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Topic: Punnet Square Question (Mendel)  (Read 17732 times)

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Offline Bekah

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Punnet Square Question (Mendel)
« on: December 08, 2007, 11:10:45 PM »
Huntington's disease is caused by a dominant allele. If one of your parents has the disease, what is the probability that you too will have the disease?

I'm not sure on this. I think either 1/4th or 1/2?

Offline Padfoot

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Re: Punnet Square Question (Mendel)
« Reply #1 on: December 08, 2007, 11:29:06 PM »
It depends on whether the parent with the desease is a heterozygote or a homozygote.

Were you told?


Offline Bekah

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Re: Punnet Square Question (Mendel)
« Reply #2 on: December 08, 2007, 11:37:26 PM »
No, the question doesn't say.  :-[

Offline Borek

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Re: Punnet Square Question (Mendel)
« Reply #3 on: December 09, 2007, 04:05:13 AM »
Neither IMHO. I would go for 5/8.

One of your parents is hh while the second is either Hh or HH. You have to write down all 8 possible outcomes.
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Offline Padfoot

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Re: Punnet Square Question (Mendel)
« Reply #4 on: December 09, 2007, 06:01:42 AM »
One of your parents is hh while the second is either Hh or HH. You have to write down all 8 possible outcomes.
If we're assuming that there is equal probability of both these genotypes, then shouldn't it be 75% (6/8) - half way between 50 and 100% since:

HH  x  hh
gives Hh (1 possible progeny)
so chance is 100% here

Hh  x hh
gives Hh,hh (2 possible progeny, equal chance of both)
so chance is 50% here

Offline Borek

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Re: Punnet Square Question (Mendel)
« Reply #5 on: December 09, 2007, 08:35:51 AM »
Sorry, my mistake. 6/8 it is. Typo when writing down combinations  :-\
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Offline Yggdrasil

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Re: Punnet Square Question (Mendel)
« Reply #6 on: December 09, 2007, 12:24:17 PM »
I wouldn't assume an equal prior probability of Hh and HH.  According to Hardy-Weinberg equilibrium, if the prevalence of the H gene is q, then the probability of an HH genotype, p(HH) is q2 and the probability of Hh, p(Hh), is 2q(1-q).

Furthermore, assuming that H is a sufficiently rare gene, q is small.  Since p(HH) ~ O(q2) and p(Hh) ~ O(q), p(Hh) >> p(HH) and you can generally ignore the contributions of homozygotes.

Offline Bekah

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Re: Punnet Square Question (Mendel)
« Reply #7 on: December 10, 2007, 09:56:42 PM »
well the question is multiple choice and gave 25%, 50%, 75% or 100% as possible answers so I am assuming I'm safe with 75%.

Offline Yggdrasil

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Re: Punnet Square Question (Mendel)
« Reply #8 on: December 10, 2007, 11:30:45 PM »
I'd say its 50%.  Individuals that are homozygous for Huntington's disease are very rare, so its safe to assume that the afflicted parent is heterozygous for the disease allele.

Offline Bekah

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Re: Punnet Square Question (Mendel)
« Reply #9 on: December 11, 2007, 01:01:10 AM »
lol...okay thats what I thought orginally...lol  :)

Offline Borek

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Re: Punnet Square Question (Mendel)
« Reply #10 on: December 11, 2007, 03:12:22 AM »
I'd say its 50%.  Individuals that are homozygous for Huntington's disease are very rare, so its safe to assume that the afflicted parent is heterozygous for the disease allele.

The question isn't clear on that. Correct answer depends on what the asker had on his mind. 50% answer can be easily challenged by HH. IMHO 75% answer is easier to defend - if you were to use more complicated approach you should be given q, if you are not, you can't ignore homozygote just because. But that's just my personal opinion, this is a question for mind-readers.
« Last Edit: December 11, 2007, 03:24:59 AM by Borek »
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Offline Yggdrasil

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Re: Punnet Square Question (Mendel)
« Reply #11 on: December 11, 2007, 12:01:54 PM »
What justification is there for a 50% probability of Hh and 50% probability of HH.  It's one of the most common logical fallacies in probability; because you have two possibility, people somehow assume that the probability of each situation must be 50%.

You can, however, justify saying that nearly all people with Huntington's disease are Hh.  We know that Huntington's disease is a rare genetic disorder.  Let's assume that Huntington's affects 1 person per 100 people (the real figure is about 1 person per 10,000 people of European descent).  Therefore, 99% of people have genotype hh.  Assuming that the population is in Hardy-Weinberg equilibrium (which is fairly valid since the HD phenotype doesn't show up until after one reaches reproductive age).  This means that:

p2 = 0.99, and
p = 0.99499

where p is the prevalence of the h allele in the population.  Therefore:

q = 1-p = 0.00501

Therefore, we can find the fraction of the population that are heterozygous, f(Hh), and the fraction of the population that are homozygotes, f(HH):

f(Hh) = 2pq = 2(0.99499)(0.00501) = 0.00997
f(HH) = q2 = (0.00501)2 = 2.51x10-5

So, among the people with Huntington's disease, the probability that an afflicted individual is heterozygous, f(Hh|Huntingtions), and the probability that an afflicted individual is homozgous, f(HH|Huntingtons), are:

f(Hh|Huntingtons) = 0.00997/.01 = 0.997
f(HH|Huntingtons) = 2.51x10-5/0.01 = 0.00251

In our example, 99.7% of the people with our "rare" disorder are heterozygotes and we can pretty safely ignore the population of homozygotes.

Offline Borek

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Re: Punnet Square Question (Mendel)
« Reply #12 on: December 11, 2007, 02:41:48 PM »
Ygg, I appreciate your effort in explaining your point. Trick is, nobody ever suggested you are incorrect - we just don't know what the teacher asks about. It is very likely that Huntington's disease is just an example used to ask for probabilities; it could be as well Hollow fang disease or something else made up. I have forwarded this question (exactly as worded) to a chemical education list. Both answers (50% and 75%) were listed as viable, and one of the comments made was:

Quote
One might say that the question writer did not think outside the box, but I think they just thought in the small box.

I think that's the best description of the problem.
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