new, complementary strand.

ii. Conservative model – the two parental strands reassociate after acting as templates for
new strands, thus restoring the parental double helix.

iii. Dispersive model – each strand of both daughter molecules contains a mixture of old and
newly synthesized DNA.

Figure 2: DNA Replication: three alternative models. (https://www.mun.ca/biology/scarr/iGen3_03-
01.html)

DNA REPLICATION
1. DNA replication is semi-conservative meaning that the replicated DNA consists of one old strand

(derived from the old molecule) and one new strand.
2. Watson and Crick’s model predicts that when a double helix replicates, each of the two daughter

molecules will have one old strand, from the parental molecule, and one newly made strand.
3. Matthew Meselson and Franklin Stahl devised a clever experiment that distinguished between the

three models (Campbell 12th Ed., page 322).
4. Their results supported the semiconservative model of DNA replication, as predicted by Watson

and Crick

Figure 3: The Watson’s an Crick Model

Rules in DNA Replication
1. Synthesis always in the 5’ to 3’ direction
2. Complementary base pairing (A=T, C=G)
3. Start at the origin of replication
4. Semi-conservative
5. RNA primers required
6. Can be unidirectional (such as in DNA of mitochondria/chloroplast) or bidirectional

i. DNA replication occurs during the Synthesis phase (S phase) of Interphase before cell
division.

ii. DNA replication begins (on DNA strands) at the origin of replication.
iii. Origin of replication : Site where the replication of a DNA molecule begins, consisting of a

specific sequence of nucleotides. (Campbell 11th ed.)
iv. Prokaryotic cells usually have only one origin of replication on each circular DNA molecule.
v. In a eukaryotic cell, the process is speeded up by having multiple origins of replication.

Origin of replication in Origin of replication in eukaryotic cell
prokaryotic cell

i. Replication begins at specific sites called origin of replication, where two parental strands
separate and form replication fork (replication bubbles); Y-shaped region.

ii. The bubbles expand laterally, as DNA replication proceeds in both directions (bidirectional).
iii. Eventually, the replication bubbles fuse, and synthesis of daughter strands is complete.
iv. When the parental DNA is unwound, new DNA is synthesized at the replication fork.
v. The overall direction of DNA synthesis must be 5’ to 3’ because the DNA polymerases only

synthesize DNA in the 5’ to 3’ direction.
vi. DNA synthesis is initiated by RNA primer.
vii. RNA Primer – a short stretch of RNA with a free 3’ end, bound by complementary base pairing

to the template strand and elongated with DNA nucleotides during DNA replication. (Campbell
11th ed.)
viii. One strand of DNA is synthesized continuously toward the replication fork known as the
leading strand.
ix. Meanwhile the other strand of DNA is synthesized discontinuously away from the replication
fork known as lagging strand.

COURSE LEARNING OUTCOMES

b) Explain the enzymes and proteins involved in DNA Replication (CLO 3)

The Enzymes & Proteins involved in DNA Replication.
1. DNA Helicase
2. Single-stranded binding proteins
3. Topoisomerase/ DNA Gyrase
4. Primase
5. DNA Polymerase I & III
6. DNA Ligase

The enzymes and its Function.
1. DNA Helicase - enzyme that is involved in unwinding the double strand DNA to produce two single

strands of DNA.
2. Single Stranded Binding Protein - to stabilize the unwound DNA strand.
3. Topoisomerase/ DNA Gyrase - enzyme that relieves tension to the DNA molecule by nicking &

cutting certain places on the sugar phosphate backbone.
4. Primase - synthesize the RNA primer.
5. DNA Polymerase III - uses the RNA primer to start adding free DNA nucleotides on the single

strand of DNA template.
6. DNA Polymerase I - remove the RNA primer and fill the gap with DNA nucleotides.
7. DNA Ligase - joins the Okazaki fragments by forming phosphodiester bonds to form continuous

DNA strands.

COURSE LEARNING OUTCOMES
c) Explain the mechanism of DNA Replication and the enzymes involved (CLO 3)

DNA replication is semi-conservative meaning that the replicated DNA consists of one old strand
(derived from the old molecule) and one new strand.

Mechanisms in DNA replication:

Step 1 : Unwinding the double helix strand of DNA.
● DNA Helicase - enzyme that is involved in unwinding the double strand DNA by breaking
hydrogen bonds, to produce two single strands of DNA. Then the Single Stranded Binding Protein
binds to stabilize the unwound DNA strand. Then topoisomerase/ DNA Gyrase - an enzyme that
relieves tension to the DNA molecule by nicking and cutting certain placed on the phosphate
backbone

Step 2 : Synthesis of RNA primer
● Primase synthesize RNA primer to initiate the process of replication by adding RNA nucleotide
using 3’ end of the parental DNA strand as a template. The new DNA strand will start synthesized
from the 3’ end the RNA primer

Step 3 : Assembling complementary strands
● DNA Polymerase III adds free DNA nucleotides to the 3’ end of RNA primer and complementary
to the bases of the template. This strand grows in the direction 5’ to 3’ since new nucleotides are
added only to the 3’ end of the growing strand.

Step 4 : Formation of leading & lagging strands
● Upon replication, two strands will be produced; Leading and Lagging strand. Leading strand:
Replications occurs continuously occur towards the replication fork. Only 1 primer is needed.
Lagging strand: Replication occurs discontinuously away from the replication fork, forming
Okazaki fragments. More than 1 primer is needed.

Step 5 : Removing the RNA primer
● Enzyme DNA polymerase I removes the RNA primer and fills in the gap with DNA nucleotides,
as well as any gaps between Okazaki fragments.

Step 6 : Joining the Okazaki fragments
● DNA Ligase joins the Okazaki fragments by forming phosphodiester bonds to form continuous
DNA strands. Two identical copies of DNA are produced.

6.3 Protein Synthesis: transcription & translation

COURSE LEARNING OUTCOMES

a) Explain briefly transcription and translation (CLO 1)
1. Transcription - The synthesis of RNA (mRNA) using the DNA as a template (Campbell 11th
ed.). Result in mRNA complementary to the gene sequence of one of the double helix
strands.
2. Translation – The synthesis of polypeptide using the genetic information encoded in mRNA
(Campbell 11th ed.) During translation; linear sequence of bases in mRNA is translated into
the linear sequence of amino acids/ polypeptide/ protein and it occurs at ribosome.

