4.10 Exercise Problems

Q1.  Mention the advantages of selecting pea plant for experiment by Mendel.

Answer: Mendel select garden pea (Pisum Sativum) for the following reasons.
(i) It is an annual plant with short life span and gives results within 3 months.
(ii) The plant is grown easily and does not require after care except at the time of pollination.
(iii) F1 hybrids are fertile.
(iv) Seven pairs of contrasting characters easily detectable.
(v) True breeding self pollination.

Q2. Differentiate between the following 
(a) Dominance and Recessive
(b) Homozygous and Heterozygous
(c) Monohybrid and Dihybrid.

Answer: (a) The difference between dominance and recessive:

DOMINANT TRAITS	RECESSIVE TRAITS
Trait that is expressed when one or both alleles are present.	Trait that is only expressed when both alleles are present.
Represented by a capital letter.	Represented by a lowercase letter.
Only one copy of the dominant allele is needed to express the trait.	Two copies of the recessive allete are needed to express the trait.
Dominant alleles are generally more common in populations.	Recessive alleles are generally less common in populations.
Expressed in both homozygous and heterozygous individuals.	Only expressed in homozygous individuals.
Dominant alleles can mask the expression of recessive alleles.	Recessive alleles can be masked by dominant alleles.
Dominant alleles segregate independently during meiosis.	Recessive alleles segregate independently during meiosis.

(b) Differences between homozygous and heterozygous individuals:

\(
\begin{array}{|c|l|l|}
\hline & \text { Homozygous } & \text { Heterozygous } \\
\hline \text { (i) } & \begin{array}{l}
\text { They have similar alleles (TT or tt) for a } \\
\text { trait. }
\end{array} & \text { Have dissimilar alleles (Tt) for a trait. } \\
\hline \text { (ii) } & \begin{array}{l}
\text { Contains either dominant (TT) or recessive } \\
\text { (tt) alleles but not both types (Tt). }
\end{array} & \begin{array}{l}
\text { Contains both-dominant and recessive } \\
\text { alleles (Tt). }
\end{array} \\
\hline \text { (iii) } & \begin{array}{l}
\text { They are true-breeding for a specific trait } \\
\text { and produce progeny having same } \\
\text { genotypes and phenotypes on selfing. }
\end{array} & \begin{array}{l}
\text { They are not true-breeding and produce } \\
\text { offspring having three genotypes and } \\
\text { mostly two or some time three phenotypes. }
\end{array} \\
\hline \text { (iv) } & \begin{array}{l}
\text { Only one type of gametes are formed either } \\
\text { ‘T’ or ‘t’ type only, not both. }
\end{array} & \begin{array}{l}
\text { Produce two types of gametes containing ‘T’ } \\
\text { as well as ‘ } \mathrm{t} \text { ‘. }
\end{array} \\
\hline
\end{array}
\)

(c) 

\(
\begin{array}{|l|l|l|}
\hline & \text { Monohybrid } & \text { Dihybrid } \\
\hline \text { (i) } & \begin{array}{l}
\text { It is a cross between two pure organisms } \\
\text { in order to study the inheritance of a single } \\
\text { pair of alleles. }
\end{array} & \begin{array}{l}
\text { It is a cross between two pure organisms of a } \\
\text { species in order to study the inheritance of two } \\
\text { pairs of alleles. }
\end{array} \\
\hline \text { (ii) } & \begin{array}{l}
\text { It produces a phenotypic monohybrid ratio } \\
\text { of } 3: 1 \text { in } \mathrm{F}_2 \text { generation. }
\end{array} & \begin{array}{l}
\text { It produces a phenotypic dihybrid ratio of } \\
9: 3: 3: 1 \text { in } \mathrm{F}_2 \text { generation. }
\end{array} \\
\hline \text { (iii) } & \text { It produces genotypic ratio of } 1: 2: 1 \text { in } \mathrm{F}_2 . & \begin{array}{l}
\text { It produces genotypic ratio of } 1: 2: 1: 2: 4: 2 \\
: 1: 2: 1 \text { in } \mathrm{F}_2 .
\end{array} \\
\hline
\end{array}
\)

Q3. A diploid organism is heterozygous for 4 loci, how many types of gametes can be produced?

