Genetic probability Question

C

its C 0.05
disease prevalence 1% = 1/100 = q2
allele frequency = q = 1/10
when disease frequency is less then 1/100 p(normal allele frequency) is considered 1
so....carrier frequency will be 2pq = 2q = 1/5
for AR disease , father must be carrier to transmit disease to child...
probability that women will marry carrier man is 1/5
probability that women will transmit her diseased allele to child is 1/2
probability that man will transmit his diseased allele to child is 1/2
now just multiply 1/5*1/2*1/2 = 1/20 = 0.05

my solution

may i suggest another solution

since 1% of that population is homozygous normal or diseased
if the woman marries a homozygous normal from that 1% she wont get a diseased homozygous baby
if she marries a homozygous diseased person from that same 1% she will get a homozygous diseased baby but for our calculation purpose we are going to neglect that homozygous diseased population since it will be negligible as it is an AR disease and it will be a small fraction of that 1%

so the probabability of that woman marrying a heterozygous carrier is 99% or 0.9
the probability of her having a homozygous diseased baby if she marries a heterozygous carrier = 1/4 or 0.25
the probability of husband having a homozygous diseased baby if he marries a heterozygous carrier = 1/4 or 0.25
multiply all the probabilities
0.9 x 0.25 x 0.25 = 0.05

The question is asking you this -

Probability of a homozygous child if someone "from the population" marries a known carrier female. Let's call her Miss X.

Now by population, it means 2 possibilities -

Situation 1 - Another carrier from the population marries Miss X. The probability of this event happening is - 2pq = 2x0.9x0.1 = 0.18 (Note: the square of q = 0.01, hence 1=0.1 and p=1-0.1 = 0.9. We can't ignore p in the equation because the prevalence is high.)
Now if this mating occurs, probability of having a homozygous child = 0.25 (draw a punnet square if you please.)
Hence combining the probabilities, final probability from situation 1 = 0.18x0.25 =0.45.

Situation 2 -A homozygous diseased male marries Miss X. Note that the question doesn't suggest that the female will only marry a phenotypically normal guy. The probability of this event happening is 0.01 (frequency of homozygotes is given in the question as 1%).
If this mating occurs, probability of having a homozygous child = 0.5.
Combined probability from situation 2 = 0.05.

Now to calculate the final answer, we must understand that EITHER of the 2 situations would give us our homozygous child. In these scenarios, the probabilities are to be added (not multiplied.)

So combining situation 1 and 2, final probability = 0.45 + 0.05 = 0.5

I know this is confusing and we probably won't be tested such complicated applications but once you read it slowly and refer to your genetics book, you would get it. I think a similar questions can be found in the Kaplan qbank. Hope that helped.