Figure 4: Overview of transcription and translation.

3. In eukaryotic cells, transcription occurs in the nucleus and translation occurs mainly at
ribosomes in the cytoplasm.

4. In prokaryotic cells, bacteria have no nucleus, both transcription and translation occurs in
cytoplasm.

a) Eukaryotic cell b) Prokaryotic cell

Figure 5: Overview: the roles of transcription and translation in the flow of genetic information.

COURSE LEARNING OUTCOMES

b) Introduce codon and its relationship with sequence of amino acid using genetic code table
(CLO 3)

Genetic codes & Codons.
i. Information is transferred in the form of codes. This is the ‘language’ of the genes, also known
as the genetic code.
ii. There are 20 different amino acids that make up proteins in the body.
iii. But there are only 4 different bases in the DNA & RNA. This would mean that one base cannot
code for one amino acid since there are 20 amino acids available.
iv. If taking only 2 bases give 16 (42) combination which does not account for all the amino acid
available.
v. A sequence of 3 bases seems the most probable since it can give 64 (43) possible combinations
of bases, this could count for the twenty amino acids available in organisms.

vi. A sequence of three bases (triplet code) in mRNA is called a codon.
vii. One codon codes for one amino acid but one amino acid may be coded by more than one

codon.
viii. Codons not only specify for protein but also responsible as codes for starting or stopping

polypeptide sequence.
a) Starting codon

➢ The starting point of synthesis is determined by the first codon/ start codon; AUG
on the mRNA

b) Stop codon / nonsense codon
➢ Three triplet bases of stop signal/ codon: UGA, UAG,UAA.
➢ The newly synthesized protein & the mRNA are released when the ribosome
encounters a stop codon.
➢ Nonsense codon because do not code for any amino acid.

Figure 6: The codon table for mRNA (Campbell, 12th ed.).

Characteristics of codon in Genetic code
1. Genetic Code is almost universal.

- Same for all organisms (bacteria, plant or animal). However, the genetic code is not quite
universal.

- e.g.: Mitochondrial genome (human, mice, cattle). UGA is read as the tryptophan.

2. The code has start & stop signals :
- There is one start codon & three stop codons.

3. Genetic Code is degenerate.
- Amino acids may be coded by more than one codon. An amino acid can be specified by more
than 1 triplet codon.
- From the Genetic Code table; there are more codons than amino acids.
- e.g. : UCU ,UCC, UCA, UCG, AGU, AGC – code for serine.

4. The Genetic Code is unambiguous
- Each triplet codon has only one meaning.
- Each triplet specifies one amino acid.

COURSE LEARNING OUTCOMES

c) Explain transcription and the stages involved (initiation, elongation and termination) in the
formation of mRNA strand 5’ to 3’ (CLO 3).

Transcription -.The process by which genetic information contained in DNA is transcribed into mRNA
using RNA polymerase. Transcription can be separated into three stages: Initiation, Elongation &
Termination

Stage 1 : Initiation
i. The promoter of a gene nucleotides typically extends several dozen nucleotide pairs
“upstream” from the start point.
ii. In eukaryotes, proteins called transcription factors recognize the promoter region or at TATA
box & bind to the promoter.
iii. After transcription factors have bound to the promoter, RNA polymerase binds to transcription
factors (at promoter) to create a transcription initiation complex.
iv. RNA polymerase then starts transcription

Stage 2: Elongation
v. RNA Polymerase adds free RNA nucleotides to the 3’ end of the growing RNA molecule as it
continues along the double helix.
vi. The new RNA molecule peels away from its DNA template and the DNA double helix reforms.

Stage 3: Termination
vii. Transcription proceeds until after the RNA polymerase transcribes a terminator sequence
(AAUAAA) in the DNA.
viii. In eukaryotes, the RNA polymerase continues transcribing for hundreds of nucleotides past the
terminator sequence, AAUAAA.
ix. At a point about 10 to 35 nucleotides past this sequence, AAUAAA, the pre-mRNA is cut from
the enzyme.

RNA processing /post-transcriptional processing.
x. Enzymes in the eukaryotic nucleus modify pre-mRNA before the genetic messages are

dispatched to the cytoplasm.

1. Alteration of pre-mRNA ends
xi. At the 5’ end of the pre-mRNA molecule is capped off with a modified form of a guanine

nucleotide forming a 5’ cap.
xii. At the 3’ end, an enzyme adds 50 to 250 adenine nucleotides, forming a poly-A-tail.
xiii. 5’ cap and poly-A-tail at mRNA also known as non-translated leader and trailer segments.

Figure 7: RNA processing: addition of the 5’ cap and poly-A tail (Campbell, 12th ed.).

2. Split Genes and RNA Splicing.
xiv. The most remarkable stage of RNA processing occurs during the removal of a large portion of

the RNA molecule during RNA splicing.
xv. Introns - Non Coding segments, lie between coding regions.

xvi. Exons - Coding segments, lie between noncoding regions.
xvii. RNA splicing removes introns and joins exons to create an mRNA molecule with a

continuous coding sequence; by complex called spliceosome
xviii. The final mRNA transcript includes coding regions, exons, that are translated into amino acid

sequences, plus the leader and trailer sequences.

Figure 8: RNA processing: RNA splicing (Campbell, 12th ed.).

COURSE LEARNING OUTCOMES

d) Explain translation and the stages involved in translation: (CLO 3)
i. Initiation
ii. Elongation (codon recognition, peptide bond formation and translocation)
iii.Termination

Translation – The synthesis of polypeptide using the genetic information encoded in mRNA involved 3
steps : Initiation, Elongation & Termination;
Step1 :Initiation

i. A small ribosomal subunit binds with mRNA
ii. Initiator tRNA which carries methionine (with anticodon – UAC) attaches to the start codon

(AUG) at mRNA
iii. The union of mRNA, initiator tRNA and small ribosomal subunit is followed by the attachment

of a large ribosomal subunit completing a translation initiation complex

Figure 9: Translation : Initiation
Step2 :Elongation
Consists of a series of three step cycles as each amino acid is added to the proceeding one.

a) Codon recognition
b) Peptide bond formation
c) Translocation

a) Codon Recognition
i. An aminoacyl-tRNA molecule with an anticodon complementary to the exposed mRNA codon
binds at A site.
ii. mRNA codon in A site forms hydrogen bonds with the anticodon of an incoming molecule of
tRNA carrying its appropriate amino acid.

b) Peptide Bond Formation
i. The large ribosomal subunit catalyzes by the action of peptidyl transferase cause the formation
of a peptide bond between the polypeptide in the P site with the newly arrived amino acid in the
A site.
ii. This step separates the tRNA at the P site from the growing polypeptide chain & transfers the
chain, now one amino acid longer, to the tRNA at the A site.

c) Translocation
i. During translocation the ribosome moves one codon ahead, the tRNA that had been in the P
site is moved to the E site & then leaves the ribosome.
ii. The tRNA with the attached polypeptide from the A site moves to the P site (the anticodon
remains bonded to the mRNA codon).
iii. The next codon is now available at the A site.
iv. The three steps of elongation continue codon by codon to add amino acids until the polypeptide
chain is completed.