Answer: For a diploid organism, which is heterozygous for 4 loci, then \(2^4\) i.e. \(2 \times 2 \times 2 \times 2=16\) types of gametes can be produced if the genes are not linked because for each heterozygous pair of genes there are two possibilities. So, for 4 pair the number of combination will be 16 gametes.

Q4. Explain the Law of Dominance using a monohybrid cross.

Answer:  When two different factors (genes) or a pair of contrasting forms of a character are present in an organism, only one expresses itself in the F, generation and is termed as dominant while the other remains unexpressed and called recessive factors (gene).
A tall (TT) true breeding plant is crossed with a dwarf ( tt ) plant. The character of height is represented by’ \(\mathrm{T}^{\prime}\) for tall’t’ for dwarf are the alternate form as character of height. The F1 hybrid ‘Tt’ is Tall, showing that tall is dominant over dwarf while dwarf remains unexpressed in F, offspring due to phenomenon of dominance by tall factor or gene. In this Tt heterozygous has tall phenotype showing T is dominant over t allele. 

Q5. Define and design a test-cross.

Answer: A test cross is a cross between an individual showing a dominant trait with the one having its homozygous recessive trait in order to know whether the dominant trait is homozygous or heterozygous. If the ratio of a test cross is 1:1, it shows that the dominant trait is heterozygous.

Q6. Using a Punnett Square, work out the distribution of phenotypic features in the first filial generation after a cross between a homozygous female and a heterozygous male for single locus.

Answer: When a heterozygous male tall plant (Tt) is crossed with the homozygous dominant female tall plant (TT), we get two types of gametes in males: half with \(T\) and a half with \(t\), and in females, we get only one type of gametes i.e., \(T\).

From the Punnett square it is seen that all the progeny in the F generation are tall ( Tt ), 50% homozygous tall (TT), and 50% heterozygous tall (Tt).

Phenotype: All tall
Genotype ratio : TT: Tt 2:2 or 1:1

Q7. When a cross is made between tall plant with yellow seeds (TtYy) and tall plant with green seed (Ttyy) what proportions of phenotype in the offspring could be expected to be
(a) tali and green.
(b) dwarf and green.

Answer: The cross between tall plant with yellow seeds (TtYy) and tall plant with green seed (Ttyy), will look like this:

Hence there will be only one plant with dwarf and green seed trait. 

Q8. Two heterozygous parents are crossed. If the two loci are linked what would be the distribution of phenotypic features in F1 generation for a dihybrid cross?

Answer: Linkage is defined as the coexistence of two or more genes in the same chromosome. If the genes are situated on the same chromosome and lie close to each other, then they are inherited together and are said to be linked genes.

For example, a cross between yellow body and white eyes and wild type parent in a Drosophila will produce wild type and yellow white progenies. It is because yellow bodied and white eyed genes are linked. Therefore, they are inherited together in progenies.

Q9. Briefly mention the contribution of T.H. Morgan in genetics.

Answer: Thomas Hunt Morgan (1866-1945), an American geneticist and Nobel Prize winner of 1933, is considered as “Father of experimental genetics” for his work on and discovery of linkage, crossing over, sex linkage, criss cross inheritance, linkage maps, mutability of genes, etc. He is called fly man of genetics because of selecting fruit fly (Drosophila melanogaster) as research, material in experimental genetics. It was largely due to his book, “The Theory of Gene”, that genetics was accepted as a distinct branch of biology. In 1910, he discovered linkage and distinguished linked and unlinked genes. Morgan and Castle (1911) proposed “Chromosome Theory of Linkage” showing that genes are located on the chromosomes and arranged in linear order. Morgan and Sturtevant (1911) found that frequency of crossing over (recombination) between two linked genes is directly proportional to the distance between the two. 1% recombination is considered to be equal to 1 centi Morgan (cM) or 1 map unit. He worked on sex linked inheritance and reported a white eyed male Drosophila in a population of red eyed and proved that gene of eye colour is located on \(X\)-chromosome. The male passed its genes on \(X\)-chromosomes to the daughter while the son gets genes on \(X\) -chromosome from the female (mother): It is called crisscross inheritance.

