Pra-U STPM Biology Term 3 2018 CB039148c

142 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 12. Another example is human height which is assumed to be controlled by two independently assorting genes. Similarly, a marriage of the shortest (assumed to be 100 cm) to the tallest (assumed to be 200 cm) would produce F1 generation of mean height of 150 cm. A dihybrid marriage of mean height of 150 cm would produce 6.25% ( 1 16) to be shortest of 100 cm or tallest of 200 cm. Therefore, contribution of each dominant allele = tallest – shortest 4 = 200 – 100 4 cm = 25 cm The marriage can be summarised as follows: P1 : A1 A1 A2 A2 × a1 a1 a2 a2 200 cm 100 cm F1 : A1 a1 A2 a2 mean height 150 cm F2 : Genotypes Ratios Phenotypes Ratios A1 A1 A2 A2 1 16 200 cm – 1 16 A1 A1 A2 a2 1 16 175 cm – 4 16 A1 a1 A2 a2 1 16 175 cm A1 a1 A2 a2 1 16 150 cm A1 A1 a2 a2 1 16 150 cm – 6 16 A1 a1 a2 a2 1 16 125 cm a1 a1 A2 A2 1 16 150 cm – 4 16 a1 a1 A2 a2 1 16 125 cm a1 a1 a2 a2 1 16 100 cm – 1 16 123 Therefore, the phenotypic ratio is 1 : 4 : 6 : 4 : 1. Linked Genes 1. Linked genes refer to two or more genes that are located in the same chromosome. 2. Each chromosome contains many genes, therefore all the genes within one chromosome are linked and they are said to be in the same linkage group. 3. Genes are linked on the autosomes or sex chromosomes such as the X chromosome. 4. Two linked genes do not follow the second law of Mendel and they do not assort independent of each other. Dihybrid cross of the genes does not produce the classical 9 : 3 : 3 : 1 phenotypic ratio.

143 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 5. A test cross of the dihybrid does not produce a phenotypic ratio of 1 : 1 : 1 : 1. 6. If there is no crossing over during meiosis, two linked genes ( AB ab ) are inherited as a single unit when test crossed, as shown below. P1: G1: F1: A B a x b A B a b a b a b a b a b B A a b a b Therefore, only two types of progeny are produced and the genes do not follow Mendel's second law of independent assortment. 7. There are two ways of getting a dihybrid of linked genes as shown below: (a) AB AB x AB AB P1: G1: ab ab ab ab (coupling phase) (b) Ab Ab Ab x Ab P1: G1: aB aB aB aB (repulsion phase) 8. If the two dihybrids are test-crossed, the gametes and the progeny are as shown in Figure 17.5: (a) AB Parental types P1: G1: ab AB ab Ab aB F1: AB ab Ab aB ab Large numbers Small numbers (Parental types) (Recombinant types) ab ab ab ab ab ab Cross-over types Exam Tips Linked genes exhibit 2 big and 2 small groups in the offspring when the dihybrid is test cross. (STPM 2014 essay question)

144 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 (b) Ab Parental types P1: G1: aB Ab aB AB ab F1: Ab aB AB ab ab Large numbers Small numbers (Parental types) (Recombinant types) ab ab ab ab ab ab Cross-over types Figure 17.5 Formation of parental and recombinant types Sex-linked Genes 1. Sex-linked genes are genes found in the sex chromosomes i.e. in the X chromosome of mammals and Drosophila or in the Z chromosome of birds and all of the chromosomes of honey bees. 2. In humans and other mammals, sex is determined by chromosomes X and Y. The male sex organ formation in the foetus is determined by a gene called factor t in the Y chromosome. This gene determines the formation of testosterone in a group of cells that form the testis. The testosterone in turns controls the formation of the male sex organs. Females have two X chromosomes and are considered homogametic, producing one type of ovum. Males have XY genotypes and are considered heterogametic, producing two types of spermatozoa, half with chromosome X and the other half with Y chromosome. An XY individual whose receptor for testosterone has mutated, cannot respond to testosterone and will develop female sex organs. 3. The symbols used to represent sex-linked genes as in Drosophila are as follows: XW – dominant allele for red eye, Xw – recessive allele for white eye. Female: XWXW – red, XWXw – red (carrier), XwXw – white Male: XWY – red, XwY – white 4. The characteristics of sex-linked genes are as follows: (a) Only one of the sexes can be a carrier of a recessive allele i.e. the individual who has two of such sex chromosomes. This is found in the females that have two X chromosomes as in mammals, Drosophila and honey bees; male birds and chickens that have two z chromosomes. (b) If a phenotype is controlled by codominant or incomplete dominant alleles, only the females have heterozygous phenotypes. For example a calico (tortoise-shell) cat is always female. XB represents black coat while XG represents ginger coat. Exam Tips If a test cross produces two different phenotypes of equally large numbers and the other two phenotypes of equally smaller groups, then the two genes involved are linked. The ones with large numbers are parental types and the others are crossing over (recombinant) types (STPM 2003 structured question). 2012

145 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 Females: XBXB – black Males: XBY – black XGXG – ginger XGY – ginger XBXG – calico (c) Reciprocal crosses always produce offsprings of different phenotypic ratios. Reciprocal crosses are two crosses; i.e. a female is crossed with a male of opposite phenotype and vice versa. Reciprocal crosses are used to test if a gene is sexlinked. For example, a red-eyed Drosophila is crossed with a white-eyed male: P1: Xw Y XWXw 1/2 red-eyed females : 1/2 red-eyed males F1: XWY XW Gametes: XWXW Red eye Xw Y White eye When a white-eyed female is crossed with a red- eyed male: P1: XW Y XWXw 1/2 red-eyed females : 1/2 white-eyed males F1: Xw Y Xw Gametes: Xw Xw White eye XWY Red eye (d) If a phenotype is determined by a recessive allele, then it is more frequently found in males because males have no heterozygous genotype. (e) Conversely, if a phenotype is determined by a dominant allele, then it is more frequently found in females because females have heterozygous genotypes. (f) If a genetic disease or mutant character, such as haemophilia is controlled by a recessive allele and if the mother is affected, then all the sons are affected. P1: XH Y XH Xh Normal females : haemophiliac males F1: Xh Y Xh Gametes: Xh Xh Haemophiliac XH Y Normal This is because the sons get the X chromosomes from the mother.

146 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 If the mother is a carrier, half the sons are affected. P1: XH Y XH Y Normal male XH XH Normal female All females normal : Males: 1/2 normal, 1/2 haemophiliac F1: Xh Y Haemophiliac male XH Yh Normal female Gametes: XH Xh carrier XH Y normal XH Xh (g) Conversely, if the genetic disease or mutant type is controlled by a dominant allele and if the father is affected, all the daughters are affected. P1: XD Y XD Xd All daughters affected Affected daughters Normal son F1: xd Y Xd Gametes: Xd Xd Normal XD Y Affected (h) Sex-linked genes usually manifest their phenotypes in an alternate generation pattern. The phenotype appears in one generation and may disappear in the next. An example is the transmission of colour blindness in humans in which XB is the dominant allele for normal and Xb is the recessive allele for colour blindness. XB P1: Gametes: Xb Y Y XB Xb Normal F2: XB Y Normal XB XB Normal P2 : Xb Y colour blind XB XB Normal XB Xb Normal F1: XB Y Normal Xb Y Colour blind XB Xb Gametes: XB Xb Normal XB Y Normal XB Colour blindness appeares Colour blindness disappeares A father cannot pass his sex-linked phenotype to his sons but pass it to his grandson through his carrier daughters.

147 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 Pedigree Analysis 1. Pedigree is a diagrammatic representation which shows the family tree of individuals whose affliction to a genetic disease or certain phenotypes is indicated, as shown below: Symbols: Generation I: II: III: Normal female Afflicted female 1 1 1 2 2 3 4 5 6 2 Normal male Afflicted male 3 4 2. From the pedigree, it is possible to determine if the affliction is caused by a recessive or dominant allele. 3. The affliction controlled by a recessive allele can be seen when a normal female married to a normal male produces at least one afflicted child, as shown below: Aa (normal) × Aa (normal) aa — afflicted 4. The affliction is controlled by a dominant allele when an afflicted couple produces at least a normal child, as shown below: Pp × Pp PP — normal Afflicted heterozygous 5. In marriages between one afflicted and one normal spouse, the dominant or recessive nature of the alleles involved cannot be confirmed. All offspring can be afflicted or non-afflicted, giving a high probability of affliction by a dominant allele or vice versa. Most probably dominant allele is involved Aa aa Aa Aa Most probably recessive allele is involved bb Bb Bb Bb

148 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 6. From the pedigree, it is possible to determine if the affliction is sex-linked i.e. the gene involved is found in the sex chromosome or autosome. 7. If the affliction is known to be caused by a recessive allele, an afflicted mother will produce a non-afflicted son. If there is such a non-affected son, it can be confirmed that the affliction is non-sex linked. Xa Y — all sons are afflicted XA Y normal Xa Xa Afflicted 8. If the affliction is known to be caused by a dominant allele, an afflicted father will produce an afflicted daughter. If there is no such daughter, it can be confirmed that the affliction is non-sex linked. XA xa — all daughters are afflicted XA Y Afflicted Xa Xa Normal 9. If the recessive or dominant nature of the afflicted allele cannot be confirmed, then the above two types of families can disprove the non-sex linked nature of the allele. The gene involved cannot be proven sex-linked. Exam Tips Remember the pattern of sex-linked inheritance from fathers to daughters and mothers to sons. Quick Check 2 1. Find out the parental genotypes of the following crosses: (a) Yellow round × green round, the phenotype of the offsprings are in the ratio of 3 8 yellow round : 3 8 green round : 1 8 yellow wrinkled : 1 8 green wrinkled. (b) Yellow wrinkled x green round, the phenotype of the offsprings are in the ratio of 1 4 yellow round : 1 4 green round : 1 4 yellow wrinkled : 1 4 green wrinkled. 2. Can you recognise a reason behind each of the following dihybrid phenotypic ratios? (a) 3 : 6 : 3 : 1 : 2 : 1 (b) 1 : 2 : 2 : 4 : 1 : 2 : 1 : 2 : 1 (c) 1 : 4 : 6 : 4 : 1 (d) 9 : 3 : 4 (e) 9 : 6 : 1

149 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 17.3 Genetic (chromosomal) Mapping Crossing-over, Parental and Recombinant Types 2016 1. Crossing-over is a process that occurs during the prophase I of meiosis. The homologous chromosomes pair and the non-sister chromatids stick together at the chromomeres and later form chiasmata where chromatids break. Further coiling and shortening of chromatids occur resulting in the breakage of chromatids. The chromatids join back with each other but may however wrongly join resulting in the exchange of chromatid parts between nonsister chromatids. 2. Therefore, in the test cross, two groups of progeny with equally large numbers and two groups with equally small numbers are obtained. The large group is called the parental type and the small number group is called the cross-over or recombinant type. Parental type as shown in Figure 17.5 is the type when there is no crossing over. The recombinant type is the type formed after crossing-over. 3. The percentage of crossing-over or the recombinant type of gametes depends on the distance between the two linked genes. It is directly proportional to the distance between the two genes. However, the distance cannot be more than 50%. Distance between Loci and Relative Position of Genes 1. The distance between two loci is equal to the percentage of crossing-over. The percentage of crossing-over is calculated based on the offspring of a test cross dihybrid, which is equal to the map distance. Map distance = number of crossing-over type offspring total offspring produced × 100% 2. For example, a dihybrid Drosophilla with normal grey body and normal long wings is test-crossed with another Drosophilla with ebony body and vestigial wings. The progeny produced are as follows: Grey body long wing – 201 Ebony body vestigial wing – 199 Grey body vestigial wing – 42 Ebony body long wing – 38 Total number of offspring – 480 Map distance = 42 + 38 201 + 199 + 42 + 38 100% = 16.67% (map units / centimorgan) Learning Outcomes Students should be able to: (a) explain crossingover and distinguish between parental and recombinant genotypes and phenotypes; (b) calculate the distance between two loci, and determine the relative position of a gene on a chromosome based on percentage of crossingover in Drosophila.

