Practice MCQ corner

Hint

\(
\text { (a) : Characters studied by Mendel are as follows: }
\)

\(
\begin{array}{|l|l|l|l|}
\hline & \text { Trait studied } & \text { Dominant } & \text { Recessive } \\
\hline \text { 1. } & \text { Plant height } & \text { Tall (T) } & \text { Dwarf (t) } \\
\hline \text { 2. } & \text { Flower position } & \text { Axial (A) } & \text { Terminal (a) } \\
\hline 3 . & \text { Flower colour } & \text { Violet (V) or }(\mathrm{W}) & \text { White }(\mathrm{v}) \text { or }(\mathrm{w}) \\
\hline \text { 4. } & \text { Pod shape } & \begin{array}{l}
\text { Full or Inflated } \\
(\mathrm{I}) \text { or }(\mathrm{C})
\end{array} & \text { Constricted (i) or (c) } \\
\hline \text { 5. } & \text { Pod colour } & \text { Green }(\mathrm{G}) \text { or }(\mathrm{Y}) & \text { Yellow }(\mathrm{g}) \text { or }(\mathrm{y}) \\
\hline \text { 6. } & \text { Seed shape } & \text { Round (R) or (W) } & \text { Wrinkled (r) or }(\mathrm{w}) \\
\hline \text { 7. } & \text { Seed colour } & \text { Yellow }(\mathrm{Y}) \text { or }(\mathrm{G}) & \text { Green }(\mathrm{y}) \text { or }(\mathrm{g}) \\
\hline
\end{array}
\)

Hint

(c) : \(
\begin{array}{|l|l|l|l|}
\hline & \text { Trait studied } & \text { Dominant } & \text { Recessive } \\
\hline \text { 1. } & \text { Plant height } & \text { Tall (T) } & \text { Dwarf (t) } \\
\hline \text { 2. } & \text { Flower position } & \text { Axial (A) } & \text { Terminal (a) } \\
\hline 3 . & \text { Flower colour } & \text { Violet (V) or }(\mathrm{W}) & \text { White }(\mathrm{v}) \text { or }(\mathrm{w}) \\
\hline \text { 4. } & \text { Pod shape } & \begin{array}{l}
\text { Full or Inflated } \\
(\mathrm{I}) \text { or }(\mathrm{C})
\end{array} & \text { Constricted (i) or (c) } \\
\hline \text { 5. } & \text { Pod colour } & \text { Green }(\mathrm{G}) \text { or }(\mathrm{Y}) & \text { Yellow }(\mathrm{g}) \text { or }(\mathrm{y}) \\
\hline \text { 6. } & \text { Seed shape } & \text { Round (R) or (W) } & \text { Wrinkled (r) or }(\mathrm{w}) \\
\hline \text { 7. } & \text { Seed colour } & \text { Yellow }(\mathrm{Y}) \text { or }(\mathrm{G}) & \text { Green }(\mathrm{y}) \text { or }(\mathrm{g}) \\
\hline
\end{array}
\)

Hint

(b) \(
\begin{array}{|l|l|l|l|}
\hline & \text { Trait studied } & \text { Dominant } & \text { Recessive } \\
\hline \text { 1. } & \text { Plant height } & \text { Tall (T) } & \text { Dwarf (t) } \\
\hline \text { 2. } & \text { Flower position } & \text { Axial (A) } & \text { Terminal (a) } \\
\hline 3 . & \text { Flower colour } & \text { Violet (V) or }(\mathrm{W}) & \text { White }(\mathrm{v}) \text { or }(\mathrm{w}) \\
\hline \text { 4. } & \text { Pod shape } & \begin{array}{l}
\text { Full or Inflated } \\
(\mathrm{I}) \text { or }(\mathrm{C})
\end{array} & \text { Constricted (i) or (c) } \\
\hline \text { 5. } & \text { Pod colour } & \text { Green }(\mathrm{G}) \text { or }(\mathrm{Y}) & \text { Yellow }(\mathrm{g}) \text { or }(\mathrm{y}) \\
\hline \text { 6. } & \text { Seed shape } & \text { Round (R) or (W) } & \text { Wrinkled (r) or }(\mathrm{w}) \\
\hline \text { 7. } & \text { Seed colour } & \text { Yellow }(\mathrm{Y}) \text { or }(\mathrm{G}) & \text { Green }(\mathrm{y}) \text { or }(\mathrm{g}) \\
\hline
\end{array}
\)