Q10. What is pedigree analysis? Suggest how such an analysis, can be useful.

Answer: Pedigree analysis is study of pedigree for the transmission of particular trait and finding the possibility of absence or presence of that trait in homozygous or heterozygous state in a particular individual. Pedigree analysis helps-
(i) in analysis of transmission of character in family over generation.
(ii) in genetic counselling of disease like haemophilia.
(iii) to identify whether a particular genetic disease is due to recessive gene or a dominant gene.
(iv) to identify the possible origin of the defective gene in the family or in a population.

Q11. How is sex determined in human beings?

Answer: In humans, there are 23 pairs of chromosomes. 22 pairs of these chromosomes do not take part in sex determination called autosomes. The 23rd pair determines the sex of an individual called allosome or sex chromosome. If it is \(X X\) then female, if \(X Y\) then male. The presence of \(Y^1\) makes a person male. Human females produce only 1 type of gamete \(22+X\). In males, it could be \(22+X\) or \(22+Y\).

Q12. A child has blood group O. If the father has blood group A and mother of blood group B, work out the genotypes of the parents and the possible genotypes of the other off springs.

Answer: If the father has blood group A i.e., \(\boldsymbol{I}^{\mathbf{A}}\boldsymbol{I}^{\mathbf{A}}\) (homozygous) and mother has blood group B i.e., \(\boldsymbol{I}^{\mathbf{B}}\boldsymbol{I}^{\mathbf{B}}\) (homozygous) then all the offsprings will have blood group AB \(\boldsymbol{I}^{\mathbf{A}}\boldsymbol{I}^{\mathbf{B}}\) and not blood group O.

Table 4.2: Table Showing the Genetic Basis of Blood Groups in Human Population
\(
\begin{array}{|l|l|l|l|}
\hline \begin{array}{l}
\text { Allele from } \\
\text { Parent 1 }
\end{array} & \begin{array}{l}
\text { Allele from } \\
\text { Parent 2 }
\end{array} & \begin{array}{l}
\text { Genotype of } \\
\text { offspring }
\end{array} & \begin{array}{l}
\text { Blood } \\
\text { types of } \\
\text { offspring }
\end{array} \\
\hline I^A & I^A & I^A I^A & \mathrm{~A} \\
\hline I^A & I^B & I^A I^B & \mathrm{AB} \\
\hline I^A & i & I^A i & \mathrm{~A} \\
\hline I^B & I^A & I^A I^B & \mathrm{AB} \\
\hline I^B & I^B & I^B I^B & \mathrm{~B} \\
\hline I^B & i & I^B i & \mathrm{~B} \\
\hline i & i & i i & \mathrm{O} \\
\hline
\end{array}
\)

Q13.  Explain the following terms with example
(a) Co-dominance
(b) Incomplete dominance

Answer: (a) Codominance: Codominance is the phenomenon of two contrasting alleles of the same gene lacking dominant recessive ratio and expressing themselves simultaneously when present together. E.g. \(A B O\) blood group system – Human blood group \(A B\) is formed when alleles of blood groups \(A\) and \(B\) are present together (IAIB). Such RBCs carry both antigen A & B showing that both die alleles are expressing their effect phenotypically & codominant.

(b) Incomplete dominance \((1: 2: 1)\): It is the phenomenon where none of the two contrasting alleles being dominant so that expression in the hybrid is intermediate between the expressions of the two alleles in the homozygous state. \(F_2\) phenotypic ratio is \(1: 2: 1\), similar to genotypic ratio. Example-In Mirabilis jalapa (Four o’clock) and Antirrhinum majus (Snapdragon or dog flower), there are two types of flower colour generation are of three types- red, pink and white flowered in the ratio of \(1: 2: 1\). The pink colour apparently appears either due to the mixing of red and white colours (incomplete dominance).