150 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 3. From the map distance, the percentage of the parental and recombinant type of gametes can be calculated. For example, (a) if the distance between two genes A/a and B/a is 20%, the types of gametes form depend on the dihybrid phase. (b) if the parent is of coupling phase, (AB/ab ), the ratio of the parental types of gametes to those crossing-over types is 40% (AB): 40% (ab): 10% (Ab): 10% (ab). (c) if the parent is of repulsion phase (Ab/aB), the ratio of the parental types of gametes to those crossing-over types is 40% (Ab): 40% (aB): 10% (AB): 10% (ab). 4. Therefore, to see if two genes are linked, a test cross of the dihybrid is done. If the offsprings are of four phenotypes with ratio 1 : 1 : 1 : 1, then the genes are not linked. If the offsprings produced have two big groups of different phenotypes and two small groups also with different phenotypes, then the two genes are linked. 5. Test cross is a technique to determine the distance between two loci based on the percentage of crossing-over between the genes involved. 6. For example, the same two loci i.e. ebony body (e+/e) and vestigial wing (vg+/vg) genes in Drosophila are considered. The procedure is as follows: (a) Initially, a dihybrid fly (say e+ vg+/e vg) is obtained and testcrossed with one of ebony body and vestigial wings (ee/vgvg). (b) The offspring produced are separated and counted based on phenotypes. Four types of offsprings are observed; two with larger numbers i.e. grey body long wing (e+ vg+/e vg) and ebony body and vestigial wing (ee/vgvg); two with small numbers i.e. grey body and vestigial wing (e+vg/e vg) and ebony body long wings (e vg+/e vg). (c) Then, the map distance is calculated based on the percentage of crossing-over type offsprings i.e. the percentage of the two small numbered groups. (d) From the map distances obtained through test crosses, the relative position of the genes along a chromosome can be determined as follows: (i) The distance between A/a, B/b = 20% while that of B/b, C/c = 10% and that of A/a, C/c = 28%. Therefore, the relative position of the genes are as shown below: A 20% 10% 28% B C or C 10% 20% 28% B A (ii) If the distance between A/a, C/c = 12%, the relative position is: A 12% 10% 20% C B or B 10% 12% 20% C A Language Check Language Check • Coupling phase = dominant alleles are on the same homologous chromosome and both recessive alleles are on the other (cis) • Repulsion phase = each homologous chromosome has one dominant and one recessive allele from the two genes (trans)

151 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 (e) The sum of two map distances i.e. between A/a and B/b; and between B/b and C/c is always bigger than the map distance of the two further loci i.e. between A/a and C/c, especially when the map distance exceeds 20%. This is because when the map distance of two genes is determined by test cross, there is a tendency to underestimate the map distance due to undetected double crossovers. 17.4 Population Genetics Concept of a Gene Pool 1. A gene pool is an aggregate of genes or gametes of a Mendelian population from which the next generation is produced. 2. The gene pool can be considered as the total genetic information possessed by reproductive members in a population of sexually reproducing organisms. 3. Genes in the pool have dynamic relationships with one another and the environment around where the organisms live. 4. Environmental factors, such as selection, can alter allelic frequencies and cause evolutionary changes in the population. 5. For easy calculation, a concept based on one gene locus is considered at one time. So, gene pool is diagrammatically represented as shown in Figure 17.6. 6. In Figure 17.6, only one gene or locus is considered. That locus consists of dominant and recessive alleles i.e. A and a. So, there are two types of gene frequencies i.e. frequency of allele A and frequency of allele a. The gene frequency is always expressed as decimal. For example, frequency of A allele is 0.5, then frequency of a allele is 1 – 0.5 = 0.5. 7. The frequencies of allele A and a depend on the genotypic frequencies of AA, Aa and aa. Hence, if the frequency of AA is very high, the frequency of A is high too. 8. From the allele frequencies, the genotype frequencies of the next generation can be calculated if we assume that random fertilisation of the gametes occurs. Let the frequency of A be p and that of a be q, in which p + q is equal to 1. The genotypes of the next generation is as shown in Table 17.7. Table 17.7 Genotype frequencies of the next generation based on the allele frequencies Male gametes Female gametes p (A) q (a) p (A) p2 (AA) pq (Aa) q (a) pq (Aa) q2 (aa) Learning Outcomes Students should be able to: (a) describe the concept of gene pool, gene/allele frequency and genotype frequency; (b) explain Hardy-Weinberg equilibrium (p2 + 2pq + q2 = 1 and P + q = 1), and calculate the gene/ allele and genotype frequencies; (c) explain the conditions for Hardy-Weinberg equilibrium to be valid; (d) describe changes in genotype frequencies in relation to evolution. Figure 17.6 A gene pool of A and a alleles A A A a a A a A A a a a A a

152 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 9. Since p + q = 1, therefore (p + q)2 = p2 (AA) + 2pq (Aa) + q2 (aa) = 1 10. The two equations are called Hardy-Weinberg equations where the population is in equilibrium. Hardy-Weinberg Equilibrium 2017 1. Hardy-Weinberg law states that after one generation of random mating, a population is in equilibrium i.e. the allelic and genotypic frequencies does not change from one generation to another. 2. However, equilibrium is only achieved based on the following conditions or assumptions: (a) The population must be large. If the population is small, chance fluctuation due to death causes rapid changes in the genotypic frequencies over a short period of time. (b) The mating must be random or panmitic. If not, homozygotes may only mate with the same type of homozygotes. This can affect equilibrium as homozygotes may increase over heterozygotes, or vice versa, thus disrupting the equilibrium. (c) There must not be any selection (artificial or natural) for or against any genotype. Every genotype has the same chance of survival. If not, the favoured ones increase and the disfavoured ones decrease very fast. (d) There must not be any migration. If there is any, emigration of the same type of genotype is balanced by that of immigration. If not, the addition or loss of certain alleles in the migrants can change the allelic frequencies in the population. (e) There must not be any mutation. If there is any, the rate of forward mutation should be equal to that of backward. Mutation rate is very small in most large populations, at a rate of one per million individually at a single locus. It only affects the allelic frequencies after many generations and only small populations are easily affected. (f) Meiosis must be normal. Every gamete formed has the same survival rate as the other. The vigour of the spermatozoa should be the same for every one of them. 3. The objectives of the law or its formula are as follows: (a) To study the changes of gene frequencies in a wild population so that the direction and rate of evolution can be determined. (b) To study the changes of gene frequencies in an artificial population such as that of a herd of cattle or a plantation of crops. (c) To plan for breeding programmes so that a large population of animals or plants can be manipulated to produce more in terms of quantity or quality.

153 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 Calculations of Allelic and Genotypic Frequencies 1. One character determined by a dominant and a recessive allele. Because the dominant genotype, say AA or Aa is indistinguishable, the population has to be assumed to be in equilibrium. Therefore, frequency of aa = q2 Frequency of a = q2 = q Therefore, frequency of A (p) = 1 – q Frequency of AA = p2 , and Aa = 2pq Example 1 9 babies from 100 births suffer from a recessive genetic disease (aa) Therefore, frequency of aa = 9 100 = 0.09 Frequency of a = 0.09 = 0.3 Frequency of A = 1 – 0.3 = 0.7 Frequency of AA = 0.72 = 0.49 Frequency of Aa = 2 × 0.7 × 0.3 = 0.42 2. One character determined by two codominant or incomplete dominant alleles. The calculation of allelic frequencies is directly from those of genotypic frequencies as each genotype is distinguishable. Example 2 Phenotype (blood group) M MN N Total = X Genotype LMLM LMLN LNLN Number D H R Frequency of LM = p = 2D + H 2X = D + 1 2H X Frequency of LN = q = 1 – p = 2R + H 2X = R + 1 2H X 2014

154 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 3. A sex-linked character in men. (a) The male genotype has only two alternatives, so the genotypic frequency is the same as allelic frequency. (b) If female allelic frequencies are not given, the allelic frequencies are assumed to be the same as those of the males. (c) Therefore, female population is assumed to be in equilibrium and from the allelic frequencies, the genotypic frequencies can be calculated. (d) If only the female genotypic or allelic frequencies are given, the male allelic frequencies can be assumed to be the same as the females. Example 3 Say the males who are colour blind is 4 100. Therefore, frequency of (Xb ) = q = 0.4 Frequency of (XB) = p = 1 – q = 0.6 As the female genotypic frequencies are not given, therefore Genotypic frequency of XBXB = p2 = (0.6)2 = 0.36 Genotypic frequency of XBXb = 2pq = 2(0.6)(0.4) = 0.48 Genotypic frequency of Xb Xb = q2 = (0.4)2 = 0.16 Exam Tips Remember Hardy-Weinberg law of equilibrium, and the two equations involved together with the conditions or assumptions required to achieve that (STPM 2009 essay question). If given the proportion or percentage of any homozygote, the allelic frequencies can be calculated using the HardyWeinberg equations by assuming the population is in equilibrium for dominantrecessive alleles (STPM 1997 essay question, 2001 structured question, 2008 essay question). The number of carrier can also be determined (STPM 2009 essay question). Quick Check 3 1. 280 college students are tested for phenylthiocarbamide (PTC) tasting. 70 of them are nontasters (tt). What is the allelic frequency of tasters? What proportion of the tasters who marry non-tasters expect only tasters children in the families? What proportion might expect some non-taster children? Changes in Gene Frequencies in Relation to Evolution 1. A population with changing gene pool in every generation brought about by many forces will cause evolution in the long run. 2. Evolution does not involve individual in its lifetime but the whole population from one generation to the next. 3. The ultimate cause of genetic variation is mutation. It creates a new allele.

