Question

This gene is X-linked. First determine the genotypes of the parents and grandfathers. RG = normal vision, rg = colorblind. Mother's father: Xrg/Y. Father's father: Xrg/Y. Mother:XRG/Xrg. Father:XRG/Y. Notice that the mother must be heterozygous for the rg allele (having normal vision but having inherited an Xrg from her father) and the father, because he has normal vision, must be XRG/Y. The fact that the father's father is colorblind is not relevant because the father will inherit his X chromosome from his mother. XRGXrg x XRGY. XRGXRG = normal daughter. XRGXrg = normal daughter. XRGY = normal son. XrgY = colorblind son. Looking at the distribution of the offspring: (a)¼ (b) ½ (c)¼ (d) zero

1)¼
2) ½
3)¼
4) zero

Answer 1

In X-linked inheritance, a cross between a heterozygous mother and a father with normal vision results in a 50% chance of sons being colorblind, while daughters will either be carriers or have normal vision but cannot be affected by color blindness.

Step-by-step explanation:

When it comes to X-linked inheritance, especially for traits like color blindness, knowing the parents' genotypes allows us to predict the likelihood of the offspring being affected. Since color blindness is a recessive trait, a person must have two copies of the allele to express the trait. For males, who have only one X chromosome, a single recessive allele will lead to color blindness. According to the given genotypes, the cross between a heterozygous mother (XRG/Xrg) and a normal vision father (XRG/Y) will result in a 50% chance of their sons being colorblind (Xrg/Y) and 0% chance for their daughters to be colorblind, as they would be either carriers (XRG/Xrg) or have normal vision (XRG/XRG).

The distribution of offspring genotypes are as follows: unaffected daughters (XRG/XRG and XRG/Xrg) will occur with a combined probability of 50%, normal vision sons (XRG/Y) will occur with a probability of 50%, and colorblind sons (Xrg/Y) will also have a 50% chance. Thus, the correct distribution for offspring is: (a) ¼ affected daughters, which is incorrect as they cannot be affected, (b) ½ carrier daughters, (c) ¼ unaffected sons, which is incorrect as they have a 50% chance, and (d) zero affected daughters since they will either be carriers or have normal vision.