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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

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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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NCERT FINGERTIPS ENGLISH-PRINCIPLES OF INHERITANCE & VARIATION-Principles Of Inheritance & Variation
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  2. Father of a child is colourblind and mother is carrier for colourblin...

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  3. A coloured man (X^(c)Y) marries a woman who is carrier for haemophilia...

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  4. A marriage between a colourblind man and a normal woman produces

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  10. Which of the following is not a example of recessive autosomal disease...

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  11. If both parents are carriers for thalassaemia, which is an autosomal r...

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  12. Select the disease which is caused by recessive autosomal genes when p...

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  14. Match column I with column II and select the correct option from th...

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  15. Which of the following trait is controlled by dominant autosomal genes...

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  16. Refer the given statements (i) Incomplete or mosaic inheritance is ...

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  18. Match column I with column II and select the correct option from the g...

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