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
- Mendel's 'law of segregation' applies to the segregation of __________ during gamete formation.
- 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 ________.
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?
- Assuming independent assortment and no recombination, the number of different combinations of maternal and paternal chromosomes in gametes of an organism with a diploid number of 12 is ________.