155 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 4. This allele may be recessive or dominant and is inherited from one generation to the next through sexual reproduction. 5. The rate of transmission to the future generation depends on many factors. One of it is environment, especially in a fast changing ones like in Galapagos Island of Equito in South America. 6. If the phenotype of an individual posseses the allele that has an advantage for survival in that particular environment, the allelic frequency will increase. 7. For example, a mutation that causes a black bird to become brown has an advantage of blending well in the brown grassland, so that predator birds like eagle may not see it. Such allele will increase very fast in future generations. 8. This is an example of natural selection proposed in Darwin’s theory of evolution, the survival of the fittest. 9. The sum of all mutations followed by natural selection in a given environment over a period of a million years will cause the evolution of a new species of organism. 10. Therefore, evolution is the result of the sum changes of allelic frequencies of many genes involved that causes the chromosome to have genes so different that they are no longer homologous to the original ones. Population genetics determines evolution. Quick Check 4 1. How do genes in a gene pool interact with one another? 2. How are genes in a gene pool affected by the environment? 3. Do dominant alleles increase faster than recessive alleles in a population? Why? 4. If there is no change in the allelic frequencies, can there be evolution? Why? 17.5 DNA Replication 1. In the 1940’s, scientists knew that chromosomes consist of DNA and protein. 2. However, it was quite a while until they were able to clarify whether the genetic material was the DNA or the protein of the chromosomes. Griffith's Experiment 1. In 1928, Fred Griffith worked on Streptococcus pneunomoniae which caused pneumonia in mammals. He grew colonies of Streptococcus pneunomoniae and noted that there were two types of colonies: the smooth (S) and the rough (R) types based on the nature of their looks. Exam Tips Remember the concept of gene pool and the relationship between population genetics and evolution. Changes in allelic frequencies are evolution. Learning Outcomes Students should be able to: (a) explain the experiments to prove DNA is the genetic material (Avery, MacLeod and McCarty experiment and Hershey and Chase experiment); (b) explain the three models of DNA replication, and interpret the experiment of Meselson and Stahl to prove the semiconservative model of DNA replication; (c) explain the mechanism of DNA replication, and the role of the enzymes involved.

156 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 2. Griffith injected live bacteria from the two different colonies into mice. Bacteria from S colony were found to kill the mouse, hence they were virulent. Bacteria from R colony did not kill the mouse, thus they were non-virulent. 3. When dead S bacteria which was boiled and cooled was injected into a mouse, the mouse lived. Similarly, the mouse came out alive when it was injected with dead R bacteria. 4. However, when dead S and live R bacteria were mixed and injected into a mouse, it was killed. Live S bacteria were seen in the blood of the dead mouse. + Mouse gets pneumonium. Smooth Pneumococcus bacteria isolated from dead mouse. Mouse gets pneumonium. Smooth Pneumococcus bacteria isolated from dead mouse. Experiment 1 Experiment 2 Experiment 3 Experiment 4 Living rough Pneumococcus bacteria/injected (R type) Living rough Pneumococcus bacteria/injected (S type) Heat-killed smooth Pneumococcus bacteria/injected (S type) Living rough (R type) and heat-killed smooth Pneumococcus (S type) bacteria/injected Mouse remains healthy Mouse remains healthy Figure 17.7  Transformation of bacteria (Griffith's experiment) 5. Griffith concluded that something in the dead S bacteria had transformed the R bacteria into S bacteria. However, he did not know what it was. 6. In 1944, Avery, MacLeod and McCarty showed that DNA isolated from S bacteria transformed R to S form. However, the notion that DNA was the genetic material was still not accepted. 7. It was not until 1952 that Hershey and Chase managed to show that DNA was the genetic material of viruses is what attacked bacteria by using T2 bacteriophage to infect E. coli. Avery, MacLeod and McCarty Experiment 1. Avery, MacLeod and McCarty carried out the experiment proving that DNA isolated from S bacteria could transform R form to S form of Streptococcus pneunomoniae. VIDEO DNA Replication

157 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 2. Avery and his colleagues carried 7 tests as shown in Figure 17.8 and summarised in Table 17.8. None None Trypsin to hydrolyse protein RNase to destroy RNA DNase to destroy DNA Smooth pneumococci (S) Extract active principle from S cells Serially dilute DNA, add medium and pretreatment to determine which part of the extract is active Pretreatment No Yes Yes Yes Yes R S S S R R S S S R S-cell extract Add rough pneumococci (R) and incubate Grow bacteria on plates for confirmation Figure 17.8 Avery et al experiment Table 17.8 Summary of Avery et al experiment Test Pre-treatment Ingredient destroyed Added with live R S found after incubation 1 None None None None 2 None None Yes Yes 3 Amylase Starch Yes Yes 4 Lipase Fat Yes Yes 5 Trypsin Protein Yes Yes 6 RNase RNA Yes Yes 7 DNase DNA Yes None 3. It was found that if the incubation mixture did not contain DNA, the R form of bacteria could not be transformed to S form.

158 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 4. This concluded that DNA transform bacteria from a noncapsulated R type to a capsulated S type. This shows that DNA is a carrier of genetic information. 5. DNA from capsulated bacteria carries a gene that controls the production of glycoprotein, the capsule material. This DNA can enter the R type bacteria and it can be used to form capsule. Thus, transformation can take place. 6. The experiment can be further simplified in Figure 17.9. RNase (destroys RNA) Type IIR bacteria Type IIIS and type IIR bacteria Type lIIS bacterial filtrate Type IIR bacteria Type IIIS and type IIR bacteria Type IIR bacteria Type IIR bacteria DNase (destroys DNA) Protease (destroys proteins) Figure 17.9 DNase destroys DNA, no S type is formed Hershey-Chase Experiment 1. Hershey and Chase used radioactive T2 bacteriophages to infect E. coli. They wanted to determine if the genes that programmed the E. coli to make T2 were in T2’s proteins or T2’s DNA. 2. Hershey and Chase cultured T2 with E. coli in the presence of radioactive sulphur (S-35). The T2’s DNA did not contain S-35 but the proteins labelled with S-35 became radioactive. 3. Hershey and Chase cultured a second batch of T2 grown in E. coli in the presence of radioactive phosphorus (P-32). The T2’s proteins did not contain P-32 but the DNA contained lots of P-32.

159 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 4. Hershey and Chase carried out Experiment 1 with T2 and S-35 labelled proteins allowed to infect E. coli. The bacteria were incubated for a day. Then, the E. coli suspensions were homogenised in Waring Blender to knock virus off of cells and then centrifuged to form pellet cells. 5. The result was S-35 was found in the pellet (pellet contained intact bacteria) and the supernatant fluid contained the T2. S-35 was found only in supernatant fluid but no S-35 proteins were found in E. coli. 6. Hershey and Chase carried out Experiment 2 with T2 labelled with P-32 DNA allowed to infect E. coli. The bacteria were incubated for a day and the E. coli suspensions were homogenized in Waring Blender to knock virus off of cells and then centrifuged to form pellet cells. Lots of DNA were labelled with P-32 inside E. coli but not in the supernatant T2 as shown in Figure 17.10. Bacteria Experiment 1 Label phage, P is an element in DNA, but not in proteins. DNA + 32P Infect bacteria with labeled viruses. Agitate in a blender to detach viruses from bacterial cells. Centrifuging forces the bacterial cells to the bottom of the tube, forming a pellet. Supernatant fluid contains the viruses. Experiment 2 Label phage, S is an element in proteins, but not in DNA. Protein coat + 35S Most of the 32P is in the pellet with the bacteria. Pellet Supernatant fluid Most of the 32S is in the supernatant fluid without the bacteria. Bacteria Figure 17.10 Hershey-Chase experiment

160 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 7. Hershey and Chase’s conclusions were that T2 injected DNA into E. coli the DNA directed the synthesis of new phage particles. Finally, the scientific community accepted DNA as the genetic material. Models of DNA Replication 1. Before Meselson and Stahl proved that DNA could replicate by a semi-conservative way, it was postulated that there are three possible models of replication i.e. conservative, dispersive and semi-conservative replication. (a) Conservative replication would leave the original DNA molecule intact and generate a completely new molecule. (b) Dispersive replication would produce two DNA molecules with sections of both old and new DNA interspersed along each strand. (c) Semi-conservative replication would produce molecules with both old and new DNA, but each molecule would be composed of one old and one new strand. 2. The different models of replication are summarised in the Table 17.9 and their progeny molecules produced are extended into the second generations. Table 17.9 Different models of replication and their progeny molecules Models of replication Conservative replication Dispersive replication Semi-conservative replication Original molecule First-generation molecules Second-generation molecule One old & one new One old & three new molecules Two dispersed Four dispersed molecules Two hybrid Two hybrid & two new molecules

161 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 Meselson and Stahl Experiment 1. Meselson and Stahl experiment was carried to prove that DNA replication is semi-conservative. Semi-conservative means that the two strands of DNA separate, each acting as a template and a new strand is synthesised to partner it. Therefore, one strand is conserved and one new strand is synthesised. 2. The experiment is summarised in Figure 17.11 below: Mix DNA into caesium chloride solution Centrifuge rotor Centrifuge at 100,000 x g until equilibrated Density Density gradient Extract DNA from cells E.coli grown many genarations in “heavy” medium Grow in “light” medium and sample after one, two, three (etc) generations 15N 14N Change media to 14N 1st generation H/H H/H H/L L/L 2nd generation 3rd generation Lighter Heavier 4th generation What Meselson and Stahl observed Figure 17.11 Summary of Meselson and Stahl experiment 3. Meselson and Stahl used E. coli bacteria that were cultured in a medium with heavy nitrogen, 15N i.e. 15N nitrate was added as a nitrogen source. 4. Bacteria that were cultured in heavy nitrogen would produce heavy DNA as the heavy nitrogen is incorporated into the nitrogenous bases of the DNA. Such DNA was isolated and found to move to the lower part of the centrifugation tube when ultracentrifuged using caesium chloride gel as shown in Figure 17.12. The heavy DNA is slightly heavier than normal DNA. 5. Then, the bacteria cultured in heavy nitrogen were transferred to a medium with normal nitrogen i.e. 14N (14N nitrate). 6. After 25 minutes, the bacteria were homogenised and the DNA was extracted and ultra-centrifuged. Bacteria would normally divide once every 20 minutes at room temperature of around 25°C and the DNA molecules would replicate. 7. The DNA was isolated and found to be neither heavy nor normal, but only one that was in between. That was a hybrid type. Heavy DNA Normal DNA Figure 17.12 The positions of DNA after ultra-centrifugation

