A man and woman, both with normal vision, have (1) a color-blind son who has a daughter of normal vision; (2) a daughter of normal vision who has one color-blind son and one normal vision son, and (3) another daughter of normal vision who has five sons, all normal. What are the probable genotypes of grandparents, children, and grandchildren? Once complete, build a pedigree to this family.
Color perception and color blindness
Colors add beauty to our world. Not only does it add aesthetic value to objects but it also helps distinguish between different objects. Human beings luckily possess the ability to perceive various primary and secondary colors in general but sadly some people are not able to distinguish between certain colors. Such people are said to be suffering from a disease or condition called color blindness. This happens due to dysfunctional color perceiving cells (the cone photoreceptors) or the absence of these cells altogether. The cones are present in the retina of the eye where these cells separately detect the different colors. Colorblind people mostly cannot distinguish between red and green colors but in some rare cases, they also have blue-yellow perception problems. In most cases, color blindness is a genetic disorder but it can also happen due to factors like aging, accidental injury, brain receptor damages. Gene therapy can be considered as a treatment option shortly once its trials are successfully done.
Answer and Explanation:
Color-blindness is an X-linked recessive trait. Due to this, for a male to be color-blind only one defective X gene is sufficient while, for a female to be color-blind both her X genes need to be defective. For this question, since both the man and woman are normal but are still having children with color-blindness, it implies that the mother must be a carrier of the defective gene.
Let us consider the alleles of the trait to be X for the normal gene and X' for the defective gene responsible for color blindness. So, among the grandparents, the man will have a genotype of XY while the woman has the genotype XX'.
For case (1), the son (parent) is color-blind which indicates that he has an X'Y phenotype. Again, his daughter is normal but we do not know whether the mother is a carrier or not. Irrespective of that, the daughter received X chromosomes from each parent, and the one she received from her father is the defective X' surely. Hence, to show a normal phenotype, she can only be a carrier of color-blindness. Her mother may have either XX' or XX genotype. In both cases, she has to be normal.
For case (2), the daughter (parent) is normal. But, one of her sons is color-blind which indicates he has an X'Y genotype. Since the Y is a male-only chromosome, it has to come from the father of the children. Which confirms that the defective X' comes from the mother. Hence, the mother is a carrier and has a XX' genotype. The other son has received the normal X gene and will have an XY genotype. The father may or may not be color-blind. His genotype cannot be predicted based on the given data.
For case (3) again the daughter (parent) is normal. Also, all of her five sons are normal. Since the Y chromosome is obtained from the father, the healthy X chromosome comes from the mother. But, the mother can be a carrier (XX' genotype) or completely normal (homozygous XX genotype). In case of being a carrier, her sons would have a 1:1 chance of being color-blind. But, since all her sons are normal. Most probably she has the XX phenotype. The father again may or may not be color-blind.
Please find the pedigree chart attached to the answer.
Learn more about this topic:
from Biology 102: Basic GeneticsChapter 6 / Lesson 3