Hint

(c) \(
\begin{array}{|l|l|l|l|}
\hline & \text { Trait studied } & \text { Dominant } & \text { Recessive } \\
\hline \text { 1. } & \text { Plant height } & \text { Tall (T) } & \text { Dwarf (t) } \\
\hline \text { 2. } & \text { Flower position } & \text { Axial (A) } & \text { Terminal (a) } \\
\hline 3 . & \text { Flower colour } & \text { Violet (V) or }(\mathrm{W}) & \text { White }(\mathrm{v}) \text { or }(\mathrm{w}) \\
\hline \text { 4. } & \text { Pod shape } & \begin{array}{l}
\text { Full or Inflated } \\
(\mathrm{I}) \text { or }(\mathrm{C})
\end{array} & \text { Constricted (i) or (c) } \\
\hline \text { 5. } & \text { Pod colour } & \text { Green }(\mathrm{G}) \text { or }(\mathrm{Y}) & \text { Yellow }(\mathrm{g}) \text { or }(\mathrm{y}) \\
\hline \text { 6. } & \text { Seed shape } & \text { Round (R) or (W) } & \text { Wrinkled (r) or }(\mathrm{w}) \\
\hline \text { 7. } & \text { Seed colour } & \text { Yellow }(\mathrm{Y}) \text { or }(\mathrm{G}) & \text { Green }(\mathrm{y}) \text { or }(\mathrm{g}) \\
\hline
\end{array}
\)

Hint

(d)

Hint

(d) \(
\begin{array}{|l|l|l|l|}
\hline & \text { Trait studied } & \text { Dominant } & \text { Recessive } \\
\hline \text { 1. } & \text { Plant height } & \text { Tall (T) } & \text { Dwarf (t) } \\
\hline \text { 2. } & \text { Flower position } & \text { Axial (A) } & \text { Terminal (a) } \\
\hline 3 . & \text { Flower colour } & \text { Violet (V) or }(\mathrm{W}) & \text { White }(\mathrm{v}) \text { or }(\mathrm{w}) \\
\hline \text { 4. } & \text { Pod shape } & \begin{array}{l}
\text { Full or Inflated } \\
(\mathrm{I}) \text { or }(\mathrm{C})
\end{array} & \text { Constricted (i) or (c) } \\
\hline \text { 5. } & \text { Pod colour } & \text { Green }(\mathrm{G}) \text { or }(\mathrm{Y}) & \text { Yellow }(\mathrm{g}) \text { or }(\mathrm{y}) \\
\hline \text { 6. } & \text { Seed shape } & \text { Round (R) or (W) } & \text { Wrinkled (r) or }(\mathrm{w}) \\
\hline \text { 7. } & \text { Seed colour } & \text { Yellow }(\mathrm{Y}) \text { or }(\mathrm{G}) & \text { Green }(\mathrm{y}) \text { or }(\mathrm{g}) \\
\hline
\end{array}
\)

Hint

(b) : Alternating form of a single gene which code for a pair of contrasting traits are known as alleles, i.e., tall and dwarf are alleles determining the height of pea plant.

Hint

(b) : The factor of an allelic or allelomorphic pair which is unable to express its effect in the presence of its contrasting factor in a heterozygote is called recessive factor or allele, e.g., the allele ‘ \(t\) ‘ in hybrid tall pea plant Tt . The effect of recessive factor becomes known only when it is present in the pure or homozygous state, e.g., tt in dwarf pea plant.