Q14. What is point mutation? Give one example.

Answer: Point mutation is a gene mutation that arises due to a change in a single base pair of DNA.
Example: Sickle-cell anaemia.
Substitution of a single nitrogen base at the sixth codon of the \(\beta\) – globin chain of haemoglobin molecule causes the change in the shape of the R.B.C. from biconcave disc to the elongated shaped, structure which results in sickle cell anaemia.

Q15. Who had proposed the chromosomal theory of the inheritance?

Answer: Chromosomal theory of inheritance was proposed by Sutton and Boveri independently in 1902. The two workers found a close similarity between the transmission of Mendelian hereditary factors (genes) and behaviour of chromosomes during gamete formation and fertilisation. They proposed that chromosomes were the carriers of the Mendelian factors. It is the chromosome and not genes which segregate and assort independently during meiosis and recombine at the time of fertilisation in the zygote. Chromosomal theory of inheritance was expanded by Morgan, Sturtevant and Bridges.

Q16. Mention any two autosomal genetic disorders with their symptoms.

Answer: Sickle cell anaemia : Haemoglobin has less 02 transport, sickle shaped RBCs etc.
Phenylketonuria : Mental retardation (due to accumulation of phenylalanine in brain), hypopigmentation of skin & hair etc.

Exemplar Section

VERY SHORT ANSWER TYPE QUESTIONS

Q1. What is the cross between the progeny of \(\mathrm{F}_1\) and the homozygous recessive parent called? How is it useful?

Answer: When a progeny of \(\mathrm{F}_1\) is crossed with the homozygous recessive parent, it is called test cross. Such a cross is useful to determine the genotype of an unknown, i.e., whether it is heterozygous, or homozygous dominant for the trait.

Q2. Do you think Mendel’s laws of inheritance would have been different if the characters that he chose were located on the same chromosome.

Answer:  If Mendel choose characters that were located on the same chromosome then Mendel would not find the law of independent assortment.

Q3. Enlist the steps of controlled cross pollination. Would emasculation be needed in a cucurbit plant? Give reasons for your answer.

Answer: Steps of controlled cross pollination: Selection of parents —> Emasculation —> Bagging —>Collection of pollen from male parent —>Dusting the pollen on stigma —>Re-bagging
Emasculation could not be needed in a cucurbit plant because it has unisexual flowers.

Q4. A person has to perform crosses for the purpose of studying inheritance of a few traits / characters. What should be the criteria for selecting the organisms? 

Answer: Criteria for selecting the organism are:
(i) Shorter life span
(ii) To produce large number of progeny-
(iii) Clear differentiation of sex/traits
(iv) Many type of hereditary variations

Q5. The pedigree chart given below shows a particular trait which is absent in parents but present in the next generation irrespective of sexes. Draw your conclusion on the basis of the pedigree.

Answer: The trait is auto some linked and recessive in nature. Both the parents are carrier (i.e., heterozygous) hence among offspring few show the trait irrespective of sex. The other off springs are either normal or carrier.

Q6. In order to obtain the \(\mathrm{F}_1\) generation Mendel pollinated a pure-breeding tall plant with a pure breeding dwarf plant. But for getting the \(\mathrm{F}_2\) generation, he simply self-pollinated the tall \(\mathrm{F}_1\) plants. Why?

Answer: Genotype of \(50 \%\) of the offspring would resemble one parent and the rest the other parent. All the \(F_1\), off springs of the cross are heterozygous so allowing self-pollination is sufficient to raise \(F_2\) offspring. Also Mendel intended to understand the inheritance of the selected trait over generations.

Q7. “Genes contain the information that is required to express a particular trait.” Explain.

Answer: The genes present in an organism show a particular trait by way of forming certain product. This is facilitated by the process of transcription and translation (according to central dogma of genetics).

Q8. How are alleles of particular gene differ from each other? Explain its significance.