162 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 8. After another 25 minutes, the second generation bacteria were similarly homogenised, their DNA was extracted and was found to contain ½ normal and ½ hybrid. 9. The third generation bacteria were similarly treated and found to contain ¾ of normal type: ¼ of hybrid type conforming a semi-conservative replication, as shown in Figure 17.13. Heavy DNA Old New Parent generation First generation Hybrid DNA Second generation 1/2 normal DNA 1/2 hybrid DNA Third generation 3/4 normal DNA 1/4 hybrid DNA Figure 17.13 DNA replicates by semi-conservation Mechanism of DNA Replication 1. The process of DNA replication is summarised in Figure 17.14. Single-strand binding protein First subunit of DNA polymerase III Helicase Primase 3' 5' 5' 3' 5' 3' 3' Second subunit of DNA polymerase III DNA polymerase I DNA ligase RNA primer Okazaki fragment Leading strand Lagging strand Figure 17.14 The mechanism of DNA replication 2013

163 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 2. The process occurs in the nucleus of the eukaryotic cell. It occurs in both prokaryotic and eukaryotic cells just before cell or nuclear division. 3. Initially, the DNA double helix unwinds after the enzyme helicase binds to specific parts of the DNA molecules. Several spots may be bound by helicase in a DNA molecule and unwind accordingly. 4. Hydrogen bonds between complementary bases are broken during the unwinding which occurs gradually. An unwinding fork of the two DNA strands is formed at the frontal end which keeps on opening forwards. Single strand binding proteins bind at the fork to keep the fork stable. 5. Complementary nucleotide pairing occurs with nucleotide triphosphates line themselves starting from the 5’ end. As such, the formation of the two new strands is different: one is faster (leading strand) and the other is slower (lagging strand). 6. Phosphodiester bonds are then formed between the lined up nucleotide triphosphates. Thus, a pyrophosphate is clipped off each time such bond is formed. 7. Both strands are used as templates. A primer is formed at the starting point of replication using ribonucleotides catalysed by primase. Later, the primer is removed and replaced by proper deoxyribonucleotides. 8 Finally, one DNA molecule produces two identical DNA molecules. In prokaryotic cells, two rings of such DNA are produced all set to be distributed to the new cells. In eukaryotic cells the two new strands are joined by special protein at one spot i.e. the centromere and the two new molecules form the chromatids of a chromosome. 9. The replication is semi-conservative with each new DNA having one ‘old’ and one ‘new’ strand. DNA polymerase catalyses the process i.e. the formation of the phosphodiester bonds. However, there are different polymerases used in the formation of each new strand of DNA. 10. Because of the anti-parallel nature of the DNA strands, small fragments called Okazaki fragments are formed in the lagging strand. This is due to the elongation of the new strand which is from 5’ to 3’ pattern. 11. Therefore, special ligase is used to join the fragments. The bonds formed are also phosphodiester bonds. Another enzyme called topoisomerase may cut the backbone of the DNA molecule in case the unwinding and the winding of the new strands get tangled. Summary Mechanism of DNA replication: 1. DNA (double helix), unwinds; 2. Hydrogen bonds between (complementary) bases break; 3. Complementary, base/ nucleotide, pair; (A-T and C-G) 4. Phosphodiester bonds form; 5. Both strands used as templates ; 6. Produces two identical DNA molecules ; 7. Semi-conservative/ each new DNA = one ‘old’ and one ‘new’ strand ; 8. Replication fork/ replication bubble(s) / Okazaki fragments,/ form due to anti-parallel nature, / activated nucleotides (3 phosphate groups) are used 9. DNA polymerase catalyses the process; 10. Helicase (unwinds), topoisomerase (cuts backbone), ligase (formation of phosphodiester bonds)

164 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 17.6 Gene Expression Beadle and Tatum Experiment 1. The experiment was to prove that one gene controls one enzyme (one gene-one enzyme hypothesis). Later, this was amended to one gene-one polypeptide hypothesis. 2. Beadle and Tatum used the wild-type fungus Neurospora, which could be cultured in a minimal medium consisting of glucose, nitrate, phosphate, sulphate and vitamin biotin. A summary of the procedure is as shown in Figure 17.15. X-ray a. Conidia irradiated b. Spores cultured in complete medium c. Mycelia crossed with opposite strain d. Ascus with 8 ascospores produced e. Ascospore dissected and cultured separately in complete medium g. Growth in minimal medium discarded h. No growth in 20 different media i. Growth in arginine (for example) indicates an enzyme for the synthesis of the amino acid is missing. minimal medium growth 1 2 3 4 20 +gly +val +asp +arg +ser f. Sample tested in minimal medium Figure 17.15 Beadle and Tatum experiment 3. Asexual conidia (asexual haploid spores) were irradiated with X-ray to induce mutation as X-ray was known to be mutagenic. 4. Then, the mutagenised conidia were cultured in complete medium consisting of glucose, nitrate, phosphate, sulphate, biotin and 20 different types of amino acids. 5. The mycelia formed were crossed with wild-type that was of opposite strain. 6. Sexual reproduction took place where asci that behaved like fruiting structures were formed. These asci were formed from diploid zygote cells that divided by meiosis followed by one mitosis to form 8 haploid ascospores within each ascus. 7. The ascospores were dissected and each was separately cultured in a complete medium. So, hundreds of tubes of complete medium were inoculated with single ascospores. Learning Outcomes Students should be able to: (a) explain the experiment of Beadle and Tatum which leads to the establishment of onegene-one-polypeptide hypothesis; (b) interpret the genetic code table, and identify the appropriate anticodon; (c) explain the characteristics of genetic code; (d) describe transcription and translation. Language Check Language Check • Conidium (singular) • Conidia (plural) • Ascus (singular) • Asci (plural)

165 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 8. Conidia from each culture were tested in a minimum medium to find out whether mutation had taken place. 9. If growth in minimal medium occured, this indicated that no mutation took place and the stock was discarded. This was due to the fungus grew like wild-type needing no amino acids supplied. The fungus could synthesised its own amino acids. 10. No growth in minimal medium indicated that mutation had taken place. This would be identified as nutritional mutant as amino acids were not being supplied. The fungus could not synthesised its own amino acids producing no growth. 11. More conidia were obtained to test what amino acid was not produced as a result of mutation. This was done by culturing the conidia from the particular nutritional mutant in 20 different media each with one amino acid added to a minimal medium. 12. The mutant fungus formed was haploid and was only able to grow in a culture of minimal medium added with only arginine. It can be concluded that one gene has mutated and one enzyme related to the synthesis of amino acid arginine was lost. Therefore, one gene controls the synthesis of one enzyme. 13. After the hypothesis was published, it was found that gene does not only control the synthesis of an enzyme but any protein. Therefore, the hypothesis was amended to become one gene-one protein hypothesis. Later, it was found that some proteins like haemoglobin were controlled by two genes, each controlled one polypeptide. Then, the hypothesis was further amended to become one gene-one polypeptide hypothesis. Genetic Code, Codons and Anticodons 1. The genetic code is a system of representation containing a code of three (triplet) specific base sequence in the DNA or codon of three specific base sequence in mRNA. Each codes for an amino acid for the synthesis of a protein during translation. The codes are shown in two forms, DNA (Table 17.10) and RNA (Table 17.11). Each is complementary to the other. Table 17.10 The genetic codes found in DNA 2nd base A G T C 1st base A AAA Phe AAG Phe AAT Leu AAC Leu AGA Ser AGG Ser AGT Ser AGC Ser ATA Tyr ATG Tyr ATT Stop ATC Stop ACA Cys ACG Cys ACT Stop ACC Trp G GAA Leu GAG Leu GAT Leu GAC Leu GGA Pro GGG Pro GGT Pro GGC Pro GTA His GTG His GTT Gln GTC Gln GCA Arg GCG Arg GCT Arg GCC Arg T TAA Ile TAG Ile TAT Ile TAC Met TGA Thr TGG Thr TGT Thr TGC Thr TTA Asn TTG Asn TTT Lys TTC Lys TCA Ser TCG Ser TCT Arg TCC Arg C CAA Val CAG Val CAT Val CAC Val CGA Ala CGG Ala CGT Ala CGC Ala CTA Asp CTG Asp CTT Glu CTC Glu CCA Gly CCG Gly CCT Gly CCC Gly 2014 2012

166 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 Table 17.11 The genetic codons found in mRNA Second letter U C A G First letter U UUU UUC UUA UUG UCU UCC UCA UCG UAU UAC UAA UAG UGU UGC UGA UGG U C A G C CUU CUC CUA CUG CCU CCC CCA CCG CAU CAC CAA CAG CGU CGC CGA CGG U C A C A AUU AUC AUA AUG ACU ACC ACA ACG AAU AAC AAA AAG AGU AGC AGA AGG U C A G G GUU GUC GUA GUG GCU GCC GCA GCG GAU GAC GAA GAG GGU GGC GGA GGG U C A G 2. The anti-codon is also consisted of three bases found in the second lobe of tRNA. Each anti-codon is also complementary to the codon of mRNA. 3. For example, the tRNA with anticodon UCU which is complementary to AGA codon in mRNA codes the amino acid arginine (Arg). Then, this tRNA can bind to arginine and carry it to the ribosome where the amino acid joins with other amino acids that are similarly brought there to form protein. 4. The relationship between the code in DNA, codon of mRNA and anti-codon of tRNA is as shown in Figure 17.16. DNA 5’ AGA 3’ non-coding strand Code 3’ TCT 5’ coding strand mRNA tRNA 5’ 5’ A-G-A U-C-U Anticodon codon 3’ 3’ C G A Arginine Figure 17.16 Relationship between code, codon and anticodon 123 123 123 123 123 123 123 123 123 123 123 123 14243 1442443 1442443 1442443 1442443 1442443 1442443 1442443 1442443 Phenylalanine (Phe) Tyrosine (Tyr) Arginine (Arg) Arginine (Arg) Glycine (Gly) Serine (Ser) Cysteine (Cys) Lysine (Lys) Aspartic acid (Asp) Glutamic acid (Glu) Asparagine (AspN) Histidine (His) Glutamine (GluN) Stop Stop Tryptophane Stop (Tryp) Leucine (Leu) Serine (Ser) Leucine (Ileu) Proline (Pro) Threonine (The) Alanine (Ala) Isoleucine (Ilu) Valine (Val) Methionine (Met)