Hint

(b) : In first filial generation or heterozygous individuals, out of the two factors or alleles representing the alternate traits of a character, one is dominant and expresses itself in the hybrid or \(F_1\) generation. The other factor or allele is recessive and does not show its effect in the heterozygous individual (Principle of dominance).

Hint

(c) : \(F_1\) progeny is the generation of hybrids produced from a cross between the genetically different individuals called parents. A cross between homozygous and heterozygous parents for a single locus will produce \(1: 1\) ratio of phenotypic features in \(F_1\) generation.

Hint

(b) : A monohybrid cross between two heterozygous individuals.

Hint

(a): The crossing between two heterozygous (Tt) plants would give phenotypic ratio of 3 tall : 1 dwarf.
If plants obtained were 500 , then the number of tall and dwarf plants will be 375 and 125 respectively.

Hint

(c) \(
\begin{array}{|c|c|c|}
\hline \text { 9. } & \text { B } & \text { b } \\
\hline \text { B } & \begin{array}{c}
\text { BB } \\
\text { Black }
\end{array} & \begin{array}{c}
\text { Bb } \\
\text { Black }
\end{array} \\
\hline \text { b } & \begin{array}{c}
\text { Bb } \\
\text { Black }
\end{array} & \begin{array}{c}
\text { bb } \\
\text { Brown }
\end{array} \\
\hline
\end{array}
\)

Hint

(b) : A cross between heterozygous long-winged flies and (homozygous) vestigial winged flies represents an example of test cross in which the exact Mendelian ratio of \(1: 1\) is obtained, i.e., 96 long-winged flies and 96 vestigial winged flies were obtained in the given cross.

Hint

(c) : In cross between two heterozygous tall pea plants the probability of production of dwarf offsprings is \(25 \%\) and tall offsprings is \(75 \%\).

Hint

(b) : When heterozygous plants are self-pollinated, \(50 \%\) of progeny would be parental type. Hence, 600 seeds would have parental genotype.

Hint

(b) : This is an example of a test cross in which a cross is made between heterozygous tall and homozygous dwarf individuals and tall and dwarf plants are obtained in same proportion.

Hint

(a) : In a test cross, an organism (pea plants) showing a dominant phenotype whose genotype is to be determined is cossed with the recessive parent instead of self-crossing. The progenies of such a cross can easily be analysed to predict the genotype of the test organism. Normal test cooss ratio for a monohybrid coss is \(1: 1\) and for a dihybrid coss is \(1: 1: 1: 1\).

Hint

(b) : In a test cross, an organism (pea plants) showing a dominant phenotype whose genotype is to be determined is cossed with the recessive parent instead of self-crossing. The progenies of such a cross can easily be analysed to predict the genotype of the test organism. Normal test cooss ratio for a monohybrid coss is \(1: 1\) and for a dihybrid coss is \(1: 1: 1: 1\).

Hint

(a) : In a test cross, an organism (pea plants) showing a dominant phenotype whose genotype is to be determined is cossed with the recessive parent instead of self-crossing. The progenies of such a cross can easily be analysed to predict the genotype of the test organism. Normal test cooss ratio for a monohybrid coss is \(1: 1\) and for a dihybrid coss is \(1: 1: 1: 1\).

Hint

(a)

Hint

(a) : The principle of segregation is the most fundamental principle of heredity that has universal application with no exception. Some workers like Bateson call the principle of segregation as the principle of purity of gametes because segregation of the two Mendelian factors of a trait results in gametes receiving only one factor out of a pair. As a result gametes are always pure for a character. It is also known as law of non-mixing of alleles and can be demonstrated through a monohybrid cross.