Answer:  Alleles of a particular gene differ from each other on the basis of certain changes (i.e., mutations) in the genetic material (segment of DNA or RNA). Different alleles of a gene increases the variability or variation among the organisms.

Q9. In a monohybrid cross of plants with red and white flowered plants, Mendel got only red flowered plants. On self-pollinating these F1 plants got both red and white flowered plants in 3:1 ratio. Explain the basis of using RR and rr symbols to represent the genotype of plants of parental generation. 

Answer: On crossing red and white flower only red colour flower appeared in the \(F_1\) generation. But the white colour flower again appear in the \(F_2\) generation which is raised out of the \(F_1\) individuals only. Mendel reasoned that there is a factor of each and every character. Accordingly, there has to be one factor (R) for red flower and other one factor ( \(r\) ) for white flower. In case, an organism possess only one copy of the gene then the possibility of reappearance of white flower in the \(\mathrm{F}_2\) generation of the given cross is not there. Also the ratio ( \(3: 1\) of red and white) indicates that each organism must possess two copies of a particular gene.

Q10. For the expression of traits genes provide only the potentiality and the environment provides the opportunity. Comment on the veracity of the statement.

Answer: \(\text { Phenotype }=\text { Genotype }+ \text { Environment }\)
\(
\quad \quad \text { (Trait) } \quad \quad \quad \quad \text { (Potentiality) } \quad \quad \text { (Opportunity) }
\)

Q11. A, B, D are three independently assorting genes with their recessive alleles a, b, d, respectively. A cross was made between individuals of Aa bb DD genotype with aa bb dd. Find out the type of genotypes of the offspring produced.

Answer: The given cross is \(A a b b d d\) x \(aa bb dd\). Accordingly the type of offspring produced would.

Q12. In our society a woman is often blamed for not bearing male child. Do you think it is right? Justify.

Answer: 50 per cent of sperms carry the \(X\) chromosome while the other 50 percent carry the \(Y\) . After fusion of the male and female gametes the zygote would carry either \(X X\) or \(Y Y\) depending on whether the sperm carrying \(X\) or \(Y\) fertilised the ovum. Sex of the baby is determined by the father and not by the mother. It is unfortunate that in our society women are blamed for producing female children and have been ostracised and ill-treated because of this false notion.

Q13. Discuss the genetic basis of wrinkled phenotype of pea seed.

Answer: Wrinkled phenotype of pea seed is due to small grain size produced by double recessive allele (bb).

Q14. Even if a character shows multiple allelism, an individual will only have two alleles for that character. Why?

Answer: A diploid organism has only two alleles of a character, although it has more than two alleles. Multiple alleles can be found only when population studies are made.

Q15. How does a mutagen induce mutation? Explain with example.

Answer: A mutagen is a physical or chemical agent that changes the genetic material, usually DNA. Different mutagens act on the DNA differently. Powerful mutagens may result in chromosomal instability, causing chromosomal breakages and rearrangement of the chromosomes such as translocation, deletion, and inversion. Ionizing radiations such as X-rays, gamma rays and alpha particles may cause DNA breakage and other damages. Ultraviolet radiations with wavelength above 260 nm are absorbed strongly by bases, producing pyrimidine dimers. Radioactive decay, such as 14C in DNA which decays into nitrogen.

SHORT ANSWER TYPE QUESTIONS

Q1. In a Mendelian monohybrid cross, the \(\mathrm{F}_2\) generation shows identical genotypic and phenotypic ratios. What does it tell us about the nature of alleles involved? Justify your answer.

Answer: In a monohybrid cross, starting with parents which homozygous dominant and homozygous recessive, F₁ , would be heterozygous for the trait and would express the dominant allele. But in case of incomplete dominance, a monohybrid cross shows the result as follows.

\(
\begin{aligned}
& \text { Phenotypic ratio -> Red : Pink : White :: 1:2:1 } \\
& \text { Genotypic ratio -> RR : Rr: rr : : 1:2:1 }
\end{aligned}
\)
Here the genotypic and phenotypic ratios are the same. So, we can conclude that when genotypic and phenotypic ratios are the same, the alleles show incomplete dominance.

