Problem 9: A color blind man and a woman with normal color vision who is also no

Question

Problem 9: A color blind man and a woman with normal color vision who is also not a carrier for color blindness are going to have children. What are the chances that their children will be color blind Parents Phenotype Genotype Father Mother What are the children's genotypes and phenotypes? Children Genotype Phenotype Boy or grl? What percentage of the children should be expected to be colorblind? What percentage of the daughters should be expected to be colorblind? What percentage of the sons should be expected to be colorblind? From which parent did the colorblind children inherit the recessive gene? What percentage of the children are carriers of the recessive allele? What percentage of the daughters are carriers of the recessive allele? What percentage of the sons are carriers of the recessive allele? - Can a man be a carrier of the red-green color blindness allele? Explain your answer:

Explanation / Answer

Please find the answers below:

According to the question, the couple is a color-blind man and a non-carrier normal woman has kids. Since color-blindness is X-related recessively inherited disease, the genotypes of the parents would be given by:

Mother: XX, Father: XcY

Thus, the Punnett square can be formulated as below:

Genotype of the children: XcX: XY:: 1:1

Phenotype: Carrier daughter : unaffected son :: 1/2:1/2

Thus, based upon this Punnett square, the answers can be found as below:

Can't a man be carrier of the disease : NO

Reason: Since the color-blindness is an X-linked recessively inherited disease, a person should carry two X-chromosomes and only one of it must be linked to color-blindness gene. Since human males carry a single X-chromosome, they can never be carrier of the disease and are affected whenever their X-chromosome carries the gene.

X X Xc XcX XcX Y XY XY

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