Figure 10: Translation: Codon recognition, Peptide bond formation & Translocation

Step3 : Termination
i. Termination occurs when one of the three stop codons reaches the A site.
ii. A release factor binds to the stop codon (UAG/ UAA/ UGA) and hydrolyzes the bond between
the polypeptide and its tRNA in the P site.
iii. This frees the polypeptide and the translation complex disassembles

Figure 11: Translation: Termination

6.3 Gene regulation and expression – lac operon

COURSE LEARNING OUTCOMES
a) Explain the concept of operon & gene regulation. (CLO 3)

Jacob & Monod’s – Operon Model (1961)
In 1961, French microbiologist Francois Jacob & Jacques Monod proposed operon model to explain
the regulation of gene expression in prokaryotes; they received a Nobel prize for this.

What is operon?
i. Operon - A genetic unit that consists of one or more structural genes that codes for
polypeptides, and promoter gene that controls transcriptional activities. (Essential Genetics 5th
ed.)
ii. Operon also know as functioning unit of genomic DNA containing a cluster of genes
under the control of a single promoter.
iii. The genes are transcribed together into an mRNA strand then translated together into enzyme;
in the cytoplasm of bacteria .
iv. Bacteria adapt to changes in their surroundings by using regulatory protein to turn groups of
genes in activated mode or inactivated mode; in response to various environmental signals.
v. When the genes in an operon are transcribed, a single mRNA is produced for all the genes in
that operon.

Lactose Operon (in Escherichia coli bacteria)
i. The first control system for enzyme production worked out at the molecular level described the
control of enzymes that are produced in response to the presence of the sugar lactose in E.
coli bacteria,
ii. The bacteria can absorb the lactose and break it down for energy or use it as a source of
organic carbon for synthesizing other compounds.
iii. Lactose metabolism begins with hydrolysis of the disaccharide into its two component
monosaccharide, glucose and galactose.
iv. The enzyme that catalyzes this reaction is called β-galactosidase.
v. The following is the pathway that leads to the production of glucose and galactose.

β-galactosidase.
Lactose ------------------------> Glucose + Galactose

vi. The operon that regulates lactose metabolism in the bacterium Escherichia coli.
vii. In bacteria, the genes coding for the enzymes of a particular pathway are clustered together

and transcribed into a single polycistronic mRNA molecule.

COURSE LEARNING OUTCOMES

b) State the components of operon (CLO 1)

An operon is a functioning unit of nucleotide sequences of DNA including an operator, a promoter, &
one or more structural genes, which is controlled as a unit to produce an messenger RNA (mRNA),
in the process of transcription by an RNA polymerase.

Figure 12: The component of operon.

c) Explain the components of lac operon and their function in E. coli (CLO 3)

i. Promoter - a sequence of DNA where RNA polymerase attaches or first binds; when a gene
is transcribed.

ii. Operator - a short sequence of DNA where the repressor protein binds, preventing RNA
polymerase from attaching to the promoter. A region between the promoter & determines the
activation of the gene transcription

iii. Structural genes - code for enzymes of a metabolic pathway; are transcribed as a unit.
- The lactose operon consists of three structural genes (lacZ, lacY, and lacA), promoter &
operator.
- If E. coli is denied glucose & given lactose instead, it makes three enzymes to metabolize
lactose.
Type of structural gene & its function:
1. lacZ gene codes for β-galactosidase.
o Function β-galactosidase. - catalyze the hydrolysis lactose to glucose &
galactose.
2. lacY gene codes for a permease.
o Function permease. - catalyze the transport of lactose into the cell.
3. lacA gene codes for enzyme transacetylase.
o Function transacetylase - catalyze the transfer of an acetyl group from acetyl-
CoA to β-galactosides.

Regulatory genes (lacI)
i. This regulatory gene (lacI) located outside of the lac operon, codes for a repressor protein that
can switch off the lac operon by binding to the operator and thus inhibits transcription.
(Campbell 12th ed. pg:418)
ii. The lac operon is said to be inducible operon because its transcription is usually off but can be
stimulated (induced) when a specific small molecule (inducer) interacts with a repressor protein
(product of regulatory gene, lacI).

COURSE LEARNING OUTCOMES

d) Explain the mechanism of the operon in the absence and presence of lactose (CLO 3)
In the lac operon, the inducer is allolactose, an isomer of lactose formed in the small amounts from
lactose that enters the cell.
In the absence of lactose (and therefore allolactose),

1. The lac repressor protein in its active shape and binds to the operator.
2. When lac repressor protein binds to the operator, it will prevent the RNA polymerase from

binding to the promoter.
3. When RNA polymerase failed to bind, there is no transcription of lacZ, lacY & lacA genes,
4. No mRNA will be produced.
5. No enzymes will be synthesized, where the enzymes β-galactosidase, permease &

transacetylase are not produced.
6. Thus, the genes of lac operon are silenced.

Figure 13: Lactose absence, repressor active, operon off.

If lactose is presence to the cell surrounding.
1. Lactose will undergo isomerization into allolactose (inducer)
2. Allolactose binds to an allosteric site of the repressor protein.
3. The binding of allolactose will cause the repressor protein to alter its shape/ change its
conformational shape. Known as inactive lac repressor protein.
4. The inactive lac repressor cannot longer bind to the operator / will detach from the operator.
5. RNA polymerase can then bind to the promoter and start transcribing the lac genes ( lacZ,
lacY & lacA) into polycistronic mRNA.
6. The enzymes β-galactosidase, permease & transacetylase are produced.