Q2. Can a child have blood group O if his parents have blood group ‘A’ and ‘B’. Explain.

Answer: If parents are heterozygous then child can have O blood group but if parents are homozygous then child cannot have O blood group.

Q3. What is Down’s syndrome? Give its symptoms and cause. Why is it that the chances of having a child with Down’s syndrome increases if the age of the mother exceeds forty years?

Answer: Down’s syndrome is a human genetic disorder caused due to trisomy of chromosome no. 21. Such individuals are aneuploid and have 47 chromosomes \((2 n+1)\). The symptoms include mental retardation, growth abnormalities, constantly open mouth, dwarfness etc. The reason for the disorder is the nondisjunction (failure to separate) of homologous chromosome of pair 21 during meiotic division in the ovum. The chances of having a child with Down’s syndrome increase with the age of the mother ( +40 ) because ova are present in females. Since their birth and therefore older cells are more prone to chromosomal nondisjunction because of various physico-chemical exposures during the mother’s life-time.

Q4. How was it concluded that genes are located on chromosomes?

Answer: Morgan confirmed Mendelian laws of inheritance and the hypothesis that genes are located on chromosomes. Morgan had discovered that eye colour in Drosophila expressed a sex-linked trait. All first-generation offspring of a mutant white-eyed male and a normal red-eyed female would have red eyes because every chromosome pair would contain at least one copy of the X chromosome with the dominant trait.

Q5. A plant with red flowers was crossed with another plant with yellow flowers. If \(\mathrm{F}_1\) showed all flowers orange in colour, explain the inheritance.

Answer: This is due to the incomplete dominance as the F, do not resemble either of the two parents and is in between the two.

Q6. What are the characteristic features of a true-breeding line?

Answer: A true-breeding line for a trait is one that, has undergone continuous self-pollination or brother-sister mating, showing a stability in the inheritance of the trait for several generations.

Q7. In peas, tallness is dominant over dwarfness, and red colour of flowers is dominant over the white colour. When a tall plant bearing red flowers was pollinated with a dwarf plant bearing white flowers, the different phenotypic groups were obtained in the progeny in numbers mentioned against them:
Tall, Red = 138
Tall, White = 132
Dwarf, Red = 136
Dwarf, White = 128
Mention the genotypes of the two parents and of the four offspring types.

Answer: The result shows that the four types of offspring are in a ratio of 1:1:1:1. Such a result is observed in a test-cross progeny of a dihybrid cross.
The cross can be represented as:
Tall & Red (Tt Rr) x Dwarf & White (ttrr)

Q8. Why is the frequency of red-green colour blindness is many times higher in males than that in the females?

Answer: For becoming colourblind, the female must have the allele for it in both her X-chromosomes; but males develop colourblindness when their sole X-chromosome has the allele for it.

Q9. If a father and son are both defective in red-green colour vision, is it likely that the son inherited the trait from his father? Comment.

Answer: Gene for colour blindness is X-chromosome linked, and sons receive their sole from their mother, not from their father. Male-to-male inheritances is not possible for X-linked traits in humans. In the given case the mother of the child must be a carrier (heterozygous) for colour blindness gene.

Q10. Discuss why Drosophila has been used extensively for genetical studies.

Answer:Morgan worked with the tiny fruit files, Drosophila melanogaster, which were found very suitable for such studies, as:

1. They could be grown on simple synthetic medium (ripe banana) in the laboratory.
2. They complete their life cycle in about two weeks.
3.  A single mating could produce a large number of progeny flies.
4. There was a clear differentiation of the sexes—the male and female flies are easily distinguishable.
5.  It has many types of hereditary, variations that can be seen with low power microscopes.

Q11. How do genes and chromosomes share similarity from the point of view of genetical studies?

Answer: Similarities between Chromosomes and Genes:

1. Both occur in pairs.
2. Both segregate at the time of gamete formation such that only one of each pair is transmitted to a gamete.
3. Independent pairs segregate independently of each other in both.

