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Person whose father is colourblind marri...

Person whose father is colourblind marries a lady whose mother is daughter of a colourblind man. Their children will be

A

All normal

B

All colour blind

C

All sons colour blind

D

Some sons normal and some colour blind

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The correct Answer is:
To solve the problem of determining the potential color vision of the children from a colorblind father and a carrier mother, we can follow these steps: ### Step 1: Understand the Genetics of Colorblindness Colorblindness is a sex-linked recessive trait, which means it is carried on the X chromosome. Males have one X and one Y chromosome (XY), while females have two X chromosomes (XX). A male will express colorblindness if his single X chromosome carries the colorblind allele (X^c). A female must have both X chromosomes carrying the colorblind allele (X^cX^c) to express colorblindness; if she has one normal allele (X^C), she will be a carrier (X^C X^c). ### Step 2: Determine the Genotype of the Father The father is colorblind, which means his genotype is X^cY. ### Step 3: Determine the Genotype of the Mother The mother is the daughter of a colorblind man. Since her father is colorblind (X^cY), her mother must have at least one normal X chromosome (X^C). Therefore, the mother’s genotype can be inferred as X^C X^c (carrier). ### Step 4: Set Up a Punnett Square Now, we can set up a Punnett square to determine the possible genotypes of their children. The father's alleles (X^cY) and the mother's alleles (X^C X^c) will be used in the Punnett square. ``` X^C X^c ---------------- X^c | X^C X^c | X^c Y | ---------------- Y | X^C Y | Y | ``` ### Step 5: Analyze the Results From the Punnett square, we can see the following possible genotypes for the children: 1. X^C X^c (female carrier) 2. X^c Y (colorblind male) 3. X^C Y (normal vision male) 4. Y (not applicable for females) ### Step 6: Conclusion The possible phenotypes of the children are: - Female children: 50% chance of being carriers (normal vision) and 0% chance of being colorblind. - Male children: 50% chance of being colorblind and 50% chance of having normal vision. ### Final Answer The children will be: - 50% chance of being normal vision females (carriers) - 25% chance of being colorblind males - 25% chance of being normal vision males

To solve the problem of determining the potential color vision of the children from a colorblind father and a carrier mother, we can follow these steps: ### Step 1: Understand the Genetics of Colorblindness Colorblindness is a sex-linked recessive trait, which means it is carried on the X chromosome. Males have one X and one Y chromosome (XY), while females have two X chromosomes (XX). A male will express colorblindness if his single X chromosome carries the colorblind allele (X^c). A female must have both X chromosomes carrying the colorblind allele (X^cX^c) to express colorblindness; if she has one normal allele (X^C), she will be a carrier (X^C X^c). ### Step 2: Determine the Genotype of the Father The father is colorblind, which means his genotype is X^cY. ...
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