[Guest]

There is an effect that only shows when someone is homozygous recessive for the trait. This phenotype shows in one certain female, who has a child with a male who does NOT show the trait. not all females get this effect, and males can also get it too. The couple has a child that shows the effect [which is being homozygous recessive]. They have another child who also shows the effect. Then they have a THIRD child who shows the effect.

What WAS the probability before it was born that the first child would show the recessive phenotype?

What WAS the probability before IT was born that the second child would show the phenotype?

What WAS the probability before IT was born that the third child would show the phenotype?

This is the answer...

The male must have at least one dominant gene because he does not show the effect that the female does. The Female is completely recessive. Crossing these genotypes:

RR x rr / Rr x rr

RR x rr gives us ALL CHILDREN BEING Rr, not showing the trait. So, this means that if the father is RR, the probability of having a recessive child is 0%. However, if the father was Rr, then there is a chance of 50% to be recessive, and 50% to be Heterozygous: a 2:4 ratio. There are 8 possibilities for the joining of genes, 4 from the RR situation and 4 from the Rr situation. There are only 2 ways the child can be recessive completely, and out of 8 ways total, that's a 1/4 chance, or 25%.

For the next children, it is different. Now, because we know the first child showed the recessive phenotype, we know the father is Rr, because if he was RR, then NONE of the children would show the trait. The cross now looks like this:

Rr x rr ONLY!

there's a 50% chance for recessive, 50% chance heterozygous. There is a 50% chance the second child will show the trait.

THE SAME IS FOR THE THIRD CHILD. 50%.

1: 25% chance - 2: 50% chance - 3: 50% chance.

[Guest]

There is an effect that only shows when someone is homozygous recessive for the trait. This phenotype shows in one certain female, who has a child with a male who does NOT show the trait. not all females get this effect, and males can also get it too. The couple has a child that shows the effect [which is being homozygous recessive]. They have another child who also shows the effect. Then they have a THIRD child who shows the effect.

What WAS the probability before it was born that the first child would show the recessive phenotype?

What WAS the probability before IT was born that the second child would show the phenotype?

What WAS the probability before IT was born that the third child would show the phenotype?

This is the answer...

The male must have at least one dominant gene because he does not show the effect that the female does. The Female is completely recessive. Crossing these genotypes:

RR x rr / Rr x rr

RR x rr gives us ALL CHILDREN BEING Rr, not showing the trait. So, this means that if the father is RR, the probability of having a recessive child is 0%. However, if the father was Rr, then there is a chance of 50% to be recessive, and 50% to be Heterozygous: a 2:4 ratio. There are 8 possibilities for the joining of genes, 4 from the RR situation and 4 from the Rr situation. There are only 2 ways the child can be recessive completely, and out of 8 ways total, that's a 1/4 chance, or 25%.

For the next children, it is different. Now, because we know the first child showed the recessive phenotype, we know the father is Rr, because if he was RR, then NONE of the children would show the trait. The cross now looks like this:

Rr x rr ONLY!

there's a 50% chance for recessive, 50% chance heterozygous. There is a 50% chance the second child will show the trait.

THE SAME IS FOR THE THIRD CHILD. 50%.

1: 25% chance - 2: 50% chance - 3: 50% chance.

This seems to be the same question that was posted he other day, but more clear this time. The issue with your answer to the problem is that you fail to realize that regardless of what you know or do not know, the probability does not change. What you did to find the probability of child 1 being recessive in either situation of the father and averaged the 2. This is acceptable math if you are estimating the probability of the first child being recessive. You must be a poker player, you are doing the odds with other payers cards considered unknown. Remember, however, no matter what your estimate is there is always a real probability when it comes to genetics. There is also a probability of whether your estimate is correct or incorrect, in this situation it happened to be conveniently the same percentage as the answer

[Guest]

Correct. So while one can estimate the chance of having Huntington's disease (which is entirely genetic) based on the parents to be 50%, it is always a 0% or 100% chance (provided the patient lives long enough).

[Guest]

Since we can deduce that the dad must be Rr doesn't the first child still have a 50% chance? I guess I understand the 25% answer if per say, they went to a genetics counselor, then it would make sense. but that still doesn't change the fact that the dad is Rr so the first child will still have a 50% chance of getting the recessive phenotype no matter what.

[Guest]

Pardon my technicality, but many phenotypes are not controlled by a single allele, or are exhibited in 100% of the people with that allele. Nowhere in the question does it state that a homozygous recessive individual MUST have the trait, just that all of those with the trait are homozygous recessive. In this case, the probability of the three children having the phenotype might not be 50%.

Edited November 10, 2009 by spoonmang

[Guest]

Pardon my technicality, but many phenotypes are not controlled by a single allele, or are exhibited in 100% of the people with that allele. Nowhere in the question does it state that a homozygous recessive individual MUST have the trait, just that all of those with the trait are homozygous recessive. In this case, the probability of the three children having the phenotype might not be 50%.

The OP stated "There is an effect that only shows when someone is homozygous recessive for the trait"

This would mean that it is a monogenic trait or at least the single allele is the crux of expression. Good observation though.

[Guest]

is that all are 50%..

The first child proved that the male was Rr.

had the male been RR then the chances are 0%

knowing the female to be rr, then the only possible contribution is r

The male being RR would contribute R always resulting in Rr (0%)

The male being Rr could contribute R or r resulting in Rr or rr (50%)

The only way there would be a 25% is if both parents were Rr resulting is 4 possible results below

rr

Rr

rR

RR

Since we know the female to be rr the answer to the first child is both 100% & 0% at the same time.

So...

(100+0) / 2 = 50

see...