Question

1. Complete a punnett square for the cross between a human female (XX) and a human male (XY). What is the chance that the parents will have a girl?
2. If the same parents have four boys, what is the probability their fifth child will be a girl?

Hemophilia is a recessive sex-linked disease carried on the X chromosome in humans.

3. Write the genotype of a woman who does not have hemophilia. ______
4. Write the genotype of a woman with hemophilia. ________
5. Write the genotype of a woman who is a carrier (heterozygous) for hemophilia. ____________
6. Write the genotype of a man who has hemophilia. _________
7. Write the genotype of a man who does not have hemophilia. __________

8. A woman who is heterozygous for hemophilia marries a normal male. What are the possible phenotypes of their children?

9. A woman who is a carrier for hemophilia marries a man with hemophilia. Could any of their children have hemophilia? If so, would the child be male or female?

Answer 1

Question 1:-

Chances of having a girl:

I have uploaded the image above showing the Punnett square between a Human male and female.

Since the chromosomes for sex determination are Xy and XX, the possible outcome out of 4 fertilization is 2 males and 2 females.

To find the probability, the formula is: Number of outcomes possible for the event of choice/ Number of total outcomes.

Since there are total 4 outcomes and 2 are for female, the possibilities of a girl are 2/4 = 1/2

Question 2:- Probability of having a fifth child as a girl:

Now in the above question, we determined that the probability of having a female child is 1/2.

That means if we talk about events in single instances, it will always be same. According to the question the couple already has 4 sons and since they already have them, that will not change the probability of having a girl. It will always remain the same, i.e, 1/2.

So the answer to this question is also 1/2.

Question 3:-

Haemophilia is a sex-linked disease and also a recessive disorder. Now in humans, Females have 2 X chromosomes and males have 1 X chromosome and 1 Y chromosome.

This, in terms of sex-linked recessive disorder, means that Females have to be homozygous for the recessive disease-causing allele while since males have only 1 X chromosome, they will never be a carrier if they do have the disease-causing allele.

So to answer questions related to these, few things to notice are:-

X^H implies the X chromosome with Haemophilia allele while X and Y are normal healthy chromosomes. Below I will write the genotype with a space between the two chromosomes. While writing them down yourself, do not include the space and write them like XX, not X X.

a) The woman who does not have haemophilia:-

Genotype - X X

Explanation:- Since the woman does not have haemophilia, we have to assume that she is completely healthy and is also not a carrier as carriers should never be assumed unless they are specifically are mentioned. So both her X chromosomes are normal and healthy.

b) The woman with haemophilia:-

Genotype: X^H X^H

Explanation:- This woman has haemophilia and since a woman has 2 X-chromosomes, this means that both her X chromosomes carry the allele for haemophilia. That makes her genotype X^HX^H.

c) A woman who is a carrier for haemophilia:-

Genotype : X X^H

Explanation:- This woman is a carrier of haemophilia and as I mentioned earlier, Haemophilia is a recessive disorder and if a woman have only one allele for haemophilia and 1 allele for normal protein, she is a carrier. Carriers do not have the disease themselves but they can pass on the diseased allele to their children.

d) Man who has Haemophilia:-

Genotype: X^H Y

Explanation:- As stated before, men only have 1 X chromosome while the other chromosome is Y chromosome. So if any man inherits a single X chromosome with haemophiliac allele, they will have haemophilia. Men are never carriers for a sex-linked disorder.

e) Man who does not have haemophilia:-

Genotype: X Y

Explanation:-

As men can never be the carrier due to them having only 1 X chromosome, a healthy man will also have a normal healthy chromosome.

Question 4:-

As the question asked, male is normal while the female is heterozygous. Heterozygous and carriers mean the same thing when we are talking about genetic conditions which are recessive.

So if we perform the Punnett, we get the following results for the phenotypes:-

Out of possible 4 children

a) 1 Female child will be a carrier

b) 1 Male child will be a haemophiliac

c) 1 Male and 1 Female child will be healthy.

Question 5:-

Carrier female with haemophiliac male:-

If we perform Punnett square for a carrier female with haemophiliac male, we get the above results of which I have attached an image.

So out of 4 possible combinations, there will be 2 children with haemophilia and out of these 2 children, 1 will be a male and 1 will be female.