Biology, 25.06.2019 05:30 hanjonez

Acouple plans to have a child. the mother is homozygous for an autosomal recessive mutation. the father is heterozygous for the same mutation. with respect to this mutation, what are the possible genotypes of any children they have? calculate the probability of each genotype's occurrence. show how you got your answer.

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Answer from: levigabehart

Answer:

The possible genotypes of the children are Hh and hh.The probability of genotype Hh = 1/2The probability of genotype hh = 1/2

Explanation:

Autosomal genes 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 mother is homozygous her genotype would be hh and the father 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 possible genotypes of the children are Hh and hh as shown in the cross.

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

Acouple plans to have a child. the mother is homozygous for an autosomal recessive mutation. the fat

Answer from: jordantay208

Answer:

The possible genotypes of the children are Hh and hh.The probability of genotype Hh = 1/2The probability of genotype hh = 1/2

Explanation:

Since the mother is homozygous her genotype would be hh and the father 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 possible genotypes of the children are Hh and hh as shown in the cross.

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

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feral cats are estimated to eat tens of millions of native animals each night in australia

Answer from: Quest

The answer is c) tons

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Acouple plans to have a child. the mother is homozygous for an autosomal recessive mutation. the fat...