Question

1) Albinism is known to be due to an autosomal recessive mutation. The first child of a couple with normal skin pigmentation was an Albino. What is the probability that the second child will also be an albino?
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2) hemophilic man marries a normal woman their offsprings will be

a) All haemophilic
b) All boys haemophilic
c) All girls haemophilic
d) All normal

Class - 12th
Chapter:- Molecular basis of inheritance.
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Answer 1

Albinism is a condition of partial loss of skin pigmentation. Vowing to the rarity of albinism, it is safe to assume that it is a recessive trait.

RRr would be the parental genotype.

Crossing over would produce off springs with the genotype,

$\begin{array}{|c|c|c|}\cline{1 - 3} & R & Rr \\ \cline{1-3} R & RR & RRr \\ \cline{1-3} Rr & RRr & RrRr \\ \cline{1-3} \end{array}$

Genotype ratio is 3:1.

The probability of second child being an albino is 25 percent.

$\rule{300}{2}$

Haemophilia is an X - linked genetic disorder.

Now, when a haemophilic man ($\sf X^hY$) marries a normal woman ($\sf XX$)

$\begin{array}{|c|c|c|}\cline{1 - 3} & X^h & Y \\ \cline{1-3} X & XX^h & XY \\ \cline{1-3} X & XX^h & XY \\ \cline{1-3} \end{array}$

We can infer that all off springs are normal.

• Female off springs would become a carrier for the haemophilic gene.
• Male off springs are normal.

Option (d) is correct.

Answer 2

$\bf{\gray{\overbrace{\underbrace{\red{1\:\:.\:\:25\%\:}}}}}$

$\bf{\gray{\overbrace{\underbrace{\red{2\:\:.\:\:(d)\:All\:are\:normal\:}}}}}$

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$\bf{\gray{\underbrace{\blue{Explanation\:\longrightarrow\:(1)\:}}}}$

[1] Here is given that, Albinism is an autosomal-recessive mutation .

☯︎ It is given that, a couple have normal skin .

➪ That means the chromosomal pattern of couple are either "AA" or "Aa" .

Where,

• AA = Homozygous condition

• Aa = Heterozygous condition

☯︎ According to the question, The first child of the couple was an Albino .

➪ i.e. the chromosomal pattern of the first child is "aa".

✨ From the above it is clearly results that, the chromosomal pattern of the couple is "Aa" .

• i.e. the couple is Heterozygous .

➪ Means, the chromosomal pattern of the mother is "Aa" and the father is "Aa" .

✨ Now, we can crossing over the both chromosomal pattern to get the probability of second child(genotype) for albino .

$\rm\begin{array}{| c | c | c |}\cline{1 - 3} Male({\downarrow})/Female{\rightarrow} & A & a \\ \cline{1 - 3} A & AA & Aa \\ \cline{1 - 3} a & Aa & aa \\ \cline{1 - 3}\end{array}$

✨ Hence, from the above crossing over we have found that,

• AA , Aa and Aa are Normal, which is 75% of total .

• Only aa is Albino, which is 25% of total .

✯ Therefore, The probability that the second child will also be an Albino is "25%" .

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$\bf{\gray{\underbrace{\blue{Explanation\:\longrightarrow\:(2)\:}}}}$

[2] Here is given that, a haemophilic man marries a normal women .

➪ The chromosomal pattern of the haemophilic man is $\bf\red{X^hY}$ .

➪ And the chromosomal pattern of the normal women is $\bf\red{XX}$ .

☯︎ Now, we can crossing over between the above two chromosomal pattern to get the probability of their offsprings for haemophilia .

$\rm\begin{array}{| c | c | c |}\cline{1 - 3} Male({\downarrow})/Female{\rightarrow} & X & X \\ \cline{1 - 3} X^h & XX^h & XX^h \\ \cline{1 - 3} Y & XY & XY \\ \cline{1 - 3}\end{array}$

✨ Hence, from the above crossing over we have found that,

• $\bf\red{XX^h}$ and $\bf\red{XX^h}$ are carrier but normal .

• And $\bf\red{XY}$ and $\bf\red{XY}$ are normal also .

✯ Therefore, Their offsprings will be normal .