Q12. What is recombination? Discuss the applications of recombination from the point of view of genetic engineering.

Answer: The formation of new combinations of genes, either by crossing over or independent assortment is called recombination. Alfred Sturtevant used the frequency of recombination between gene pairs on the same chromosome as a measure of the distance between genes and ‘mapped’ their position on the chromosome. Today genetic maps are extensively used as a starting point in the sequencing of whole genomes as was done in the case of the Human Genome Sequencing Project.

Q13. What is artificial selection? Do you think it affects the process of natural selection? How?

Answer:A process in the breeding of organisms by which the scientist chooses to only those forms having certain desirable characteristics is called artificial selection.

• Natural selection is the differential survival and reproduction of individuals due to differences in phenotype. In natural selection, the difference in reproductive success based on a particular trait is driven by natural processes like predators, weather conditions, and environmental constraints.
•  Artificial selection is imposed by humans on other organisms. It affects the process of natural selection through change in environmental conditions.

Q14. With the help of an example differentiate between incomplete dominance and co-dominance.

Answer: 

incomplete dominance	co-dominance
Effect of one of the two alleles is more conspicuous.	The effect of both the alleles is equally conspicuous.
It produces a fine mixture of the expression of two alleles.	There is no mixing of the effect of the two alleles.
The effect in hybrid is intermediate of the expression of the two alleles, e.g., flower colour in Mirabilis jalapa Red, Pink, and White.	Both the alleles produce their effect independently, e.g., AB blood group.
Alleles show quantitative effect. One dominant allele produces half and two dominant alleles produce full phenotype.	There is no quantitative effect of the alleles.

Q15. It is said, that the harmful alleles get eliminated from population over a period of time, yet sickle cell anaemia is persisting in human population. Why?

Answer: The harmful alleles get eliminated from population over a period of time, yet sickle cell anaemia is persisting in human population because SCA is a harmful condition which is also a potential saviour from malaria. Those with the benign sickle trait possess a resistance to malarial infection. The pathogen that causes the disease spends part of its cycle in the red blood cells and triggers an abnormal drop in oxygen levels in the cell. In carriers, this drop is sufficient to trigger the full sickle-cell reaction, which leads to infected cells being rapidly removed from circulation and strongly limiting the infection’s progress. These individuals have a great resistance to infection and have a greater chance of surviving outbreaks. This resistance to infection is the main reason the SCA allele and SCA disease still exist. It is found in greatest frequency in populations where malaria was and often still is a serious problem.

LONG ANSWER TYPE QUESTIONS

Q1. In a plant tallness is dominant over dwarfness and red flower is dominant over white. Starting with the parents work out a dihybrid cross. What is standard dihybrid ratio? Do you think the values would deviate if the two genes in question are interacting with each other?

Answer: 

Yes, the ratio will deviate if the two genes are interacting with each other. In this condition the genes do not independently assort with each other. So, ratio will be changed from 9:3:3:1.

Q2. a. In humans, males are heterogametic and females are homogametic. Explain. Are there any examples where males are homogametic and females heterogametic?
b. Also describe as to, who determines the sex of an unborn child? Mention whether temperature has a role in sex determination.

Answer: (a) The term homogametic and heterogametic refer to the organism depending upon whether all the gametes contain one type of sex chromosome (Homo = same) or two different types of sex chromosomes (Hetero = different). Humans show XX/XY type of sex determination, i.e. Females contain two copies of \(X\) chromosome and males contain one \(X\) and one \(Y\) chromosome. Therefore, ova produced by females contain the same sex chromosome, i.e. X . On the other hand the sperms contain two different types of chromosomes, i.e. \(50 \%\) sperms have \(X\) and \(50 \%\) have \(Y\) chromosome open from half the autosomes (Meiosis). Therefore, the sperms are different with respect to the composition of sex chromosome. In case of humans, females are considered to be homogametic while males are heterogametic. Yes, there are examples where males are homogametic and females are heterogametic. In some birds the mode of sex determination is denoted by ZZ (males) and ZW (females),

(b) As a rule the heterogametic organism determines the sex of the unborn child. In case of humans, since males are heterogametic it is the father and not the mother who decides the sex of the child. In some animals like crocodiles, lower temperature favour hatching of female offsprings and higher temperatures lead to hatching of male off springs.