167 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 Characteristics of genetic code 1. The genetic code is a triplet code. Three bases in the DNA code for one amino acid in the polypeptide. There are four different bases i.e. A, C, G and T. There are 20 amino acids. Therefore, less than three bases in a code cannot code for all 20 different amino acids. The total number of codes is 64. 2. It is degenerative. There are more codes than amino acids. Therefore, some amino acids are coded by more than one code. For example, isoleucine is specified by three codons AUU, AUC and AUA. Codons that specify the same amino acid are called synonyms. The genetic code is called degenerate because synonymous codons generally differ only in the third base. 3. It is non-overlapping. This means that the base in the code is read once and codes for one amino acid. No overlapping of reading is done. TACGCACAT is read as TAC-GCA-CAT, and would not be read otherwise. 4. It is collinear with the amino acid sequence in the polypeptide. This means that the first code codes for the first amino acid and so on. It follows a linear sequence in both DNA and polypeptide. 5. It is comaless. No base acts as the border between two codes. Every three bases are read one after another without any base left out. 6. There are three ‘nonsense’ codes. These three codes act as ‘full stop’. There are found at the end of a gene marking the end of the codes for the series of amino acids in a polypeptide. The three codes: ATT, ATC and ACT in DNA are stop codes directing the termination of transcription. Meanwhile, the three codons: UAA, UAG and UGA in mRNA are stop codons directing the termination of translation. 7. It is ‘wobbly’. This means that the third base usually is not so important in a code. If the third base is replaced by another base, it still codes the same amino acid. For example, CAA, CAG, CAC and CAT code the amino acid valine. Protein Synthesis Genetic information flows from DNA to mRNA then to protein. The process can be divided into transcription, activation of amino acids, translation and formation of functional protein. Transcription 1. Transcription is the process of mRNA formation using DNA as a template to ‘copy’ the genetic information of one or more genes. The process is summarised in Figure 17.17. Summary Characteristics of genetic code: 1. Triplet code 2. Degenerative 3. Non-overlapping 4. Collinear 5. Comaless 6. Have 3 ‘nonsense’ codes (UAA, UAG, UGA) 7. Wobbly

168 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 DNA strands still joined Ribonucleotide triphosphates RNA polymerase adds complementary DNA strand. Nucleotides are only added to the 3’ end of the RNA molecule. Non-coding strand of DNA G 5’ 3’ 5’ cap Exons The remaining coding segments, or exons are joined together. 5 Spliced exons Intron Exon 3’ tail Intron 3’ tail Noncoding introns are removed from the RNA strands. A cap and tail are added to the 5’ and 3’ ends of the new RNA of molecule. 5’ cap The cap can then be recognised by a nuclear pore complex, allowing the mRNA to leave the nucleus. A A A A Coding strand of DNA mRNA strands being assembled Cleavage site T A C C G G G T G T A C G G G G T G CCGGUA C A C A U G C C C C A C A T G G C C A C A T C G C C C C A C 5’ 5' 5’ 3’ 3’ mRNA 3’ (poly-A) tail consists of hundreds of A nucleotides. 5’ cap consists of a single G nuclotide 3’ A U G G C C C A C A U G C C C C A C A U G G C C C A C A U G C C C C A C 6 4 1 3 Various proteins bind to a sequence near the 3’ end of the pre-mRNA molecule. 2 Figure 17.17 Transcription process 2. Initially, uncoiling of DNA occurs when the uncoiling enzyme helicase, binds to a particular section of the DNA, leaving the single nucleotide chain open to be copied. The process is also helped by other proteins that are called transcription factors and co-activators. 3. Another enzyme RNA polymerase causes the arrangement of ribonucleotide triphosphates along one of the strands, the coding strand (also known as ‘sense’ or ‘template’ strand). 4. The transcription starts from the 3’ end of the DNA coding strand but the mRNA starts to form from 5’ to 3’ end. Base pairing between the ribonucleotide triphosphates and the base sequence of coding strand occurs; A with T, C with G, G with C and U with A.

169 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 5. The base sequence in the mRNA is complementary to the DNA copied strand and exactly the same as that of the other ‘noncoding’ strand as shown below: DNA ATGTTTAAA TAA non-coding strand TACAAATTT ATT coding strand mRNA AUGUUUAAA UAA 6. The ribonucleotides are then linked by phosphodiester bonds to form mRNA. A pyrophosphate is cut from the ribonucleotide triphosphates each time a bond is formed. 7. The first product of transcription differs in prokaryotic cells from that of eukaryotic cells. In prokaryotic cells, the product is mRNA, which needs no post-transcriptional modification. However, in eukaryotic cells, the first product is called primary transcript, that needs post-transcriptional modification (capping with 7-methyl-guanosine, tailing with a poly-A tail) to give hnRNA (heterophil nuclear RNA). hnRNA then undergoes splicing of  introns  (non-coding parts of the gene) via  spliceosomes  to produce the final mRNA. 8. The process takes place in the nucleus, the finished mRNA moves across the nuclear envelope into the cytoplasm. Activation of amino acids 1. This is the process where amino acids are individually bound to a specific tRNA forming amino acyl-tRNA in the cytoplasm. 2. There are 61 types of tRNA, each with a specific anticodon. 3. Specific tRNA with a particular anticodon sequence only attaches to a specific amino acid whose code is complementary to the tRNA anticodon. 4. The attachment is determined and catalysed by specific enzyme ligase or synthetase (synthase-USA). 5. For example, phenylalanine is coded by UUU. Therefore, only tRNA with anticodon AAA can be linked and carried the phenylalanine. 6. The linkage process is as follows: Phenylalanine + ATP Phenylalanine-AMP + P-P (pyrophosphate) Phenylalanine-AMP + tRNA Phenylalanine-tRNA + AMP 7. Both the amino acid and tRNA are then activated. This particular tRNA carries the amino acid say phenylalanine, to the ribosome. Exam Tips Remember how the genetic information in DNA is used to form sequence of amino acids in protein. (STPM 2001 essay question). Also remember detailed description of transcription (STPM 2009 long essay) and translation.

170 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 Translation 1. This is the process in which the base sequence in mRNA is translated into a sequence of amino acids in a polypeptide. 2. Initially, mRNA forms a complex with the ribosome subunits as shown in Figure 17.18. Ribosome Ribosome sub-units P A + + 5’ 3’ A U G A C G G A U 5’ 3’ A U G A C G G A U Figure 17.18 The formation of initiation complex 3. The first and second codons of the mRNA attach to the small ribosome subunit and occupy the P and A sites of the large ribosome subunit respectively. 4. Two tRNAs, each carrying an amino acid with anti-codons complementary to the mRNA codons, are then attached to the P site and the A site respectively. 5. A peptide bond is formed between the two amino acids (Figure 17.19), which are brought closer together. The process is catalysed by a ligase. Ribosome Termination codon Peptide bond forms A U U A A U G U Met A C G G A C A U G C Thr U A C Asp Figure 17.19 The formation of dipeptide 6. A relative movement occurs in which the third mRNA codon is shifted to the A site and its first codon is shifted out, releasing the first tRNA. The second codon now sits on the P site as shown in Figure 17.20.

171 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 Ribosome Relative movement Peptide lengthening A U U A A U G U Met G G A C G A U C C A Asp U A C Thr Figure 17.20 The elongation of polypeptide 7. A peptide bond is formed between the second and third amino acid lengthening the polypeptide. 8. Another relative movement occurs and the process is repeated with further lengthening of the polypeptide until the second last codon occupies the A site as shown in Figure 17.21. mRNA Ribosome A U G U U A A Met A C G G A C A C C A C G U G His Completed polypeptide Thr Asp His Figure 17.21 The end of translation 9. The last codon is a ‘nonsense’ codon indicating a ‘full stop’. No amino acyl-tRNA can get into the A site. The final relative movement would liberate the completed polypeptide, mRNA, tRNA and the ribosomes subunits as shown in Figure 17.22. mRNA A U G U U U U A A Met C A G A C G G A C A C A G C A C G U G Completed polypeptide Ribosome subunits freed Thr Asp His His Figure 17.22 The final release of completed polypeptide

172 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 Formation of functional protein 1. As for simple proteins, the polypeptide formed can automatically secondarily coil to form fibrous proteins such as collagen. Some can further coil to form tertiary globular proteins such as enzymes or antibodies that are functionally active. 2. As for conjugated proteins, a prosthetic group would have to be added to the polypeptide catalysed by an enzyme. For example, myoglobin is a polypeptide that forms a covalent bond with a haem group before it can be used to store oxygen in the muscle. 3. As for proteins with quaternary structure, the polypeptide automatically coils to form the globular subunit. Two or more of such identical or different subunits are bound automatically together to form a single large functional unit such as haemoglobin for oxygen transport. 4. Other proteins like zymogen and pro-insulin, are inactive when they are formed. A part of it is cut by enzymes or hydrolysed by acids such as pepsinogen before it can be active as proteinase. These proteins have to be produced in an inactive forms so as not to destroy the cells that produce them, before they can function outside the cells. Likewise, pro-insulin has part of its polypeptide cut off before it can lower blood glucose content. 17.7 Regulation of Gene Expression The Lactose (lac) Operon – An Inducible Operon 1. The lactose operon is as shown in Figure 17.23. Regulator gene Promoter Structural gene Operator Lac operon I lac Z lac Y lac A Figure 17.23 The lactose operon and its components 2. The components are regulator gene (I), promoter (P), operator (O) and structural genes that code for enzymes required to metabolise lactose. 3. The regulator gene is adjacent to the operon. (a) It is the DNA segment that can be transcribed into an mRNA molecule and then translated into an protein called a repressor. The repressor is active; it binds to the operator and blocks the binding of RNA polymerase to stop (switch off) the transcription of the structural genes. Learning Outcomes Students should be able to: (a) define repressor, inducer, negative control in lac operon and constitutive enzyme; (b) describe the components of lac operon, and explain its mechanism.