Figure 14: Lactose presence, repressor inactive, operon on.

7.0 MUTATION

COURSE LEARNING OUTCOMES

7.1 MUTATION CLASSIFICATION AND TYPES

a) Define mutation (CLO1)
b) Classify mutation to

i. gene/ point mutation
ii. chromosomal mutation (CLO1)
c) State two types of mutation:
i. spontaneous mutation
ii. induced mutation (e.g. exposure to mutagens) (CLO1)
d) Define mutagen. (CLO1)
e) State types of mutagen:
i. physical (e.g. UV rays and gamma rays)
ii. chemical (e.g. colchicine and ethidium bromide) (CLO1)

7.2 GENE MUTATION

a) Define gene mutation (CLO1)
b) State the four types of gene mutation. (CLO1)
c) Explain four types of gene mutation: (CLO3)

i. base substitution
ii. base insertion
iii. base deletion
iv. base inversion

d) Explain base substitution (e.g. sickle cell anaemia as missense mutation) (CLO3)
e) State the effect of base substitution (missense, nonsense and silent mutation) and base insertion and

base deletion (frameshift mutation). (CLO1)
f) Explain base insertion and base deletion as frameshift mutation. (CLO3)

7.3 CHROMOSOMAL MUTATION

a) Define chromosomal mutation. (CLO1)
b) State two types of chromosomal mutation: (CLO1)

i. changes in chromosomal structure/ chromosomal aberration.
ii. changes in chromosomal number
c) Explain changes in chromosomal structure/ chromosomal aberration. (CLO3)
d) Explain types of chromosomal aberration: (CLO3)
i. translocation
ii. deletion (segmental deletion) (e.g. cri du chat)
iii. inversion
iv. duplication
e) Explain alteration of chromosome number. (CLO3)
f) State the types of the alteration: (CLO1)
i. aneuploidy
ii. euploidy/ polyploidy
g) Explain aneuploidy. (CLO3)
h) State aneuploidy effect on autosomal chromosome (Monosomy 21 and Trisomy 21) and sex
chromosome (Klinefelter syndrome and Turner syndrome). (CLO1)
i) Explain autosomal abnormalities and their effects: (CLO3)
i. Monosomy (Monosomy 21)
ii. Trisomy (Down syndrome/ Trisomy 21)
j) Explain sex chromosomal abnormalities: (CLO3)
i. Klinefelter syndrome (47, XXY)
ii. Turner syndrome (45, XO)
k) Explain euploidy/ polyploidy: (CLO3)

i. autopolyploidy
ii. allopolyploidy

7.1 MUTATION CLASSIFICATION AND TYPES

a) Define mutation (CLO1)

MUTATION

Mutation - A change in the nucleotide sequence of an organism’s DNA or in the DNA or RNA virus.
(ref: Campbell 11th ed.)

• Occurs randomly at gene or chromosome.
• Mutations create genetic diversity (variation) among individuals.
• Mutation leads to change in genotype & phenotype, that is different from its parent.
• Can be passed from generation to generation (inheritable) if it occurs in gamete cells.
• Mutation in somatic cells is passed on to daughter cell by mitosis.

b) Classify mutation to: (CLO1)

i. gene/ point mutation
ii. chromosomal mutation

CLASSIFICATION OF MUTATION

c) State two types of mutation: (CLO1)
i. spontaneous mutation
ii. induced mutation (e.g. exposure to mutagens)
TYPES OF MUTATION

d) Define mutagen (CLO1)
e) State types of mutagen: (CLO1)

i. physical (e.g. UV rays and gamma rays)
ii. chemical (e.g. colchicine and ethidium bromide)

MUTAGEN
Mutagen – A chemical or physical agent that interacts with DNA & causes a mutation.
(ref: Campbell 9th ed.)

TYPES OF MUTAGEN

Mutant
• An organism that carries a mutated gene or chromosome.
• Or an individual that undergoes changes in base pair sequence or chromosome number or structure.
• Change in genotype is associated with a change in phenotype.

7.2 GENE MUTATION

a) Define gene mutation (CLO1)
GENE MUTATION

The changes in the nucleotide sequence of genetic material (DNA) that represents a gene.
• Gene mutation can cause mistakes in base pairing during DNA replication or misreading of
the genetic code during translation phase of protein synthesis.
• Lead to the change in amino acid sequences, thus may change the proteins/enzymes
produced.
• Effect of mutation; protein not functioning as normal// unable to function.
• When the gene mutation involves only a single base pair, it is called a point mutation.

b) State the four types of gene mutation. (CLO1)
c) Explain four types of gene mutation: (CLO3)

i. base substitution
ii. base insertion
iii. base deletion
iv. base inversion

TYPES OF GENE MUTATION

i) Base substitution
● One nucleotide/nucleotide pair is
replaced by another nucleotide in
the genetic material (DNA).

ii) Base insertion
● One or more nucleotides are
inserted in the genetic material
(DNA)/gene.
● Resulting in frameshift mutation.
● Cause the change of all amino acid
at and after the point of insertion

iii) Base deletion
● One or more nucleotides
disappear/ remove/ delete in the

genetic material.
● Resulting in frameshift mutation.
● Cause the change of all amino acids

at and after the point of deletion.

iv) Base inversion

● More than one nucleotide or small

region of DNA breaks off & rotates

1800 before rejoining the DNA.

● Causes change in the particular

codons in mRNA (transcription).

● Amino acid sequence is also

changed (translation).

● Results in the changes on the

particular amino acids / inversion of

amino acids sequence / produced

truncated / short polypeptide.

● Cause minor phenotypic

abnormalities.

b) Explain base substitution (e.g. sickle cell anaemia as missense mutation). (CLO3)

c) State the effect of base substitution (missense, nonsense and silent mutation) and base
insertion and base deletion (frameshift mutation). (CLO1)

EFFECT OF BASE SUBSTITUTION- MISSENSE, NONSENSE AND SILENT MUTATION

Figure show a normal DNA strand that i. Missense mutation
act as template during protein
synthesis. The template DNA will
produce single strand mRNA (through
transcription) and normal polypeptide
chain (through translation).

● Changes in the base
sequence will result in
changes of codon.

● Changes in the codon lead
to changes in the amino
acid sequence.

