One parent is heterozygous and one is homozygous dominant for a gene that causes DIDS. These two parents have children (F1) The alleles for this gene are S (dominant) and s(recessive). To have this disease, the child has to be homozygous recessive (ss). What are the chances of a child in the F1 generation having a child (f2) that has the disease if the second parent in the F1 generation (from outside the family) is heterozygous for SIDS?

Question

One parent is heterozygous and one is homozygous dominant for a gene that causes DIDS. These two parents have children (F1) The alleles for this gene are S (dominant) and s(recessive). To have this disease, the child has to be homozygous recessive (ss). What are the chances of a child in the F1 generation having a child (f2) that has the disease if the second parent in the F1 generation (from outside the family) is heterozygous for SIDS?

Answer 1

Your information is contradictory. First, you say, "One parent is heterozygous and one is homozygous dominant for a gene that causes DIDS." This leads me to think that SIDS is dominant.

Later you say, "To have this disease, the child has to be homozygous recessive (ss).", which indicates that it is recessive. Which is it?

Assuming that it is recessive — one parent has SS and the other has Ss, half of the children will have Ss, but none will have SIDS. If either from this half marries an Ss, there will be a 1/4 probability of having an ss with SIDS. The probability of both/all events occurring is found by multiplying the probability of the individual events.

I hope this helps. If not, repost with a clearer definition of the problem. Thanks for asking.

Answer 2

well one parent is heterozygous (Ss) and the other is homozygous dominant (SS)They will have children: SS Ss SS and Ss.What are the chances of those children having children with SIDS with a parent who is heterozygous (Ss)? Does that help because I got .625% even through I don't really understand the problem.

Answer 3

To determine the chances of a child in the F2 generation having the disease, we need to consider the genotypes of both parents in the F1 generation and the possible combinations of alleles that can be passed on to the F2 generation.

In this scenario, one parent in the F1 generation is homozygous dominant (SS) while the other parent is heterozygous (Ss). The alleles for the gene causing DIDS are S (dominant) and s (recessive), and for a child to have the disease, they need to be homozygous recessive (ss).

Let's break down the possible genotypes in the F2 generation:

1. SS x Ss:
- In this case, all offspring will have the genotype Ss, and none will have the disease (ss genotype). Therefore, the chance of a child in the F2 generation having the disease is 0%.

2. SS x ss:
- In this case, all offspring will have the genotype Ss, and none will have the disease (ss genotype). Therefore, the chance of a child in the F2 generation having the disease is 0%.

3. Ss x Ss:
- In this case, there is a 25% chance of a child having the ss genotype and thus having the disease. This is because the possible combinations are SS (25%), Ss (50%), and ss (25%).

4. Ss x ss:
- In this case, there is a 50% chance of a child having the ss genotype and thus having the disease. This is because the possible combinations are Ss (50%) and ss (50%).

So, overall, the chances of a child in the F2 generation having the disease would be the combined probabilities from cases 3 and 4. Adding the probabilities together, we get:

0.25 + 0.5 = 0.75

Therefore, the chances of a child in the F2 generation having the disease when the second parent in the F1 generation is heterozygous for DIDS is 75%.