173 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 (b) Its regulatory effect also depends on the allosteric nature of the specific repressor protein. The protein has an allosteric site that can bind with an inducer, allolactose. Then, the repressor comes off from the operator. The RNA polymerase binds to the promoter and switches on the transcription of the structural genes. (c) Therefore, this is a negative contril system. If there is no lactose, the regulatory gene functions to produce a repressor that binds to the operator and stops the synthesis of the enzymes for lactose metabolism. However, if lactose is present, it detaches the repressor from the operator to start the production of necessary enzyme for its metabolism. So, no lactose will result in no enzyme and with lactose, enzymes will be produced. 4. The promoter is the first part of the operon. (a) It is the segment of DNA that allows the binding of RNA polymerase so that transcription of structural genes can occur. (b) Therefore, a single promoter is used to control the production of three enzymes required for lactose breakdown. It is an efficient and fast way to control several related reactions at one time. (c) In detail, it consists of three parts, as shown in Figure 17.24. (i) In the presence of lactose but low glucose (ii) In the presence of lactose but high glucose (high cAMP), lac genes are transcribed. (low cAMP), lac genes are not transcribed. I lac Z DNA Regulator gene Promoter CRP binding site RNA polymerase Active CRP Inactive CRP cAMP Inactivated repressor Operator RNA polymerase I lac Z DNA Promoter CRP binding site Inactive CRP Inactivated repressor Operator Figure 17.24 The structure and function of promoter (d) The promoter has another site called cAMP receptor protein (CRP) or catabolite activator protein (CAP) binding site. Once this site is bound with CRP or CAP, it is activated for the letter binding of RNA polymerase. This is true when there is lactose but low glucose, resulting in more cAMP binding to the protein. The presence of glucose does not enable the system to produce the enzymes, even in the presence of lactose. • Repressor is a protein produced from regulator gene (I) and it binds to the operator gene (O) to stop enzymes transcription so no enzymes of lac is produced. • Inducer is allolactose, an isomer of lactose that inactivates the inducer of lac operon so transcription and production of lac operon enzymes result in the metabolism of lactose. • Negative control is when a substrate (lactose) is present, enzymes for lactose metabolism is produced. When there is no lactose, no enzymes are produced for its metabolism. Info Bio

174 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 (e) Therefore, its function is to provide a site for the RNA polymerase to bind and start the transcription of the three enzymes for the metabolism of lactose. 5. The operator is part of the promoter. (a) It is the DNA segment that allows the binding of an active repressor. (b) Once the operator is bound, the repressor blocks the binding of RNA polymerase to the promoter making it impossible for the transcription of structural genes. (c) Therefore, its function is to act as a switch. It switches off the production of enzymes when attached with a repressor, blocking the binding of RNA polymerase. It switches on when it is not attached to a repressor. 6. The structural genes are lac Z, lac Y and lac A. (a) Lac Z codes for β-galactosidase, which is the same as lactase and hydrolyses lactose into glucose and galactose. (b) Lac Y codes for permease, a transport protein in the plasma membrane, that transports the lactose across the membrane into the cell where β-galactosidase hydrolyses it into glucose and galactose for cell use. (c) Lac A codes for transacetylase, an enzyme whose role in lactose metabolism is unknown. 7. In the absence of lactose, no enzymes are produced for its metabolism (Figure 17.25). I RNA polymerase Bind to operator No transcription of structural genes – no enzymes are produced DNA mRNA 5‘ 3‘ Protein Active repressor lac Z lac Y lac A Regulator gene Promoter Structural genes Operator Figure 17.25 No lactose, no enzymes are produced (a) The regulatory gene is transcribed into a mRNA molecule that is translated into protein, an active repressor. (b) The repressor binds to the operator and blocks the RNA polymerase from binding to the promoter. (c) When the RNA polymerase cannot bind to the promoter, there is no transcription of the enzymes and the enzymes are not produced. (d) Thus, the cell saves energy by not producing unnecessary enzymes for a non-existent respiratory substrate.

175 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 8. In the presence of lactose, the three enzymes are produced. (a) The lactose binds to the repressor, inactivating it and causing it to detach from the operator. (b) RNA polymerase can bind to the promoter and transcription of structural genes starts. (c) Thus, a polycistronic mRNA is produced, which is translated to produce β-galactosidase, permease and transcetylase, as shown in Figure 17.26. I RNA polymerase RNA polymerase can move across DNA mRNA 5‘ mRNA 5� 3� 3‘ Protein Allolactose Inactive repressor Enzymes are produced β -galactosidase Permease Transacetylase lac Z lac Y lac A Figure 17.26 With lactose, enzymes are produced (d) Therefore, the cell only produces the enzymes to transport and to break down lactose when there is lactose. Lactose is an inducer for the production of enzymes. The process of control is called enzyme induction. Tryptophan (trp) Operon – A Repressible Operon 1. The tryptophan operon (Figure 17.27) is an operon that is involved in the control of synthesis of amino acid tryptophan. Exam Tips Remember the meaning and components of lactose operon, the function of each component and what occurs in the absence and presence of lactose. (STPM 1996 structured question and 2003 essay question) I Regulator gene Promoter Structural genes trp operon Operator P O trpE trpD trpC trpB trpA Figure 17.27 Tryptophan operon and its components 2. It consists of a regulator gene, a promoter, an operator and five structural genes. 3. The regulator gene is far away from the operon not adjacent to the operon. (a) It has its own promoter and is transcribed to form mRNA that is translated into a repressor protein. (b) The repressor protein is inactive so, it does not bind to the operator. Language Check Polycistronic mRNA = an mRNA that codes for more than one polypeptide.

176 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 (c) The repressor is also an allosteric protein which has a site for tryptophan binding. After binding with tryptophan, it becomes activated and can then bind with the operator. 4. The promoter is the first part of the operon. It is the binding site for RNA polymerase before transcription of structural genes occur. 5. The operator is part of the promoter. It is the binding site or switch for the binding of repressor. Once it is bound, it blocks off the attachment of RNA polymerase. 6. The structural genes consists of five genes i.e. trp E, trp D, trp C, trp B and trp A. Each of the genes codes for an enzyme. Each enzyme converts a precursor (chorismic acid is the first precursor) to another, finally forming amino acid tryptophan, as shown in Figure 17.28. Genes Enzymes: Precursor E (chorismic acid) tryptophan E D D C B A C B A trp E trp D trp C trp B trp A Figure 17.28 Schematic diagram to show tryptophan synthesis 7. Tryptophan is always required as an amino acid in protein synthesis. 8. The five constitutive enzymes involved in tryptophan synthesis are required all the time. 9. When there is no tryptophan, the enzymes are produced. (a) The regulatory gene is transcribed into an mRNA molecule that is translated to form an inactive repressor. (b) The inactive repressor cannot bind to the operator and does not block the RNA polymerase from binding to the promoter. (c) The RNA polymerase can transcribe the five structural genes to form a polycistronic mRNA, which is then translated into five enzymes, as shown in Figure 17.29. Summary Lac operon Trp operon Inducible Repressible Adjacent regulatory gene Distant from regulatory gene Regulatory gene produces active repressor Regulatory gene produces inactive repressor 3 structural genes 5 structural genes Controls catabolic pathway of breakdown of lactose Controls anabolic pathway of tryptophan synthesis Switched on when susbtrate inactivates the repressor Switched on when no product activates the repressor Lactose as inducer Tryptophan as co-repressor Often switched off Rarely switched off Constitutive enzyme is an enzyme that is produced all the time such as enzymes controlled by the trp operon. Its production can be repressed if the synthesis of related product tryptophan is in excess. Info Bio Regulator gene Promoter I trpE trpD trpC trpB trpA RNA polymerase Operator RNA polymerase moves across DNA mRNA 5 5 3 3 Protein Inactive repressor are produced Enzymes: E D C B A Polycistronic mRNA is transcribed Figure 17.29 In the absence of tryptophan, enzymes are produced

177 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 (d) Thus, all the five enzymes are synthesised at the same time to produce tryptophan. 10. When there is an excess of tryptophan produced or it is added from the outside, the enzymes are not produced. (a) The tryptophan acts as a corepressor and binds to the inactive repressor, forming an active repressor-corepressor complex. (b) The active complex can bind to the promoter, blocking the RNA polymerase from transcribing the five structural genes. (c) No polycistronic mRNA is produced and so, no enzymes are produced for the synthesis of tryptophan, as shown in Figure 17.30. I trpE trpD trpC trpB trpA RNA polymerase block Attached Active repressorcorepressor complex No transcription of polycistronic mRNA – no enzymes are produced DNA mRNA 5‘ 3‘ Protein tryptophan (co-repressor) Figure 17.30 In the presence of tryptophan, no enzymes are produced (d) Therefore, the cell saves energy by not synthesising excess amino acid when there is enough of it. Only when cell runs low on tryptophan, the production is resumed. (e) This is an end-product repression, a process that prevents its production at the transcription level by its excessive production. 11. The trp operon is also operating a negative feedback control. When there is little or no tryptophan, a substrate of another anabolic reaction, the operon is not repressed; so more tryptophan is produced. When there is an excess of tryptophan, the operon is repressed so no new tryptophan is produced. The production will resume when tryptophan is low again. Quick Check 5 1. What is the advantage of having a lactose operon to E. coli? 2. Why is lactose not immediately used up like glucose, when the bacteria are transferred to a medium that contains glucose and lactose? Exam Tips Remember the tryptophan operon and what happens in the absence and presence of tryptophan.

178 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 17.8 Mutation Types of Gene Mutations Substitution 1. It involves the replacement of a base with another. An example is the replacement of a base in the sixth code in the DNA of β gene of haemoglobin causing sickle cell anaemia. This causes the original amino acid, glutamic acid being replaced by valine. 2. The valine residue cannot fit neatly in the haemoglobin structure, causing the haemoglobin molecule to bind with one another. The molecules form chains inside the red blood cell, causing it to be distorted to become sickle shape. The mutation is as shown in Figure 17.31. Original β gene Mutated β gene DNA CTT DNA CAT mRNA GAA mRNA GUA Normal β polypeptide Mutated β polypeptide - - - - - - - - glu - - - - - - - - - - - - - - - - val - - - - - - - - Figure 17.31 Substitution mutation in β haemoglobin gene 3. Base pair substitutions can result in three types of mutations: (a) Silent mutation (i) A change in one base pair has no effect on the protein produced by the gene. (ii) This is allowed for by the redundancy in the genetic code i.e. its degenerate nature. (iii) E.g. both GGC and GGU code for the amino acid glycine. Thus, the mutation is “silent,” i.e. causes no change in the final protein product. (b) Missense mutation (i) A change in one base pair causes a single amino acid to change in the resulting protein. (ii) The result is called “missense” since the code is now different. (iii) E.g. GGC which codes for the amino acid glycine is changed to AGC codes a different amino acid serine. (iv) The effect of a missense mutation on the protein is unpredictable. (v) A missense mutation can cause disease like sickle cell anaemia. The effects of a change in one base pair alters one amino acid in the resulting haemoglobin protein, causing red blood cells to take on a strange shape. Learning Outcomes Students should be able to: (a) describe the different types of gene mutation with examples of its consequences (substitution – sicklecell anaemia, insertion/ addition – frameshift mutation, deletion – frameshift mutation and thalassaemia major and inversion); (b) differentiate missense, nonsense and silent/ neutral mutations; (c) describe the four structural changes in chromosomes (duplication, deletion, inversion and translocation); (d) describe the changes in chromosome number, including the definition of non-disjunction; (e) describe the consequences of nondisjunction in relation to meiosis; (f) explain and give examples of different types of aneuploidy (monosomy and trisomy); (g) explain and give examples of different types of euploidy: diploid and polyploid, including autopolyploidy and allopolyploidy). Sickle cell anaemia can be considered as a lethal recessive (s) genetic disease. SS and Ss are non-lethal. ss is lethal. It can be considered as codominant allele HS verse HA normal allele. HSHS has all the red blood cells sickleshaped and usually dies. HAHA is normal. HAHS has half the red blood cells sickle-shaped and anaemic. Info Bio

