Colour Blindness Probability: X-linked Recessive Inheritance
Red-green colour blindness is an X-linked recessive trait. This means the gene responsible is located on the X chromosome, and an individual needs two copies of the recessive allele (on both X chromosomes for females) to be affected. Males only need one copy (on their single X chromosome) to be affected.
Genotype Determination
- Alleles: Let $X^B$ represent the allele for normal vision and $X^b$ represent the allele for colour blindness.
- Father: The man is phenotypically normal. His genotype is $X^B Y$.
- Mother:
- She is phenotypically normal, meaning she has at least one $X^B$ allele.
- She has a colour-blind brother. His genotype is $X^b Y$. He inherited the $Y$ from his father and $X^b$ from his mother. This indicates the mother's mother was a carrier ($X^B X^b$).
- Since the question's mother inherited one X chromosome from her mother and one from her father (who is normal, $X^B Y$), and she received $X^b$ from her mother (to have a colour-blind brother), her genotype must be heterozygous: $X^B X^b$.
Punnett Square Analysis
We need to find the probability of having a colour-blind son ($X^b Y$) from the cross between the mother ($X^B X^b$) and the father ($X^B Y$).
Possible offspring genotypes:
- From mother $X^B$: Can give $X^B$ (to daughter) or $Y$ (to son).
- From mother $X^b$: Can give $X^b$ (to daughter) or $Y$ (to son).
- From father $X^B$: Can give $X^B$ (to daughter) or $Y$ (to son).
The possible combinations are:
- Daughters: $X^B X^B$ (normal), $X^B X^b$ (normal carrier)
- Sons: $X^B Y$ (normal), $X^b Y$ (colour-blind)
Probability Calculation
The possible genotypes for sons are $X^B Y$ and $X^b Y$. Each occurs with equal probability.
- Probability of having a son: $\frac{1}{2}$
- Probability of the son inheriting the $X^b$ allele from the mother: $\frac{1}{2}$
The probability of having a colour-blind son ($X^b Y$) is the probability of having a son AND that son inheriting $X^b$ from the mother:
$ P(\text{Colour-blind son}) = P(\text{Son}) \times P(X^b \text{ from mother}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} $
Converting the fraction to a percentage:
$ \frac{1}{4} = 0.25 = 25\% $
Therefore, the probability of having a colour-blind son is 25%.
Important Questions from Mendelian Inheritance
The allele associated with albinism in humans is recessive ($c$). The probability that an albino male ($cc$) and a carrier female ($Cc$) will have an offspring with normal skin pigmentation is _________.
(Round off to one decimal place)
- The blood group of the mother is A⁺ and that of the father is AB⁺. Which of the following statements is/are correct?
- Fabry disease in humans is a X-linked disease. The probability (in percentage) for a phenotypically normal father and a carrier mother to have a son with Fabry disease is ________.
A pedigree of an inheritable disease is shown below.
- In tomato plant, red (R) is dominant over yellow (r) for fruit color and purple (P) is dominant over green (p) for stem color. Fruit color and stem color assort independently. The number of progeny plants of different fruit/stem colors obtained from a mating are as follows:
Red fruit, purple stem - 145
Red fruit, green stem – 184
Yellow fruit, purple stem - 66
Yellow fruit, green stem - 47
What are the genotypes of the parent plants in this mating?
