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A coloured man (X^(c)Y) marries a woman ...

A coloured man `(X^(c)Y)` marries a woman who is carrier for haemophilia `(XX^(h))`. Which of the following is true for their progenies ?

A

`25 %` female progenies carry the genes foe both haemophilia and colourblindness

B

`25 %` male progenies carry only the gene for haemophilia

C

`25 %` female progenies carry only the gene for colourblindness

D

all of these

Text Solution

AI Generated Solution

The correct Answer is:
To solve the question regarding the progenies of a colored man (X^cY) and a woman who is a carrier for hemophilia (XX^h), we will follow these steps: ### Step 1: Identify the Genotypes - The colored man has the genotype X^cY (where X^c represents the allele for colorblindness). - The woman is a carrier for hemophilia, so her genotype is XX^h (where X^h represents the allele for hemophilia). ### Step 2: Determine Gametes - The man can produce two types of gametes: X^c and Y. - The woman can produce two types of gametes: X and X^h. ### Step 3: Create a Punnett Square - Set up a Punnett square to determine the possible combinations of alleles from the parents. ``` X X^h ----------------- X^c | X X^c | X X^h | ----------------- Y | Y X | Y X^h | ----------------- ``` ### Step 4: Analyze the Progeny From the Punnett square, we can derive the following combinations: 1. Female: XX^c (carrier for colorblindness) 2. Female: XX^h (carrier for hemophilia) 3. Male: XY (healthy male) 4. Male: XY^h (hemophilic male) ### Step 5: Determine the Proportions - The progeny can be categorized as follows: - 25% females are carriers for colorblindness (XX^c). - 25% females are carriers for hemophilia (XX^h). - 25% males are healthy (XY). - 25% males have hemophilia (XY^h). ### Step 6: Conclusion Based on the analysis, we can conclude: - 25% of female progenies carry the gene for both hemophilia and colorblindness. - 25% of male progeny carry only the gene for hemophilia. - 25% of female progenies carry only the gene for colorblindness. - Therefore, all the statements provided in the options are correct. ### Final Answer The correct answer is **Option D: All of these statements are correct.** ---

To solve the question regarding the progenies of a colored man (X^cY) and a woman who is a carrier for hemophilia (XX^h), we will follow these steps: ### Step 1: Identify the Genotypes - The colored man has the genotype X^cY (where X^c represents the allele for colorblindness). - The woman is a carrier for hemophilia, so her genotype is XX^h (where X^h represents the allele for hemophilia). ### Step 2: Determine Gametes - The man can produce two types of gametes: X^c and Y. ...
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Knowledge Check

  • If a haemophilic man marries a carroes women than which of the following holds true for their progenies ?

    A
    50 % daughters are carrier and 50 % are haemophilic
    B
    All the daughters are haemophilic
    C
    All sons are haemophilic and all daughters are normal
    D
    All sons normal, all daughters carriers
  • If x+y=90^(@) , which of the following must be true?

    A
    cosx=cosy
    B
    sin x=-sin y
    C
    tan x=cot y
    D
    sin x+cos y=1
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