179 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 The strange-shaped red blood cells cause a variety of phenotypic effects in the individual: this is an example of pleiotropy. (vi) However, a missense mutation is not always bad. It could create a protein that improves reproductive success or survivability. Natural selection could then help increase the frequency of this new allele in the population. (c) Nonsense mutation (i) A change in a single base pair creates a stop code e.g. ACT, ATC or ATT. (ii) Because this kind of mutation creates a stop signal in the middle of a normally functional gene, the resulting protein is almost always nonfunctional, hence the term “nonsense” mutation. (iii) E.g. TTC which codes for lysine when changed to ATC results in the formation of a stop code. Insertion or addition 1. It involves the addition of one or more nucleotides to the DNA. If one or two nucleotides are added, the code will be misread from the point of addition onwards, causing frameshift mutation. 2. The protein formed cannot perform its function. The mutation causes the changing of dominant allele to form recessive allele of a genetic disease. If three nucleotides are added, then an extra amino acid is inserted in the polypeptide formed. The addition of an extra amino acid may not change the function of a protein. 3. Addition may also cause the formation of termination code known as ‘nonsense’ mutation, resulting in the formation of a short polypeptide that cannot function. The mutation is as shown in Figure 17.32. One A is added One U is added 123 123  Original gene:AUGUUUGGGCGU Addition of one nucleotide AUGUAUUGGGCGU Polypeptide: met-phe-gly-arg met-tyr-trp-ala Frame is shifted  Original gene:AUGUUUGGGCGU Addition of three nucleotides AUGAAAUUUGGGCGU Polypeptide: met-phe-gly-arg met-tyr-phe-gly-arg (One extra amino acid)  Original gene:AUGUUUAAACGU Addition of one nucleotide AUGUUUUAAACGU Polypeptide: met-phe-lys-arg and produces ‘nonsense’ mutation met-phe-stop Three A's are added Figure 17.32 Insertion mutations Summary Gene Mutations: 1. Substitution CTT → CAT e.g. sickle cell anaemia 2. Insertion CTT → ACTT e.g. transmit mutation 3. Deletion CTT → TT e.g. frameshift mutation and thalassemia 4. Inversion CTT → TTC no e.g. but common

180 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 Deletion 1. It involves the losing of one or more nucleotides within a gene. If one or two nucleotides are lost, it results in frameshift mutation. An example occurs in the β chain of haemoglobin in which nonfunctional β subunits are formed, resulting in thalassemia and impairing in oxygen-carrying capacity. 2. Deletion can also cause ‘nonsense’ mutation. The mutation is as shown in Figure 17.33. One G and one U are deleted 123 One U is added  Original gene:AUGUUUGGGCGU Deletion of one nucleotide AUGUUGGGCGU Polypeptide: met-phe-gly-arg met-leu-gly-val Frame is shifted  Original gene:AUGUUUGGGCGU Deletion of three nucleotides AUGGGGCGU Polypeptide: met-phe-gly-arg met-gly-arg  Original gene:AUGUUUAAACGU Deletion of two nucleotide AUUUAAACGU Polypeptide: met-phe-lys-arg and produces ‘nonsense’ mutation Ile-stop Three U's are deleted Figure 17.33 Deletion mutations Inversion 1. It involves breaking of phosphodiester bonds in two locations and the DNA strand in the middle is inverted when joined back. This is common during the prophase I of meiosis. This results in the code within the inverted part misread and produced no functional protein. The mutation is as shown in Figure 17.34. Breaking Original gene: AUGUUUGGGCGU Inversion AUGGGGUUUCGU Polypeptide: met-phe-gly-arg met-gly-phe-val Figure 17.34 Inversion mutation Chromosomal Mutations Chromosomal mutations are divided into two types which are changes in chromosome structures and changes in chromosome numbers. Changes in chromosome structures These changes occur spontaneously, but the frequency is increased by ionising radiations and chemical mutagens. Usually, they occur in the prophase I of meiosis when DNA breaks and rejoins. There are four types of changes to the chromosome structure, namely deletion, addition, translocation and inversion. Exam Tips Remember the meanings and examples of gene mutations such as substitution, insertion, deletion and inversion. You must be able to explain why deletion or addition of base in the most serious of gene mutation and the type of mutation that occurs, for e.g. sickle cell anaemia. Thalassemia can be considered as codominant allele HT . HTHT – thalassemia major, lethal HAHA – normal. HAHT – thalassemia minor, anaemic Info Bio

181 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 Deletion 1. It is also known as deficiency involving the loss of a chromosome segment. 2. Usually, more than one gene is lost as a small portion of chromosome or chromatid containing many genes are lost. 3. The loss may be at one end or in the middle, as shown in Figure 17.35. Normal chromosome: A Deleted chromosomes: Pairing: B C D Break E F G A B C D E F G C D E C D E A B F G B F C E D A G A B F G Figure 17.35 Loop formation during pairing of two unequal length chromosomes 4. The effect of deletion may not be serious because another gene in the homologous chromosome is present. However, if the deleted alleles are dominant, leaving alleles that are recessive and deleterious, the consequences are serious or lethal. 5. Expression of a recessive allele due to deletion of the dominant allele is termed pseudodominance in structural heterozygotes. Many dominant characters in Drosophila are associated with small deletions in heterozygous conditions such as the Minute, Delta, and Notch loci. 6. Small deletions can be homozygous viable and produce recessive phenotypes; however, larger deletions are often lethal in structural homozygotes. 7. No crossovers can be detected in a deleted region; this produces no recombinant type. 8. Deletions often cause pollen sterility in plants, as the pollens are haploid. 9. In animals, they are lethal to males if it occurs on the X chromosome.

182 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 10. Only large deletions can be detected by karyotype analysis or in bivalents which are observed during meiosis I. 11. Pairing in meiosis causes a loop formation, as shown in Figure 17.35. 12. An example is cri-du-chat syndrome in humans that is resulted from an observable deletion of part of the short arm of chromosome 5, as shown in Figure 17.36. This results in severe mental retardation, physical abnormalities, mewing, cat-like cry during infancy and early death. Chromosomes Deletion Figure 17.36 Deletion of human chromosome 5, producing cri-du-chat syndrome Duplication 1. It is a duplication or addition of a chromosome segment. 2. It may or may not be lethal. Usually the presence of more than one of the same copy of gene is harmless. Only the duplication of special genes may bring about lethality. 3. The length involved varies considerably. 4. It is thought to be caused by the repairing process during slipped pairing. 5. Unequal length of homologous chromosomes resulted may be observable during pairing in meiosis, where a loop is formed, as shown in Figure 17.37. A Normal chromosome: Duplicated chromosome: B C D E F A B C C D E D E F Pairing A B C D E F A B C C D E D E F Figure 17.37 Loop formation during pairing of two unequal length chromosomes 6. Unequal numbers of sequence repeats may lead to not properly formed bivalents resulting in fertility reduction, especially in pollens. 7. There is a tendency for duplications to build up in number. This is now known to be a very important force in mutation.

183 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 8. An example is in Drosophila where duplications in a certain region of X chromosome produce Bar eye, a reduction in the number of facets in the eye. More duplications reduce the eye from oval to rod shape. 9. It contributes to extra DNA in the chromosome, where beneficial gene mutations may occur. Inversion 1. It is the breaking of a chromosome at two points and the rejoining of the excised segment after turning 180 degrees from the original orientation. 2. The formation is common during prophase I of meiosis, when chromosomes coil and shorten. The formation is as shown in Figure 17.38. 3. During pairing with a normal chromosome, a loop is formed, as shown in Figure 17.39. DF A A B B G G C C H H I I E F E D Inverted chromosome Normal chromosome Figure 17.39 Pairing in inverted chromosome and normal chromosome 4. No genetic material is lost when an inversion takes place, but there can be phenotypic changes due to position effects of genes that are brought together or further apart. 5. Therefore, there are changes in map distances, resulting in the increasing or decreasing of certain allele combinations. 6. Small inversion can reduce crossing-over. 7. The loop formation during pairing does not favour crossing-over. 8. Longer inversion can bring about semi-sterility as any crossingover occurs in the loop during prophase I resulting in the loss of chromosome due to the inability of acentric and dicentric chromosomes formed to move. 9. This is the most common type of chromosomal aberration in plants or animals. About one in 50 people has a visible chromosome inversion. This abnormality usually has no visible effect on phenotype. A B G C H I E F D A I B G C H E F D A B G F E D C H I Figure 17.38 Formation of inversion

184 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 Translocation 1. It is the transfer of chromosome segments from one part to another or between non-homologous chromosomes. 2. Simple translocations are shown in Figure 17.40. An example in human is the translocation of chromosome 21 to other chromosome, which also produces Down syndrome features. Chromosome 21 Breaks Translocated Chromosome 9 Figure 17.40 Simple translocation 3. Reciprocal translocations are shown in Figure 17.41. Breaks Reciprocal Translocated Figure 17.41 Reciprocal translocation 4. They reduce pollen viability as pollens are haploid and need to germinate in the stigma. 5. They also reduce crossing-over due to abnormal pairing of homologous chromosomes. 6. The chromosomes can be observed to form rings during anaphase I. Oenothrea has all its chromosomes reciprocally translocated, inhibiting crossing-over and produces little variation in its offspring. 7. Reciprocal translocations bring about position effects where genes are brought to new locations and expressed differently. 8. For example, a white-eye recessive allele in Drosophila, when transferred from chromosome X to chromosome 4 is expressed differently with white facets in the red-eye. 9. In human, reciprocal translocation between chromosome 22 and that of 9 causes chronic myelogenous leukemia. Exam Tips Remember the meaning of chromosomal mutations, the various types and how each is formed. (STPM 2000 essay question) They consist of aneuploidy, autopolyploidy, allopolyploidy, deletion, addition, inversion and translocation.

