
What is the phenotype of the offsprings born to a woman with normal vision (homozygous) and a colour blind man?
A) All the son and daughter are with normal vision.
B) All the sons and daughters are colorblind.
C) All the sons are colorblind and daughters are with normal vision.
D) All the sons are with normal vision and daughters are color blind.
Answer
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Hint: Color blind trait is encoded by gene that is present on the X chromosome. That’s why it is a sex linked trait. It is a recessive trait because females are affected only when she has both trait encoding genes. The male has a hemizygous condition. Hemizygous refers to the single X chromosome condition.
Complete answer:
Let XC encode normal eye traits in individuals and it is dominant over Xc allele. Xc gene in homozygous recessive conditions (Xc Xc) encodes colorblind traits in females. Whereas male is affected in case of homozygous condition having Xc chromosome in genotype. That’s why the affected male genotype is Xc Y.
The woman is normal in case of two genotypes i.e. XC Xc (carrier) and XC XC (Homozygous). Carrier has one Xc allele in genotype. Whereas homozygous has both normal alleles in normal vision females.
It is given that a woman is homozygous for normal color vision. That’s why the genotype of woman is XC XC.
The man is colorblind. That’s why the genotype of man is Xc Y.
Parents: XC XC (Homozygous normal woman) * XcY (Color blind man)
XC XC genotype bearing woman produces XC gametes only by the process of meiosis. Whereas Xc Y genotype bearing man produces Xc and Y gametes by the process of meiosis.
Punnett square for F1:
That’s why normal vision bearing progenies are produced whether they have male or female sex. That’s why all son and daughter bear normal vision. Answer is “Option A”.
Note: Color blind is a trait which affects eye development. In this case, a person is not able to differentiate between red and green colors. The colorblind trait is encoded by X linked gene. This is the case of X-linked recessive inheritance.
Complete answer:
Let XC encode normal eye traits in individuals and it is dominant over Xc allele. Xc gene in homozygous recessive conditions (Xc Xc) encodes colorblind traits in females. Whereas male is affected in case of homozygous condition having Xc chromosome in genotype. That’s why the affected male genotype is Xc Y.
The woman is normal in case of two genotypes i.e. XC Xc (carrier) and XC XC (Homozygous). Carrier has one Xc allele in genotype. Whereas homozygous has both normal alleles in normal vision females.
It is given that a woman is homozygous for normal color vision. That’s why the genotype of woman is XC XC.
The man is colorblind. That’s why the genotype of man is Xc Y.
Parents: XC XC (Homozygous normal woman) * XcY (Color blind man)
XC XC genotype bearing woman produces XC gametes only by the process of meiosis. Whereas Xc Y genotype bearing man produces Xc and Y gametes by the process of meiosis.
Punnett square for F1:
| Gametes | Xc | Y |
| XC | XC Xc ( carrier- normal daughter) | XC Y (Normal son) |
That’s why normal vision bearing progenies are produced whether they have male or female sex. That’s why all son and daughter bear normal vision. Answer is “Option A”.
Note: Color blind is a trait which affects eye development. In this case, a person is not able to differentiate between red and green colors. The colorblind trait is encoded by X linked gene. This is the case of X-linked recessive inheritance.
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