Hint

(a) 

Hint

(b) : The inheritance of flower colour in the Antirninum majus (snapdragon or dog flower) is an example of incomplete dominance. It is the phenomenon in which neither of the two alleles of a gene is completely dominant over the other. In a cross between true-breeding red-flowered (RR) and true-breeding white-flowered plants ( \((\pi)\), the \(F\) plants obtained were pink (Rr) coloured. When the \(F_1\) plants were selfpollinated, the \(F_2\) generation resulted in the ratio, 1 (RR) Red: 2 (Rr) Pink: 1 (rr) White. The phenotypic ratios had changed from the normal \(3: 1\) dominant : recessive ratio to \(1: 2: 1 . F_2\) phenotypic ratio is \(1: 2: 1\), similar to genotypic ratio. \(R\) was not completely dominant over r and this made it possible to distinguish Rr (pink) from RR (red) and Ir (white).

Hint

(b) : \(F_2\) phenotypic ratio is \(1: 2: 1\), similar to genotypic ratio.

Hint

(c) : Incomplete dominance is reported in flowers of Mirabilis jalapa (Four o’clock). When red flowers are crossed with pink flowers, \(50 \%\) red flowers and \(50 \%\) pink flowers are obtained.

Hint

(d) : The cross between blue (BW) and black (BB) forms would produce offspring in ratio of 1 black : 1 blue.

Hint

(d) : In human beings \(A B O\) blood groups are controlled by gene \(I\) which has three alleles \(P^A, I^8\) and \(i\). The six possible genotypes are \(A^A A^A, A^A \beta^\beta, \mu_i, \beta^B \|^B, B^B\) and \(i i\). The phenotypes which occur by these genotypes are A ,B ,AB and O.

Hint

(c) : In human beings \(A B O\) blood groups are controlled by gene \(I\) which has three alleles \({ }^A, I^B\) and \(i\). The six possible genotypes are \(A^A A, A^A A^B, A_i, I^B \|^\beta, B^B\) iand \(i\). The phenotypes which occur by these genotypes are A , B , AB and O .

Hint

(b) : When both parents have blood group A B there offsprings can have blood groups A , B or A B but no 0 as allele \(i\) is not present in any of the parents.

Hint

(b) : Coat colour in short-horned cattle is an example of codominance. If a cattle with red coat is crossed with a cattle with white coat, the \(F_1\) hybrids possess neither red nor white coat colour, but have roan coat colour, where red and white patches appear separately. The effect is produced due to juxtaposition of small patches of red and white colour. Hence the alleles which are able to express themselves independently when present together are called codominant alleles and the inheritance pattern is called co-dominance. On inbreeding, the roan hybrids produce three types of cattle – red, roan and white in the ratio of \(1: 2: 1\).

Hint

(c) : Coat colour in short-horned cattle is an example of codominance. If a cattle with red coat is crossed with a cattle with white coat, the \(F_1\) hybrids possess neither red nor white coat colour, but have roan coat colour, where red and white patches appear separately. The effect is produced due to juxtaposition of small patches of red and white colour. Hence the alleles which are able to express themselves independently when present together are called codominant alleles and the inheritance pattern is called co-dominance. On inbreeding, the roan hybrids produce three types of cattle – red, roan and white in the ratio of \(1: 2: 1\).

Hint

(d) : ABO blood group system in human beings is an example of co-dominance and multiple alleles. Co-dominance is a phenomenon in which alleles do not show dominance-recessive relationship and are able to express themselves independently when present together.
More than two alternate forms of a gene present on the same locus are called multiple alleles and the mode of inheritance in these alleles is called multiple allelism. Human beings have six genotypes and four blood group phenotypes-A, \(\mathrm{B}, \mathrm{AB}\) and O .

Hint

(b) : Given, that \(Y\) is dominant for yellow fur and \(y\) is recessive for grey fur, the cross between Yy and \(Y y\) will produce ratio of \(1 \mathrm{YY}: 2 \mathrm{Yy}: 1 \mathrm{y}\). But as \(Y\) is lethal in homozygous condition, hence ohenotvpic ratio would be 2 yellow \((\mathrm{Yy}): 1\) grey (yy).

Hint

(d)

Hint

(a)

Hint

(a) : Gametes produced by AaBb parent would be \(25 \% \mathrm{AB}\), \(25 \% \mathrm{aB} 25 \% \mathrm{Ab}\) and \(25 \% \mathrm{ab}\).