● In other words, it is a type of
base substitutions that
change one amino acid to
another one in the

polypeptide (still code for an
amino acid, but not the
correct amino acid), during
translation.
● Example effect is sickle cell
anaemia.

Sickle cell anaemia (defective in
erythrocyte)

● Normal people have normal

haemoglobin that consist of

2α & 2 β polypeptide chains.

● The replacement of base

thymine (T) to base adenine

(A) at the 17th nucleotide DNA

for the β-chain of

haemoglobin/ CTT to CAT.

● changes the codon GAA into

GUA.

● Glutamic acid is replaced with

valine in both β-polypeptide

chains.

● Caused abnormal

haemoglobin allele (Hb-S) in

erythrocyte.

● Erythrocyte with Hb-S are

sickle-shape/ deformed

shape of red blood cell

● That shape is not efficient for

transporting oxygen

throughout the cell.

● Caused blood to clot in small

blood vessel

● The HBS produces pain &

eventually damaging organs.

Symptom of sickle cell anaemia
● The person suffers physical
weakness/fatigue/headache,
pale skin, chest pain, short
breath.
● Fatal form of anaemia.
● Insufficient supply of oxygen
to the organs.
● Organ damage.
● Physical weakness / fatigue.
● Periodic pain occurs in the
chest, abdomen, bones and
joints due to blockage of tiny
blood vessels by sickle-
shaped RBC.

ii. Nonsense Mutation

● Occurs when base

substitution results in

changing a codon into stop

codon.

● It causes translation to be

terminated prematurely (stop

codon; UAA, UAG, UGA).

● The earlier in the gene

sequence that this mutation

occurs, the more truncated

(made shorter) the

polypeptide/protein product

becomes & more likely that it

will be unable to function.

iii. Silent Mutation

• A change in base sequence
transforms one codon to
another codon that still
codes/translates the same
amino acid.

• Called silent; do not cause
changes in their amino acid
product because they have
no effect on the encoded
protein (because of
redundancy in the genetic
code).

Mutated haemoglobin is resistant to malaria. How it occurs?

The malaria parasite life cycle involves two hosts. During a blood meal, a malaria-infected female
Anopheles sp. a mosquito inoculates sporozoites into the human host. Sporozoites infect liver cells and
mature into schizonts, which rupture and release merozoites.

The body destroys the sickle red blood cells rapidly. Plasmodium sp./ malaria parasites cannot
thrive / survive in abnormal / sickle cell haemoglobin / HbS.

The sickle cells have membranes, stretched by their unusual shape, that become porous and leak
nutrients that the parasites need to survive and the faulty cells eventually get eliminated quite fast by the
organisms, destroying the parasite along the way.

e) Explain base insertion and base deletion as a frameshift mutation. (CLO 3)

EFFECT OF BASE INSERTION AND BASE DELETION – FRAMESHIFT MUTATION

● Frameshift mutation is the effect of the

base insertion & base deletion.

● In gene mutation; inserts or deletes one or

more nucleotides (that are not multiples of

threes) at the DNA nucleotide sequence

change mRNA strand

(transcription).

● It also changes/shifts the reading frame

(the grouping of the codons) during

translation.

● Gene will be transcribed in the wrong

three base groups (codon).

● Codons will be translated differently & will

abrupt the coding sequence of amino acids.

Thus, codons will code different amino

acids.

● This will change all amino acids at and

after the point of mutation.

● Many of these deletions/insertions start in

the middle of a codon.

● Originally, the stop codons (UAA, UAG,

UGA) will not be read or may be created at

an earlier site.

● Because of frameshift mutation, the protein

produced may be abnormally short,

abnormally long, or contain the wrong

amino acids. It will be most likely not

functional.

● Effect of frameshift mutation ~ usually

harmful to humans.

● E.g.: Thalassaemia

7.3 CHROMOSOMAL MUTATION

a) Define chromosomal mutation. (CLO1)
● Changes in chromosomal structure (chromosomal aberration) & changes in
chromosome number (aneuploidy & euploidy / polyploidy).
● To cause changes in characteristics of the particular organism.

b) State two types of chromosomal mutation: (CLO1)
i. changes in chromosomal structure/ chromosomal aberration.
ii. changes in chromosomal number

c) Explain changes in chromosomal structure/ chromosomal aberration. (CLO3)
CHROMOSOMAL ABERRATION

• Changes to chromosomal structure result in abnormalities in the structure of
chromosomes.

• The breakage & re-joining of chromosome parts result in four structural changes within or
between chromosomes.

d) Explain types of chromosomal aberration: (CLO3)
i. translocation
ii. deletion (segmental deletion) (e.g. cri du chat)
iii. inversion
iv. duplication

CHROMOSOMAL ABERRATION (CHANGES IN CHROMOSOMES STRUCTURE)

TYPES DESCRIPTION

i. Translocation
● Occurs when a chromosomal
segment / section (or a region of a
chromosome) breaks off & attach/
rejoining another region; either the
same chromosome or another non-
homologous chromosome.
● Chromosomal material is maintained,
but in different arrangements after a
translocation.
● 2 types of translocation:
● Intrachromosomal translocation - A
chromosomal fragment breaks off &
joins within the same chromosome.
● Interchromosomal translocation - A
chromosomal fragment breaks off &
joins to a non-homologous
chromosome.
● Two types for interchromosomal
translocation:

1) Reciprocal translocation
(non-Robertsonian translocations)
● Chromosomal segments are
exchanged between two non-
homologous chromosomes.
● The results are two derivative non
homologous chromosomes; one
becomes extra-long & other one
becomes extra-short.
● Changes in position of the genes
involved.

● Chromosomal translocations have

been implicated in certain diseases.

● E.g: Chronic myelogenous

leukaemia (CML) or Philadelphia

diseases.

● In CML a portion of chromosome 22
has switched places with a small
fragment from a tip of chromosome 9.

2) Non reciprocal translocation
(Robertsonian translocations)
● Fragment rearrangement between
two non-homologous chromosomes,
that result from the fusion of the entire
long arms of two acrocentric
chromosomes (chromosomes with
centromere near their ends).

● Most common involve the five
acrocentric human chromosome
pairs such as chromosomes of 13, 14,
15, 21 and 22.

