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1 answer:
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<u>Answer</u>:
- The possible genotypes of the children are Hh and hh.
- The probability of genotype Hh = 1/2
- The probability of genotype hh = 1/2
<u>Explanation</u>:
- <em>Autosomal genes</em> are the genes present on any of the chromosomes except the sex chromosomes.
- An autosomal recessive mutation is the mutation in an autosomal gene which only expresses itself in the phenotype in homozygous condition and affects both males and female equally.
- For the given question, let the autosomal gene be represented by H (dominant) and h (recessive)
- According to the question,
Since the <em>mother</em> is homozygous her genotype would be hh and the <em>father</em> is heterozygous so his genotype would be Hh.
The following cross is made to know the genotypes of the children :
hh X Hh--------> Hh Hh hh hh
So the <em>possible genotypes of the children are Hh and hh as shown in the cross. </em>
Note : An image is attached representing the cross in a Punnett square.
- The probability of an event = No. of favourable outcomes/total no. of outcomes
The probability of occurrence of genotype Hh would be = 2/4 = 1/2
Similarly, the probability of occurrence of genotype hh would be 2/4 = 1/2