Hint

(a)

Hint

(c) : Number of gametes \(=2^n\) where \(n\) is number of heterozygous loci. Thus, gametes produced by a diploid organism could be \(2^4=16\).

Hint

(d) : Plant \(H\) is formed by fusion of gametes \(y R\) and \(Y R\) and hence has the genotype YyRR. Plant \(N\) is formed by fusion of gametes YR and yR and hence will have the same genotype as plant N i.e., YyRR.

Hint

(d) : Cross involving two contrasting character is called dihybrid cross or a two factor cross. The two factors of each trait assort at random and independent of the factors of other traits at the time of meiosis (gametogenesis) and get randomly as well as independently rearranged in the offspring producing both parental and new combinations of traits. This forms the basis of the law of independent assortment given by Mendel.

Hint

(a) : Cross involving two contrasting character is called dihybrid cross or a two factor cross. The two factors of each trait assort at random and independent of the factors of other traits at the time of meiosis (gametogenesis) and get randomly as well as independently rearranged in the offspring producing both parental and new combinations of traits. This forms the basis of the law of independent assortment given by Mendel.

Hint

(b) : Gregor Johann Mendel (1822-84) is known as Father of Genetics. Mendel’s work remained un-noticed and unappreciated for some 34 years. In 1900, three workers independently rediscovered the principles of heredity already worked out by Mendel. They were Hugo de Vries of Holland, Carl Correns of Germany and Erich von Tschermak of Austria.

Hint

(d) : Gregor Johann Mendel (1822-84) is known as Father of Genetics. Mendel’s work remained un-noticed and unappreciated for some 34 years. In 1900, three workers independently rediscovered the principles of heredity already worked out by Mendel. They were Hugo de Vries of Holland, Carl Correns of Germany and Erich von Tschermak of Austria.

Hint

(b): Chromosomal theory of inheritance believes that chromosomes are vehicles of hereditary information which possess Mendelian factors or genes and it is the chromosomes which segregate and assort independently during transmission from one generation to the next. Chromosomal theory of inheritance was proposed by Walter Sutton and Theodore Boveri independently in 1902. But it was later modified and expanded by Morgan, Sturtevant and Bridges.

Hint

(b): Chromosomal theory of inheritance believes that chromosomes are vehicles of hereditary information which possess Mendelian factors or genes and it is the chromosomes which segregate and assort independently during transmission from one generation to the next. Chromosomal theory of inheritance was proposed by Walter Sutton and Theodore Boveri independently in 1902. But it was later modified and expanded by Morgan, Sturtevant and Bridges.

Hint

(b)

Hint

(c)

Hint

(d) : Given that recombinant percentage is \(7 \%\) and \(5 \%\) therefore, total recombinants would be \(7+5=12 \%\). It is known that one map unit is the distance that yields \(1 \%\) recombinant chromosomes. Hence distance between two non-allelic genes is 12 map units.

Hint

(a) : The physical distance between two genes determines both the strength of the linkage and the frequency of the crossing over between two genes. The strength of the linkage increases with the closeness of the two genes. On the other hand the frequency of crossing over increases with the increase in the physical distance between the two genes.

Hint

(c) : Linked genes are those genes which do not show independent assortment but remain together because they are present on the same chromosome. In linkage there is a tendency to maintain the parental gene combination except for occasional crossovers.

Hint

(d) : Linkage or genetic or chromosome map is a linear graphical representation of the sequence and relative distances of the various genes present in a chromosome. The first chromosome maps were prepared by Sturtevant in 1911 for two chromosomes and in 1913 for all the four chromosomes of Drosophila.

Hint

(b) : The distance between genes is measured by map unit. \(1 \%\) crossing over between two linked genes is known as 1 map unit or centi Morgan (cM). \(100 \%\) crossing over is termed as Morgan (M) and \(10 \%\) crossing over as deci Morgan (dM).