● The most frequent between
chromosomes pair:

● 13 & 14, 14 & 21, 21 & 22.
●
• Robertsonian translocation between

chromosome:
• 13 & 14 - Patau syndrome.
• 14 & 21 or 21 & 22 - Trisomy 21 or

Down syndrome.
• Down syndrome is caused in a minority

of cases (5% or less) by a
Robertsonian translocation of the
chromosome 21 long arm onto the long
arm of chromosome 14.
ii. Deletion
• Removal/loss of one segment of
chromosome, containing one or more
genes when the chromosome breaks
at two points.
• The remaining segments of the
chromosomes will join again & become
shorter.
• This results in the loss of certain genes
& can cause abnormalities & serious
genetic diseases.
• If deletion occurs only in one of the
homologous pairs, the allele present
on the other pair will be expressed
phenotypically.

• If deletion affects genes on both
homologous chromosomes, the effect
is harmful / lethal / genetic imbalance.

• An example in humans, the deletion of
a small part of the short arm of
chromosome 5 causes Cri-du-chat
syndrome (‘cry of the cat syndrome’).

• Characteristic of Cri du chat
syndrome individuals:
✔ mental retardation.
✔ small head.
✔ unusual facial features.
✔ a cry like the mewing of a
distressed cat.
✔ fatal in infancy and early
childhood (short lifespan).

iii. Inversion
• A rearrangement of a chromosome
segment, where a segment of
chromosome is break off, inverted
(rotated) 180o and rejoined within a
chromosome.
• Usually do not cause any
abnormalities, as long as the
rearrangement is balanced with no
extra/missing genetic information.

• 2 types of inversion: -
✔ pericentric – include the
centromere.
✔ paracentric – does not
include centromere.

• No change in the genotype but
phenotype may change.

• Also known as the position effect.

iv. Duplication
• Repetition of one or more segments of
chromosome; giving the additional sets of
genes.
• Chromosomes becomes longer due to
duplication.

e) Explain alteration of chromosome number. (CLO3)

● Changes in the number of chromosomes.
● Alterations of chromosome number cause some genetic disorders.

f) State the types of the alteration: (CLO1)
i. aneuploidy
ii. euploidy/ polyploidy

There are two types of alteration:
• The loss or gain of a single chromosome, a condition called aneuploidy.
• The increase in entire haploid sets of chromosomes, a condition called euploidy/ polyploidy.

g) Explain aneuploidy. (CLO3)
• Condition where the cell of diploid organism (2n) gains or loses 1 or more individual
chromosomes. (2n + 1) or (2n -1).
• The most common cause of aneuploidy is nondisjunction during meiosis I and II.
• It may occur:
• in autosomes.
• in sex chromosome - give effect to sex chromosomal abnormalities.

Nondisjunction
● Homologous chromosomes fail to separate and move properly to opposite poles during
anaphase meiosis I.
● Alternatively, sister chromatids fail to separate and move properly to opposite poles during
anaphase meiosis II.
● Occur due to meiotic spindle fibres problem; spindle fibres cannot
form during cell division.
● This process can affect production of gametes:
● Some gametes produced have an extra chromosome (n+1) and other gametes have a
chromosome missing (n-1).
● If nondisjunction occurs during meiosis I:
(a)Half the daughter cells/2 gametes produced have an extra chromosome (n+1) while the
other half/ 2 gametes have a chromosome missing (n-1).
● If nondisjunction occurs during meiosis II:
(b) One daughter cell/gamete produced have an extra chromosome (n+1), while the other
one gamete has a chromosome missing (n-1) and 2 gametes are normal (n).

h) State aneuploidy effect on autosomal chromosome (Monosomy 21 and Trisomy 21) and sex
chromosome (Klinefelter syndrome and Turner syndrome). (CLO1)

● Offspring results from fertilization of a normal gamete with one of these gametes will have an
abnormal chromosome number or aneuploidy.

• During fertilization the fusion of gametes occur between:

1. Gametes with chromosome (n+1) &
normal gamete (n),

✔ Produced embryo with

chromosome (2n+1).

✔ Example: Trisomy

Autosome (e.g.: Trisomy 21/

Down syndrome).

Sex chromosome (e.g.:

Klinefelter syndrome)

✔ Trisomy: refers to the gain of an

extra chromosome to a diploid

chromosome set.

✔ Individuals are called trisomic.

2. Gametes with chromosome (n-1) &
normal gamete (n),

✔ Produced embryo with

chromosome (2n-1).

✔ Example: Monosomy

Autosome (e.g.: Monosomy 21).

Sex chromosome (e.g.: Turner

syndrome

✔ Monosomy: refers to the loss of

one chromosome from a diploid

chromosome set.

✔ Individuals are called monosomic.

i) Explain autosomal abnormalities and their effects: (CLO3)

i. Monosomy (Monosomy 21)
ii. Trisomy (Down syndrome/ Trisomy 21)

AUTOSOMAL ABNORMALITIES DESCRIPTION

Monosomy (Monosomy 21):

2n-1/ 45 chromosomes

● A condition of monosomy 21
– lack of one chromosome

at chromosome 21.

● Occurs due to nondisjunction

during meiosis I or meiosis II

(anaphase I or anaphase II)

for chromosome 21.

● Produce abnormal gametes

contain 22 chromosomes

(n-1).

● A type of autosomal trisomy.

● Arise by the fusion of normal

gamete (n) or gamete with

23 chromosomes & a

gamete with a lack of one

chromosome number 21

(n - 1).

● Producing individuals with

chromosome number

2n – 1 (45 chromosome).

Symptom of monosomy 21 :
● Short distance between eyes
● Large ears
● Contracted muscle.
● Small head and jaw.
● Asymmetric face.

(a-c) Demonstrates the observed features of Monosomy-21

Trisomy (Trisomy 21):

(2n+1/ 47 chromosome)
• A condition of trisomy 21 –

extra of one chromosome

at chromosome 21.
• Occurs due to nondisjunction

of chromosome 21 during

meiosis I or meiosis II

(anaphase I or anaphase II).
• Produce abnormal gametes

contain 24 chromosomes

(n+1).
• A type of autosomal trisomy.
• Arise by the fusion of normal

gamete (n)/ gamete with 23

chromosomes and a gamete

with an extra of one

chromosome number 21 (n

+ 1).
• Producing individuals with

chromosome number 2n + 1/

47 chromosomes.

