Hardy Weinberg problem

  • Thread starter colton4286
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  • #1
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Hardy Weinberg problem plz help

If 4% of a population in Hardy Weinberg equilibrium expresses a recessive trait, what is the probability that the offspring of 2 individuals who do not express the trait will express it?

I did what I could and got an answer. My work is below. Could you please check my work and comment? Thanks.

rr = expresses recessive trait = 4% = 4/100
Rr, RR = doesn't express recessive trait = 100-4= 96% = 96/100

q = 2(4) + 96 = 104/200 = 0.52

The punnett square Rr X Rr can only result in offspring who will express the recessive trait so
rr would equal 0.52*o.52= 0.270
 

Answers and Replies

  • #2
tiny-tim
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q = 2(4) + 96 = 104/200 = 0.52

The punnett square Rr X Rr can only result in offspring who will express the recessive trait so
rr would equal 0.52*o.52= 0.270
Hi colton4286! :smile:

I don't understand what you've done here … why did you add 8? :confused:

Hint: you know RrxRr is the only possiblity that can produce rr. So …

i) what is P(rr|RrxRr)?
ii) what is P(RrxRr|(not rr)x(not rr))? :smile:
 
  • #3
epenguin
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I would from the HW law work out the r allele frequency
The punnett square Rr X Rr can only result in offspring who will express the recessive trait so
I confess it is the first time I met the word punnett in this context, but I looked it up and that statement I think is just not true, so start again.


rr would equal 0.52*o.52= 0.270
For all quantitative calculations I strongly recommend the habit of plausibility, qualitative (e.g. something > or < something else) and when possible order-of-magnitude checks, as we do not have such an inbuilt instinct for just formulae. It will help and prevent persisting with mistakes when you realise whether a result is reasonable or not.

Here you know (are given) the frequency of expressing the trait when you know nothing about parents.
Then, when you do know that the parents do not express the trait, that shifts the frequency you expect in the offspring - in what direction?

The trait is not all that frequent, so the r gene frequency is here not what you call rare but not all that high. So is its expression in your case likely to be rare, not all that different from when you don't know the parents, or pretty frequent as you have concluded?
 

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