Color blindness is a sex-linked recessive trait. A mother with normal color vision and a color blind father have a color blind daughter. Which can you conclude on the offspring?

The question is incomplete as it does not have the option which are:

A). All of their daughters must be color blind ...

B). Some of their sons can have normal color vision ...

C). Their daughters might carry a gene for color blindness.

D). All of their sons will carry a gene for color blindness.

Answer:

Option-B and C are correct

Explanation:

The gene for color blind is located on the X chromosome which is passed on to the progeny during sexual reproduction.

Since the disease is a recessive trait therefore in mothers the gene expresses themselves when both the chromosome contains the allele that is during homozygous condition.

In males, since one Y chromosome is present therefore only one copy of the chromosome carrying the allele will cause disease.

The heterozygous females (XcX) - the carrier of the disease The homozygous females (XcXc) - the affected females. The boy with genotype - (XcY) will be affected males The boy with (XY) will be normal males.

In the given question, the cross between the normal color vision genotype - XcX and affected male that is XcY is performed.

Xc X

Xc XcXc XcX

Y XcY XY

This shows that the their daughter might carry gene for color blindness (XcX) and their son ca have normal vision (XY).

Thus, Option-B is the correct answer.

Answer: The daughter will be color blind

Explanation:

Sex-linked characters such as colour vision are determined by a gene located on the X-chromosome.

Now the daughter will be colour blind because she inherited the recessive allele for colour blind vision that was masked in her mother (that is why she had normal colour vision), but expressed in father (reason for his being colour blind).

Thus, her colour blind vision means she possess a homozygous allelic pair for the sex-linked character, since her both parents contributed X-chromosome each.