Hint

Hint

(d) : Increase in distance between two genes increases the frequency of crossing over while closeness of the genes reduces the chances of crossing over. Since A and d and a and D are located far away, maximum crossing over would occur between them.

Hint

(b) : Mendel did not have any knowledge about linkage and incomplete dominance.

Hint

(a) : As per linkage experiments carried out by Morgan, the two linked genes do not always segregate independently of each other and \(F_2\) ratio deviated very significantly from 9:3:3:1 ratio (expected when two genes are independent). Hence, if linkage was known at the time of Mendel, he would not have been able to explain law of independent assortment.

Hint

(b) : As per linkage experiments carried out by Morgan, the two linked genes do not always segregate independently of each other and \(F_2\) ratio deviated very significantly from 9:3:3:1 ratio (expected when two genes are independent). Hence, if linkage was known at the time of Mendel, he would not have been able to explain law of independent assortment.

Hint

(b) : The phenotypic ratio of monohybrid \(\mathrm{F}_2\) generation in case of Mirabilis jalapa is 1 Red : 2 Pink : 1 White, due to incomplete dominance.

Hint

(b) : Types of gametes \(=2^n\), where \(n\) is number of heterozygous loci. Thus, man will produce \(2^2=4\) types of gametes and woman will produce \(2^3=8\) types of gametes.

Hint

(a)

Hint

(b)

Hint

(b) : The ability of a gene to have multiple phenotypic effects because it influences a number of characters simultaneously is known as pleiotropy and the gene is called pleiotropic gene. In human beings pleiotropy is exhibited by syndromes called sickle cell anaemia and phenylketonuria.

Hint

(d) : The ability of a gene to have multiple phenotypic effects because it influences a number of characters simultaneously is known as pleiotropy and the gene is called pleiotropic gene. In human beings pleiotropy is exhibited by syndromes called sickle cell anaemia and phenylketonuria.

Hint

(c)

Hint

 (b) : The ability of a gene to have multiple phenotypic effects because it influences a number of characters simultaneously is known as pleiotropy and the gene is called pleiotropic gene. In human beings pleiotropy is exhibited by syndromes called sickle cell anaemia and phenyiketonuria.

Hint

(a)

Hint

(a) : In \(X O\) type and \(X Y\) type of sex determining mechanisms, males produce two different types of gametes, either with or without X -chromosome (XO type), or some gametes with X -chromosome and some with Y-chromosome (XY type). Such type of sex determination mechanism is designated to be the example of male heterogamety. In both, females are homogametic and produce \(X\) type of gametes in both the cases and have \(X X\) genotype.

Hint

(b) : In \(X O\) type and \(X Y\) type of sex determining mechanisms, males produce two different types of gametes, either with or without X -chromosome (XO type), or some gametes with X -chromosome and some with Y-chromosome (XY type). Such type of sex determination mechanism is designated to be the example of male heterogamety. In both, females are homogametic and produce \(X\) type of gametes in both the cases and have \(X X\) genotype.

Hint

(b) : In \(Z W-Z Z\) type of sex determination, the male has two homomorphic sex chromosomes (ZZ) and is homogametic, and the female has two heteromorphic sex chromosomes (ZW) and is heterogametic.

Hint

(b) : The possibility of having a girl or boy child is equal i.e., \(50 \%\), as \(50 \%\) male gametes are \(Y\) type and \(50 \%\) are \(X\) type. Fusion of egg with X type sperm will produce a girl child.

Hint

(d)

Hint

(b) : In humans, number of autosomes are \(2 n=44\) or 22 pairs regardless of the sex.

Hint

(c)

Hint

(d) : Haploid-diploid mechanism or Haplodiploidy is a unique phenomenon in which an unfertilised egg develops into a male and a fertilised egg develops into a female. Therefore, the female is diploid (2n) and the male is haploid ( \(n\) ). Eggs are formed by meiosis and sperms by mitosis. Fertilisation restores the diploid number of chromosomes in the zygote which gives rise to the female. If the egg is not fertilised, it will still develop but into a male (arrhenotoky). It is seen in hymenopterous insects, such as bees, wasps, saw flies and ants.