Characteristics of Down

syndrome such as :
● Mentally retarded
● Protruding furrowed tongue
● Flat and rounded face
● Flattened nose
● Slanted eyes
● Wide/broad forehead
● limbs/hands are often short
and stubby
● most infertile/sterile/sexually
underdeveloped
● usually short life span
● at higher risk for
infections/disease such as
leukaemia & Alzheimer’s

diseases
● heart defect
● respiratory problems

j) Explain sex chromosomal abnormalities: (CLO3)

i. Klinefelter syndrome (47, XXY)
ii. Turner syndrome (45, XO)

SEX CHOROMOSOMAL DESCRIPTION
ABNORMALITIES

● Abnormalities in the sex
chromosome number.

● Causes by non-disjunction
during spermatogenesis or
oogenesis at meiosis I or
meiosis II.

● Any extra copies of the sex
chromosome can cause
developmental errors.

Nondisjunction during

oogenesis:

● Women carry two sex

chromosome, X

chromosome

● If nondisjunction

happened during meiosis

I and II:

Some gametes might not

carry any X chromosome;

(O), others might carry two

X chromosomes; (XX).

● If nondisjunction

happened during meiosis

I and II:

Some gametes might not

carry any X

chromosome; (O), others

might carry two X

chromosomes; (XX) &

others might carry one X
chromosome (X).

Nondisjunction during

spermatogenesis:

● Men carry different sex

chromosome, normal

sperm have chromosome X

or Y.

● If non disjunction occur

during meiosis I, sperm

will have the abnormal sex

chromosome:

XY or O (not have X or Y

chromosome)

• If non disjunction during
meiosis II, sperm will have
the abnormal sex
chromosome:
XX, YY or O (not have X or
Y chromosome)

Klinefelter syndrome (47
chromosomes, XXY)
How does it occur?

● Causes by non-disjunction
at:

✔ Meiosis I or
Meiosis II during
oogenesis.

✔ Meiosis I during
spermatogenesis.

● If nondisjunction happens
during meiosis I or
meiosis II of oogenesis,
an abnormal gamete with
two X chromosomes (XX)
is produced.

● When this abnormal egg/
XX gamete combines with
normal sperm with Y
chromosome/ Y gamete
during fertilization, XXY
zygote is formed.

● This condition is called
Klinefelter syndrome.

● If nondisjunction happens
during meiosis I of
spermatogenesis, an
abnormal gamete with XY
chromosome (XY) is
produced.

● When this abnormal sperm/
XY gamete combines with
a normal egg with X
chromosome/ X gamete
during fertilization, XXY
zygote is formed.

● Male individual with

Klinefelter syndrome has

extra copies of the sex

chromosome, X can cause

developmental errors.

● The number of

chromosomes for this

individual is 44 + XXY.

Characteristics
● Sterile.
● Testes and prostate glands
are abnormally small,
produce small amounts of
testosterone,
● Have breast enlargement.
● Have long arms and legs.
● They may have feminine
characteristics such as soft
voice, and may have
subnormal intelligence.
● Others have a mild mental
retardation or learning
disability.

Turner syndrome (45, XO)

How does it occur?
● Causes by non-disjunction

at:

✔ Meiosis I or

Meiosis II during

oogenesis.

✔ Meiosis I during

spermatogenesis.

● If non-disjunction

happened during meiosis I

or meiosis II of

oogenesis, an abnormal

gamete without sex

chromosome is produced

(O).

● When this abnormal egg/ O

gamete combines with

normal sperm with X

chromosome/ X gamete

during fertilization, XO

zygote is formed.

● This condition is called

Turner syndrome.

● If non disjunction happened
during meiosis I of
spermatogenesis, an
abnormal gamete without
sex chromosome is
produced.

● When this abnormal sperm/
O gamete combines with a
normal egg with X
chromosome/ X gamete
during fertilization, XO
zygote is formed.

● Female individual with

Turner syndrome has lost a

copy of the sex

chromosome, X can cause

developmental errors.

● The number of

chromosomes for this

individual is 44 + XO.

Characteristics
● They are females who are
usually short.
● Broad-necked (webbed
neck).
● Sterile because lack of
menstruation and sex
organ does not mature, that
do not undergo changes
during puberty.
● Most have normal
intelligence.

k) Explain euploidy/ polyploidy: (CLO3)

i. autopolyploidy
ii. allopolyploidy

EUPLOIDY/ POLYPLOIDY DESCRIPTION
• The mutation that causes
Ploidy= The number of sets of chromosomes in a cell or an
an organism to undergo organism.
Example:
polyploidy/euploidy & have Haploid/monoploid : a set of chromosomes without their pair,
n.
more than two complete Diploid : 2 sets of homologous chromosomes, 2n.
Polyploid : multiple sets of chromosome pairs, 3n, 4n,…..
sets of chromosomes. 3n= 3 sets of chromosomes
• Polyploid is individual 4n= 4 sets of chromosomes

having more than two

complete sets of

chromosomes; >3n, 4n, 5n,
………etc
• Polyploid organism is

naming based on the total

sets of chromosomes
present:

• Occur as a result of

nondisjunction of all the

chromosome pairs,

producing gametes with

two sets of chromosomes

(2n).
• Euploidy/ Polyploidy has

two types:

✔ Autopolyploidy.

✔ Allopolyploidy.

TYPES OF EUPLOIDY/POLYPLOIDY

AUTOPOLYPLOIDY

● Increasing the genome

(complete set of

chromosomes) more than

two chromosome sets

derived from the same

species (single species).
● Occur due to non –

disjunction of all

chromosomes during

meiosis.

● Can arise from: a

spontaneous, naturally

occurring or induced

genome doubling of a

single species (by using

chemicals like colchicine). This autopolyploid mutant can reproduce with itself (self-
● Important in economic pollination) or with other tetraploids.

value is that autopolyploid

plants produce flower and

fruit bigger than normal

diploid plants.

An individual can have more than two sets of chromosomes from
a single species if a failure in meiosis results in a tetraploid (4n)
individual.

Allopolyploidy

● Allopolyploidy is the

increasing genome Mechanism 1- Chromosome doubling

(complete set of

chromosomes) of more

than two chromosomes

sets derived from a

different species through

hybridization and doubling

of genomes of different

species.

● F1 hybrids produced from
different species are

usually sterile.

● Because the haploid set of

chromosomes from one

species cannot pair during

meiosis with the haploid set

of chromosomes from the

other species.

● Gamete cannot produce.

