The probability of having a haemophilia carrier daughter by a haemophilia carrier mother and a normal father is:
50%
Haemophilia is a genetic disorder that affects blood clotting. It is an X-linked recessive trait. This means the gene responsible for haemophilia is located on the X chromosome, and two copies of the recessive allele (one on each X chromosome) are needed for a female to be affected, while only one copy on the single X chromosome is needed for a male to be affected.
Let's denote the allele for normal blood clotting as $\text{X}^\text{H}$ and the allele for haemophilia as $\text{X}^\text{h}$.
We are given that the mother is a haemophilia carrier. A carrier female has one normal allele and one recessive haemophilia allele. Her genotype is therefore $\text{X}^\text{H}\text{X}^\text{h}$.
The father is normal. Since haemophilia is X-linked, a normal male has the normal allele on his X chromosome and a Y chromosome. His genotype is $\text{X}^\text{H}\text{Y}$.
To find the probability of their children inheriting these traits, we can use a Punnett square. The mother can produce gametes with either the $\text{X}^\text{H}$ or the $\text{X}^\text{h}$ chromosome. The father can produce gametes with either the $\text{X}^\text{H}$ or the Y chromosome.
| Gametes | $\text{X}^\text{H}$ (Father) | Y (Father) |
|---|---|---|
| $\text{X}^\text{H}$ (Mother) | $\text{X}^\text{H}\text{X}^\text{H}$ | $\text{X}^\text{H}\text{Y}$ |
| $\text{X}^\text{h}$ (Mother) | $\text{X}^\text{H}\text{X}^\text{h}$ | $\text{X}^\text{h}\text{Y}$ |
The Punnett square shows the four possible genotypes for the offspring, each with an equal probability of $1/4$ or 25%:
From the analysis, the four possible outcomes are a normal daughter, a normal son, a carrier daughter, and an affected son. Each outcome has a $1/4$ probability.
The probability of having a carrier daughter among all possible offspring is $1/4$, which is 25%.
However, genetics problems sometimes ask for the probability within a specific sex. Let's look at the female offspring only:
There are two possible outcomes for a daughter. Out of these two outcomes, one is a carrier daughter ($\text{X}^\text{H}\text{X}^\text{h}$).
Therefore, the probability of a daughter being a carrier is $1$ out of $2$, which is $1/2$ or 50%. Based on the provided options and common interpretations of such questions in exams aiming for the 50% answer, the question is likely asking for the probability of having a carrier daughter *among the female offspring*.
Thus, the probability of having a haemophilia carrier daughter in this scenario is 50%.
| Trait | Inheritance Pattern | Key Features |
|---|---|---|
| Haemophilia | X-linked recessive | Gene is on the X chromosome. Recessive allele causes the condition. Males are more frequently affected (only need one copy). Females can be carriers (one normal, one recessive allele). Affected females need two copies of the recessive allele. |
Which of the following is a recessive trait for garden pea plant?
Which of the following pair of contrasting traits was not studied by Mendel?
Failure of chromatids to segregate during cell division cycle results in:
Select the correctly matched pair about sickle cell anaemia:
Genotype: Phenotype:
(A) HbA HbA : Diseased phenotype
(B) HbA HbS : Diseased phenotype
(C) HbS HbS : Diseased phenotype
(D) HbS HbA : Carrier of disease
Choose the correct answer from the options given below:
Match List-I with List-II:
| List-I | List-II |
|---|---|
| (A) Down's Syndrome | (I) Absence of a copy of X chromosome |
| (B) Klinefelter's Syndrome | (II) Presence of additional copy of X chromosome |
| (C) Turner's Syndrome | (III) Mutation of X chromosome |
| (D) Colour blindness | (IV) Trisomy of 21st chromosome |
Choose the correct answer from the options given below.