185 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 Changes in chromosome numbers There are two types of changes, namely aneuploidy and euploidy. Aneuploidy 1. Aneuploidy is the addition or deletion of one or more chromosomes from a complete set of chromosomes in a cell or organism as shown in Table 17.12. Table 17.12 Terms Meaning Aneuploidy 2n ± chromosome • Monosomy • 2n – 1 chromosome • Trisomy • 2n + 1 chromosome • Tetra-, penta-, … • 2n + 2, 2n + 3, … 2. It is caused by nondisjunction, i.e. inability of usually one bivalent or chromosome to separate during meiosis or mitosis. 3. It is divided into different types and examples that occur in humans as shown in the Table 17.13. Table 17.13 Types of aneuploidy in humans Types and chromosome Chromosome formula Frequency at birth Main phenotypic chracteristics Monosomic X chromosome: Turner syndrome (45, X) 2n – 1 1 5000 • Female with under-developed ovaries and uterus • Usually sterile • Short stature of about 1.5 m • Webbing of skin in the neck region Trisomic autosome: Down's syndrome (47, 21) 2n + 1 1 700 • Typical facial look with slanting eyes • Mentally retarded, always severe • Short stature with poor skeletal development • Broad hand with straight palm crease Trisomic autosome: Patau syndrome (47, 13) 2n + 1 1 5000 • Mentally retarded and deaf • Cleft lip or palate • With extra finger • Minor muscle seizures • Heart abnormalities Trisomic autosome: Edward's syndrome (47, 18) 2n + 1 1 4000 to 1 18000 • Many inborn organ abnormalities • Mentally retarded • Low set abnormal eyes • Small nose and mouth • Dies in the first 6 months Klinefelter syndrome (47, XXY) trisomy (48, XXXY) tetrasomy (48, XXYY) double trisomies (49, XXXXY) pentasomy (50, XXXXXY) hexasomy 2n + 1 2n + 2 2n + 1 + 1 2n + 3 2n + 4 1 500 • Male with small testes • Usually sterile • With developed breast • Feminine voice • Usually tall with long limbs • Usually mentally retarded Trisomic X chromosome: Triple X syndrome 2n + 1 1 700 • Female with not well-developed sex organs • Usually sterile • Usually mentally retarded

186 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 4. In meiosis, it can occur in meiosis I or meiosis II. If it occurs in meiosis I, it is as shown in Figure 17.42. Half of the gametes formed is n + 1 and the other half is n – 1. If the n + 1 gamete is fertilised by a normal gamete, then a trisomy (2n + 1) is formed. If a n – 1 gamete is fertilised by a normal gamete, then a monosomy (2n – 1) is formed. 5. In mitosis, it gives rise to one daughter cell that is 2n + 1 and the other is 2n – 1. If meiosis followed that of mitosis, then half of the gametes are n + 1 and the other half is n – 1. It produces a mosaic of aneuploid cells (called chimera), if it occurs in the zygotic or apical meristem cell division. 6. Aneuploidy can involve autosomes or sex chromosome. In human, aneuploidy results in genetic syndrome as many characters are affected. More obvious characters are physical deformities, retardation in growth, mental retardation and sterility. Sterility is caused by the inability of the individual to have normal meiosis, as there is no partner or proper pairing of homologous chromosomes. n n n n Gametes Meiosis I Meiosis II Number of chromosomes (a) Normal meiosis (2n = 4) n + 1 n + 1 n – 1 n – 1 Gametes Meiosis I Meiosis II Nondisjunction Number of chromosomes (b) Nondisjunction occurs in meiosis I n + 1 n – 1 n n Gametes Meiosis I Meiosis II Number of chromosomes Nondisjunction (c) Nondisjunction occurs in meiosis II Figure 17.42 Formation of aneuploid gametes in meiosis Exam Tips Remember chromosomal mutation resulting changes in number and Klinefelter syndrome. (STPM 2015 structured question)

187 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 2n 2n 2n 2n Nondisjunction 2n + 1 2n – 1 (a) Normal mitosis (2n = 4) (b) Nondisjunction of one chromosome Figure 17.43 How nondisjunction occurs in mitosis 7. A common example is Down's syndrome. It occurs often in children born from mothers of older than thirty-five years. It involves the trisomy of chromosome 21; the individual has three of them. Children suffering from Down's syndrome have typical facial looks with slanting eyes, shorter stature, mentally retarded, sterile and prone to diseases, so they have shorter lifespans. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 X Y Figure 17.44 Down syndrome 8. Other not so common autosomal trisomies are Patau syndrome (trisomy of chromosome 13) and Edward's syndrome (trisomy of chromosome 18).

188 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 9. Aneuploidy involving sex chromosomes is common. A common example is Turner syndrome, a monosomy in chromosome X. The individual has 45 chromosomes (45, X), with only one X chromosome and is short, stout and sterile. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 X Figure 17.45 Turner syndrome 10. Another example is Klinefelter syndrome (47, XXY). The individual has 47 chromosomes with an extra X chromosome. He has some female features, especially enlargement of breasts caused by high oestrogen level. He is mentally retarded, sterile and has a short lifespan. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 XX Y Figure 17.46 Klinefelter syndrome 11. Other not so common examples include Triple X syndrome (47, XXX) and XYY syndrome. Triple X syndrome is slightly different from Turner syndrome the sufferer is not so muscular. Men of XYY syndrome are aggressive, and the majority of them land themselves in jail as they have the tendency to commit crimes. They are all sterile. Exam Tips Remember the meaning of aneuploidy together with three examples of this type of mutation i.e. Down syndromes (STPM 1998 essay question), Klinefelter and Turner syndromes.

189 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 Euploidy or polyploid 1. Euploidy is a term to describe a nucleus, cell, or organism that has an exact multiple of haploid number (n) of chromosomes, for example, diploidy (2n), triploidy (3n), tetraploidy (4n), and higher polyploidy (3n, 4n, 5n, ……). There are as shown in Table 17.14. 2. Polyploidy has more than the normal two sets of chromosomes. They are divided into two different types, namely autopolyploidy and alloplolyploidy. Autopolyploidy 1. Autopolyploidy have extra chromosome sets that originated from the same genome or same species. 2. Common examples are triploid (3n), tetraploid (4n), pentaploid (5n), hexaploid (6n) and octaploid (8n). 3. They occur mostly in plants as polyploidy in animals seldom survive. This is because the sex determination mechanism in animals is upset by changes in the number of sex chromosome to that of autosome ratio. Exceptions of triploid animals that survive are species of salamander and brine shrimp. Human liver and kidney polyploid cells can survive. 4. They can be formed during meiosis or mitosis, as shown in Figure 17.47. In meiosis (2n = 4), non disjunction in meiosis II Replication (2n = 8) 2n = 4 Disjunction Nondisjunction Pairing In mitosis (2n = 4), Replication (2n = 8) 2n = 8 Nondisjunction Metaphase Figure 17.47 Formation of unreduced gametes or cell Table 17.14 Terms Meaning Euploidy Multiple of n • Diploidy • 2n • Triploidy • 3n • Tetra-, penta-, … • 4n, 5n, … • Polyploidy • 3n, 4n, 5n, … 5. In meiosis, it is formed by nondisjunction of the whole chromosome sets. Therefore, unreduced gametes or diploid gametes are formed. 6. Fertilisation between unreduced gametes and normal gametes produces triploids, while fertilisation between two unreduced gametes produces tetraploids. 7. Higher ploidies of more than tetraploid are produced from fertilisations of further unreduced gametes. An example is found in Chrysanthemum with monoploid number of 9 but with tetraploid, hexaploid, octaploid and decaploid. 2012

190 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 8. Most examples of autopolyploids in agriculture and horticulture are artificially induced by colchicine. It inhibits the formation of spindle fibres, resulting in unreduced gametes in meiosis or chromosme doubling after mitosis in buds or new shoots. 9. Autopolyploids have the following characteristics in plants: (a) They grow faster, more luxuriant, with more leaves, stems and roots. (b) They have bigger leaves, stems, roots, flowers and fruits. (c) The colour of their leaves is greener and their flowers are darker in colour. (d) However, there is an optimum ploidy, after which if further increased, the reverse effect of smaller size and weaker growth are produced. (e) They are mostly sterile, though some polyploids of even number are fertile. This is because pairing of homologous chromosomes is not proper during meiosis for the formation of gametes. (f) There is an advantage to have sterile fruit plants, as they are seedless like those of water-melons, guavas, bananas, citrus and grapes. (g) Propagation of sterile polyploid is by vegetative means of grafting. Allopolyploids 1. Allopolyploids are plants with at least four sets of chromosomes (tetraploid) in which each of the two sets come from different species. Therefore, their genome is the sum of those of the original species. They are also known as amphidiploids. 2. An example of allopolyploid is a species of North American grass, Spartina anglica that is formed, as shown in Figure 17.48. Spartina maritima × Spartina alteniflora (2n = 60, AA) (2n = 62, BB) Spartina townsendii (sterile) Spartina anglica (allopolyploid/amphidiploid) (2n = 122, AABB) Chromosome doubling Figure 17.48 Formation of allopolyploid/amphidiploid (a) Spartina townsendii is sterile because their chromosomes are not homologous and cannot pair during prophase I of meiosis. The products of meiosis cannot survive due to genetic imbalance. (b) In the course of evolution, the chromosome number is doubled to form Spartina anglica, then each chromosome has a partner and normal meiosis occurs. Chromosome doubling can occur from nondisjunction of the whole set of chromosome during meiosis, forming unreduced gametes.

191 Biology Term 3 STPM Chapter 17 Inheritance and Genetic Control 17 Chromosome doubling can also occur in mitosis, forming new shoots which can then become fertile. The shoot can then reproduce by sexual reproduction, forming normal seeds. (c) Spartina anglica becomes a new species as it is isolated from its two ancestral species. It cannot freely breed with ancestral species to produce fertile offspring. Thus, in just two generations a new species is formed, a fastest way of forming new species in evolution. Triticum monococcum (2n = 14, AA) Sterile hybrid (allodiploid = 14, AB) Chromosome doubling Triticum searsii (2n = 14, BB) Triticum turgidum (allotetraploid, 2n = 28, AABB) Triticum tauschii (2n = 14, CC) Sterile hybrid (allotriploid = 21, ABC) Chromosome doubling Triticum aestivum (allohexaploid, 2n = 42, AABBcc) Figure 17.49 Formation of present-day wheat 3. In wheat, Triticum turgidum is allopolyploid (4n = 28) which originated from T. monococcum. and T. searsii. The present-day bread wheat, T. aestivum (6n = 42) originated from a cross of T. turgidum and T. tauschii (2n = 14), as shown in Figure 17.49. 4. Allopolyploids have specific characteristics, consisting of characteristic mixtures of its two ancestral species. 5. Many cereals, orchids, cotton, fruit and horticulture species are artificially created through this form of hybridisation. Quick Check 6 1. Explain how XYY individual arise. 2. Explain why tetraploid is sometime sterile. 3. Explain how sterile hybrids can be induced to become fertile. 4. Explain why genes transferred to a new chromosome produce different phenotypes.