● However, the hybrids may

be able to reproduce

asexually (vegetative

propagation) to colonize an

area.

● Various mechanisms can

transform a sterile hybrid

into a fertile polyploidy.

✔ Hybrids undergo

chromosome

doubling.

✔ Or hybrids can

interbreed with

either parent

species.

✔ Hybrids undergo the

chromosome doubling or

can interbreed with either

parent species to transform

a sterile hybrid into a fertile

polyploidy.

✔ Important in producing new

species, and important in

economic value such as

producing flowers & fruits

bigger, more resistant

compared to normal diploid

plant.

Example: Avena strigosa
2N = 2X = 14
where N = gametic number of chromosomes

X = ploidy number

- Chromosome number varies by species.

- Genomic (X) number is a set of different chromosomes.

Species Ploidy Somatic Gametic Genomic

chromosome chromosome chromosome

number number number

(2N) (N) (X)

Spinach 2X 12 6 6

Corn 2X 20 10 10

Potato 4X 48 24 12

Strawberry 8X 56 28 7

Mechanism 2- Interbreed with either parent species

● Unreduced (diploid) gametes fuse with normal haploid
gametes to form triploids.

● Triploids are normally sterile, but can contribute to
fertile hybrid formation by themselves producing
occasional,

● unreduced triploid gametes that can back-cross with a
normal haploid gamete

● to form fertile hybrids (new species).

The new species has a chromosome number equal to the sum
of chromosomes in the two parent species.



8.0 RECOMBINANT DNA
TECHNOLOGY

COURSE LEARNING OUTCOMES

8.1 RECOMBINANT DNA TECHNOLOGY

a) Define recombinant DNA technology (CLO1)
b) Define and explain the tools used in recombinant DNA technology: target DNA (gene of interest),

restriction enzymes, DNA cloning vector, host cell & modifying enzymes (DNA ligase). (CLO1)
c) Explain restriction enzyme & examples of enzymes that produce sticky ends.

(EcoRI: G AATTC) & blunt ends (SmaI: CCC GGG). (CLO3)
d) Explain the characteristics of plasmid as cloning and expression vector. (CLO3)
e) Explain the characteristics of E. coli as host cell (bacteria) and its characteristics. (CLO3)
f) Explain modifying enzyme and its function: (CLO3)

i. DNA ligase for DNA ligation
ii. Taq polymerase for DNA amplification using PCR

8.2 METHODS IN GENE CLONING

a) Overview using diagram to show the steps in gene cloning by using plasmid (CLO1)
b) Describe the steps in gene cloning by using plasmid as the vector: (CLO3)

i. Isolation of gene
ii. Cleave/cut
iii. Insertion
iv. Transformation and amplification
v. Screening (blue/white screening)

8.3 APPLICATION OF RECOMBINANT DNA TECHNOLOGY

a) Briefly explain applications of Recombinant DNA Technology in mass production of insulin using cDNA.
(CLO3)

b) Describe the steps in production of insulin using cDNA. (CLO3)

8.1 RECOMBINANT DNA TECHNOLOGY

a) Define recombinant DNA technology (CLO1)

TERMINOLOGIES DEFINITION
Recombinant DNA The DNA that contains genes from different sources that have been
combined by the techniques of ‘genetic engineering
Genetic engineering The process of changing the genetic make-up of an organism by
introducing DNA from another organism.
Recombinant DNA The technology of preparing recombinant DNA in vitro by cutting up
Technology DNA molecules & splicing together fragments from more than one
organism.

b) Define and explain the tools used in recombinant DNA technology: target DNA (gene of interest),
restriction enzymes, DNA cloning vector, host cell & modifying enzymes (DNA ligase). (CLO1)

No Tools Description Diagram

1 Target DNA (gene of - The DNA must be enzymatically cleaved

interest) to form shorter fragments.

- The target gene/ gene of interest can be

Definition - fragment of isolated using 2 methods,

chromosomal DNA to i) by cutting the gene from a complete

be cloned. chromosome

ii) by producing a complementary DNA

(cDNA).

2 Restriction Enzymes - Isolated from bacterial cells.

- Restriction sites are a DNA sequence EcoRI

Definition: An containing the cleavage site that is cut by a

endonuclease that particular restriction enzyme.

recognize & cleave

DNA molecule at

specific nucleotide

sequences (restriction

site).

3 DNA Cloning Vectors - Vector is an agent/ small DNA molecule
- Derived from a plasmid/ bacteriophage/
cosmid/ YAC.
- Carries foreign DNA fragments/ gene of
interest to be introduced into a host cell.

4 Host cell - Vector molecule cannot replicate in test

tube & it has to be taken up by a cell inside

Definition: cells in which it can then replicate.

which recombinant

DNA can be replicated

or recombinant

protein can be

produced.

5 Modifying enzyme - Example of modifying enzyme is DNA

ligase and Taq polymerase.

Definition: enzyme use

in modifying the DNA - DNA ligase used to join DNA fragments

strand. and plasmid by catalyses the formation of

phosphodiester bonds. This process is

called DNA ligation.

DNA ligase

c) Explain restriction enzyme & examples of enzymes that produce sticky ends.
(EcoRI: G AATTC) & blunt ends (SmaI: CCC GGG). (CLO3)

Characteristics of restriction enzyme:
1) Each restriction enzyme is very specific.
2) Recognizing a particular short DNA sequence / restriction site & cutting both DNA strands at precise

points within this restriction site.
3) Different restriction enzymes have different restriction sites.
4) Cut the sugar-phosphate backbones of DNA by breaking down phosphodiester bonds between

nucleotides.
5) Restriction site are palindromic sequence.

• Palindromic sequence is the same base sequence in opposite directions on both DNA strands.
6) Example of palindromic sequence of restriction site:

MODES OF ACTION OF RESTRICTION ENZYMES

Sticky Ends Blunt Ends

• Restriction enzyme that leaves a staggered cut with • Restriction enzymes that make a simple cut
single-stranded ends. across both strands at a single point, leaving,
what are called blunt ends.
• The unpaired base of a DNA fragment can form

hydrogen bonds with compatible of the sticky

end of another fragment; cut up by the same enzyme.

Example:

1.

2.

3.

d) Explain the characteristics of plasmid as cloning and expression vector. (CLO3)
Characteristics of cloning vectors:
1) Able to accept foreign DNA in multiple cloning sites (MCS).