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wel
3 years ago
8

7. Normal color vision in the human is due to a dominant gene, C, and is x-linked. The recessive form of the gene, c, causes the

individual to be red-green color blind. A color blind woman marries a man with normal color vision. What are the possible phenotypes of children with respect to vision?
Biology
2 answers:
Vladimir79 [104]3 years ago
6 0

Answer:

50% will be normal and 50% will be carriers

Explanation:

Color blindness is a reduced ability to distinguish between certain colours.

The condition is often inherited. Other causes include certain eye diseases and medication. More men than women are affected.

If such a carrier woman with normal vision (heterozygous for colour blindness) marries a normal man (XY), the following progeny may be expected in the generation: among the daughters, 50% is normal and 50% are carriers for the diseases; among sons, 50% are colour blind and 50% are with normal vision.

olganol [36]3 years ago
4 0

Colour blindness is an example of a sex-linked characters.

Sex-linked characters are characters controlled by genes located on the X-chromosome.

The genes for these characters are recessive.

The male has only one gene for a sex-linked character (X^cY).

A female has two genes for sex-linked characters. One in each of the sex chromosomes. If she is homozygous (X^cY^c) for the recessive allele the trait will be expressed.

If she is heterozygous (X^CX^c) she will be normal but carry the abnormal trait involved. She is known as a carrier.

If a colour blind woman marries a man with a normal vision, All their male children will be colour blind(X^cY) because they will inherit if from their mother as the 'Y' chromosomes is empty of the trait.

All their female children will have a normal vision (X^CX^c).

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Answer:

Option C

Explanation:

Given ,

A1 allele is carried by 70 % people

Let us assume A1 s dominant genotype

This means p= \frac{70}{100} = 0.7

Thus, frequency of allele in the given population is 0.7

It is also given that the population is in Hardy-Weinberg equilibrium thus

p+q=1\\0.7+q=1\\q = 1-0.7\\q= 0.3

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q^2\\= (0.3)^2\\= 0.09

As per Hardy-Weinberg's second equation of equilibrium

p^{2} + q^{2} + 2pq = 1\\0.49+0.09+2pq = 1\\2pq = 1-0.49-0.09\\2pq= 0.42

Hence, option C is correct

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