A person whose father is colour blind marries a lady whose mother is daughter of a colour blind man. Their children will be:
Correct answer: B. Some sons normal and some colour blind
- A. All sons colour blind
- B. Some sons normal and some colour blind
- C. All colour blind
- D. All daughters normal
Explanation
In question where the genotype of the other parent is not mentioned then that should be considered as normal. (i) To find out the genotype of person. His father is colour blind. genotype = Xc Y ... (I) His mother is normal XX ... (ii) Xc Y × XX Offsprings Xc X, Xc X, XY, XY As all sons will be normal therefore the genotype of the person will be XY. ...(iii) (ii) To find out the genotype of lady Father's genotype - XY ... (iv) Her mother is a daughter of colourblind father and normal mother. Xc Y × XX X cX, Xc X, XY, XY So the mother of lady would be carrier having genotype X c X ... (v) Performing cross between (iv) and (v) to find out lady's genotype. XY × Xc X Xc X, XX, Xc Y, XY As 50% daughter are carrier and 50% daughter are normal. So the lady can be normal or carrier having genotype XX, X c X respectively. ... (vi) Now considering both the genotype of the lady and the genotype of the person, the result would be as follows. XY × X c X XY × XX Xc X, XY, XX, Xc Y XX, XX, XY, XY About cases show that if mother (lady) is carrier then options (a) and (c) are not true. Option (b) is true and option (d) all daughters normal (though phenotypically) is also true. If mother is normal then options (a), (b) and (c) are not true and option (d) is true so from the cases, it is concluded that option (d) is true.
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Mendel's laws explain segregation, independent assortment and the prediction of genetic ratios using alleles, genotypes and phenotypes. The chapter also covers linked genes, recombination and crossing over during meiosis, then applies pedigree reasoning to X-linked recessive traits, which show different inheritance patterns in males and females.
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