Bio Destroyer Genetics and HW questions need help

[Pennpaki]

Hey guys,

I was going over Bio from destroyer and I just can't figure out these two questions even after having read the descriptions.

" A certain autosomal recessive condition is seen in approximately 1 in every 50 people, but they are symptom free. If a male and female are both carriers of this gene, what is the probability that they have a diseased child?"

The answer is: 0.01%

I honestly have no idea how they got to this. They say each parent has 2% of a chance to carry the gene... but i thought each parent is a carrier? *confused*



Second Question
"The frequency for the allele for an autosomal recessive trait is 1 in 2500 in white Americans. What is the heterozygote frequency?"

The answer they gave is 4%. Can someone explain how to use HW here? I would reaallly appreciate it. Thanks

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[UndergradGuy7]

Second question if I remember, I think is wrong. He mixed up allele and trait. So his work and answer is wrong.

It should be 1/2500 = the q (but he put q^2 I think). It is q since he says that's the frequency of the allele.

So p + q = 1
1-q=p
p=0.9996

Heterozygote is 2pq = 2(0.9996)(4E-4) = 7.9968E-4 or 0.079968%

 

[lala05]

Second Question
"The frequency for the allele for an autosomal recessive trait is 1 in 2500 in white Americans. What is the heterozygote frequency?"

The answer they gave is 4%. Can someone explain how to use HW here? I would reaallly appreciate it. Thanks

Hardy Weinberg Equilibrium: p^2 + 2pq + q^2 = 1 AND p + q = 1
p^2 = frequency of homozygous dominant genotype, q^2 = freq of homozygous recessive, and 2pq = freq of heterozygotes
--> we are given q^2 = 1/2500 = 0.0004 so q = 0.02
--> since q + p = 1, then p = 1 - q --> p = 0.996
--> 2pq = (2)(0.996)(0.02) = 0.03984 x 100% = approximately 4%

 

[Pennpaki]

Thanks for all the responses.. I get the second question now but wat bout the 1st one?

" A certain autosomal recessive condition is seen in approximately 1 in every 50 people, but they are symptom free. If a male and female are both carriers of this gene, what is the probability that they have a diseased child?"

The answer is: 0.01%

 

[nainaaz]

Alrite reading the question properly..it says probablity of certain autosomal rcessive condition seen is 1/50=0.02 but are SYMPTOM FREE...meaning it has one normal dominant allele (heterozygous condition) Aa
so parents who are carries(Aa*Aa) will have (0.02*0.02)*1/4 probability of having a diseased child.

 

[Pennpaki]

Lets say dominant autosomal allele represents A
Recessive condition here represents recessive allele (a) and its probability is 1/50=0.02)

so for the diseased child recessive alleles has to be herited from each parent..(a+a)..0.02+0.02=0.04

For parents who are carriers(Aa), probability of autosomal recessive trait (aa) is 1/4 ( AA1:Aa2:aa1)

therefore 0.04/4=0.01

Hope it helps

hey Im sorry but I guess I'm not gettin the first part... You said that the probability two carriers have a diseased child is 1/4th which I understand.

But how did you get 0.04 ...as in why did you add the freq of a... do you think you/ anyone think they could elaborate further please? Thanks

 

[Pennpaki]

bump

 

[RCT PC CRN]

Part 1:

Step 1 - Find freq of carriers (heterozygous) = 2pq = 2*0.02*0.98 = 0.04 - this is your probability of being a carriers
Step 2 - Now out of those heterozygous carriers, getting recessive homozygous is 1/2*1/2
Step 3 - Answer is combining those 2 probability which is 0.04*1/2*1/2 = 0.01 %

Part 2:

q^2 = 1/2500 => q = 1/50 => p = 49/50

heterozygote frequency = 2pq = 2*49/50*1/50 = 1/25 = 0.04 or 4%