Q3. A normal visioned woman, whose father is colour blind, marries a normal visioned man. What would be probability of her sons and daughters to be colour blind? Explain with the help of a pedigree chart. 

Answer: 

All daughters are normal visioned; 50% of sons are likely to be colour blind.

Q4. Discuss in detail the contributions of Morgan and Sturvant in the area of genetics.

Answer: Experimental verification of the chromosomal theory of inheritance by Thomas Hunt Morgan (Father of experimental genetics) and his colleagues, led to discovering the basis .for the variation that sexual reproduction produced. Morgan worked with the tiny fruit files, Drosophila melanogaster, which were found very suitable for such studies.

• Morgan carried out several dihybrid crosses in Drosophila to study genes that were sex-linked. The crosses were similar to the dihybrid crosses carried out by Mendel in peas. For example, Morgan hybridised yellow-bodied, white-eyed females to brown-bodied, red-eyed males and intercrossed their F, progeny.
• He observed that the two genes did not segregate independently of each other and the F2 ratio deviated very significantly from the. $9: 3: 3: 1$ ratio (expected when the two genes are independent). Morgan and his group knew that the genes were located on the X chromosome and saw quickly that when the two genes in a dihybrid cross were situated on the same chromosome, the proportion of parental gene combinations were much higher than the non-parental type.
• Morgan attributed this due to the physical association or linkage of the two genes and coined the term linkage to describe this physical association of genes on a chromosome and the term recombination to describe the generation of non-parental gene combinations.
• His student Alfred Sturtevant used the frequency of recombination between gene pairs on the same chromosome as a measure of the distance between genes and ‘mapped’ their position on the chromosome. Today genetic maps are extensively used as a starting point in the sequencing of whole genomes as was done in the case of the Human Genome Sequencing Project.

Q5. Define aneuploidy. How is it different from polyploidy? Describe the individuals having following chromosomal abnormalities.
a. Trisomy of \(21^{\text {st }}\) Chromosome
b. XXY
c. XO

Answer:Failure of segregation of chromatids during cell division cycle results in the gain or loss of a chromosome(s), called aneuploidy. – Difference between aneuploidy and polyploidy

• Failure of segregation of chromatids during cell division cycle results in the gain or loss of a chromosome(s), called aneuploidy. Failure of cytokinesis after telophase stage of cell division results in an increase in a whole set of chromosomes in an organism and,this phenomenon is known as polyploidy.
• Polyploidy occurs due to altering set of chromosome number such as \(2 n, 3 n, 5 n\), whereas aneuploidy occurs due to altering particular chromosome or part of a chromosome such as \(2 n+1\) (trisomic) and \(2 n-\) 1 (monosomic).
• Aneuploidy can be seen in human as genetic disorders; for example, Tuner syndrome, Klinefelter syndrome and Down syndrome, whereas polyploidy is common in plants.

(i) Down’s Syndrome (Mongolism)

• The cause of this genetic disorder-is the presence of an additional copy of the chromosome number 21 (trisomy of 21) due to non-disjunction of chromosomes during sperm or ova formation.
• The affected individual is short statured with small round head, furrowed tongue and partially open mouth. Palm is broad with characteristic palm crease.
• Physical, psychomotor and mental development is retarded.

(ii) Klinefelter’s Syndrome

• This genetic disorder is also caused due to the presence of an additional copy of X -chromosome resulting into a karyotype of 47, XXY.
• Such an individual has overall masculine development, however, the feminine development (development of breast, i.e., Gynaecomastia) is also expressed. Such individuals are sterile male.

(iii) Turner’s Syndrome

Such a disorder is caused due to the absence of one of the X chromosomes, i.e., 45 with XO. Such females are sterile as ovaries are rudimentary besides other features including lack of other secondary sexual characters.