Hint

(c)

Hint

(d)

Hint

(a) : The given figure represents karyotype of a person suffering from Down’s syndrome.

Hint

(a) : Turner’s syndrome is due to monosomy \((2 n-1)\). It occurs by the union of an allosome free egg \((22+0)\) and a normal \(X\) sperm or a normal egg and an allosome free sperm \((22+0)\). Turners’ syndrome (most common type of female genetic disease) having \(X 0\) genotype is caused by the absence of \(X\) chromosomes in female. These are sterile females with poorly developed ovaries, small uterus and underdeveloped breasts. They have webbed neck and broad chest. The individual has \(2 n=45\) chromosomes \((44+\) XO) instead of 46 .
Klinefelter’s syndrome occurs by the union of an abnormal \(X X\) egg and a normal \(Y\) sperm or a normal \(X\) egg and abnormal XY sperm. The individual has \(2 n=47\) chromosomes \((44+X X Y)\). Such persons are sterile males with undeveloped testes, mental retardation, sparse body hair, long limbs and with some female characteristics such as enlarged breast, i.e. gynaecomastia. It is considered that more the X chromosomes, greater is the mental defect. As the syndrome has two X chromosomes, one Barr body is seen in this case.

Hint

(d) : Any extracellular physical or chemical factor which can cause mutations or increase the frequency of mutations in organisms is called mutagen. They include temperature and high energy radiations such as X -rays, ultra-violet rays and gamma rays.

Hint

(c)

Hint

(b)

Hint

(a) : In pedigree analysis, solid symbol shows affected individuals.

Hint

(c)

Hint

(a)

Hint

(d)

Hint

(a)

Hint

(d) : The given trait cannot be sex-linked as sex-linked traits follow criss-cross inheritance and in the given pedigree, no criss-cross inheritance is being followed. The trait exhibited in pedigree chart is autosomal recessive and appears in case of marriage between two heterozygous individuals ( \(\mathrm{Aa} \times \mathrm{Aa}=3 \mathrm{Aa}+1 \mathrm{aa}\) ), a recessive individual with hybrid ( \(\mathrm{Aa} \times\) aa \(=2 \mathrm{Aa}+2 \mathrm{aa})\) and two recessives (aa \(\times\) aa \(=\) all \(a a\) ). It expresses its effect only in pure or homozygous state. If the trait had been controlled by dominant gene, then one of the parent must have possessed the dominant gene and hence the disease.

Hint

(a)

Hint

(d)

Hint

(b)

Hint

(a) : When a haemophilic man ( \(X^h Y\) ) marries a carrier woman \(\left(X X^h\right)\), then \(50 \%\) daughters are carriers and \(50 \%\) are haemophilic.

Hint

(b) : Haemophilia is genetic disorder due to the presence of a recessive sex linked gene \(h\), carried by \(X\)-chromosome.

Hint

(c) : When a haemophilic man ( \(X^{\dagger} Y\) ) marries a normal woman \((X X)\), carrier girls \(\left(X X^h\right)\) and normal boys \((X Y)\) are produced.

Hint

(b) : If the father of a child is colourblind \((X \subset Y\) ) and mother is carrier of colourblindness (XXC), then probability of the child being colourblind is \(50 \%\).

Hint

(d) : When a colourblind man \(\left(X^c Y\right.\) ) marries a woman who is carrier for haemophilia (XXh), then 25 % female progenies will carry genes for both diseases, 25 % female progenies will carry only the gene for colourblindness, 25 % male progenies would carry only the gene for haemophilia and 25 % male progenies would be normal.

Hint

(a) : Colourblindness is a recessive sex-linked trait. Man is colourblind therefore his genotype is \(X \subset Y\) and woman is normal so her genotype is \(X X\).

Hint

  (c) : Since colourblindness is a sex-linked recessive trait and males just have one X chromosome, they can never be the carriers. Males will always express the disease/phenotype.