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## PLEASE HELP! Genetics HW: Hardy-Weinburg Question
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## PLEASE HELP! Genetics HW: Hardy-Weinburg QuestionI would really, really appreciate some help on this question on this homework I have been trying sooo hard to get this stuff down it's just not happening! PLEASE..shed some light on this for me.
Question: Sickle-cell anemia is coded for by the S allele (i.e., S codes for the dominant trait and s for the recessive). If 19% of an equatorial population is born with sickle-cell anemia (ss), what percentage of the population will be more resistant to malaria? What percentage of the population would suffer from malaria, but not sickle-cell anemia? If malaria was removed from the environment, how would this influence the frequency of the recessive sick-cell allele? I really appreciate it if anyone responds...it is due tomorrow and this the hardest question so if I could see all the work from this one I am hoping i can do the rest! Thanks. - laghmani786
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**Posts:**6**Joined:**Tue May 06, 2008 12:18 pm
## Re: PLEASE HELP! Genetics HW: Hardy-Weinburg QuestionYou can solve it the hard way or the easy way, whichever you prefer. To use the Hardy-Weinberg Equation you need to find—directly or indirectly—the allele frequencies among the study population. Here, you’re assuming S is a simple dominant gene for the sickle cell trait and s is the wild-type recessive allele. What you’ve been given is the proportion of the genotypes that are homozygous recessive (ss) in the population. What you need to find is the sum of the proportions of genotypes that are homozygous (SS) and heterozyous (Ss) for the sickle cell trait. These will be the individuals with resistance to malaria.
s^2 = 0.19 is the genotype frequency for the recessive trait. From that, you can calculate s = sqrt(0.19) which is the recessive allele frequency. Under the assumptions of a Hardy-Weinberg equilibrium the probabilties of the allele frequencies have to equal unity. That would apply no matter how many alleles are in the system, but here they’ve kept it simple and restricted the discussion to 2 alleles. So: S + s = 1. You’ve already calculated s, so now you know S. At this point you know all you need to know to finish the calculation. The proportions of genotypes that are (SS) is S^2, and the proportion of genotypes that are heterozygous (Ss) is 2Ss. and the proportion of resistant individuals (anyone who carries the dominant sickle cell allele) is S^2 + 2Ss. But there’s an easier way: under the Hardy-Weinberg assumption the proportions of types of individuals also has to add to unity. The exact statement of the law depends on the number of alleles, but for this case it is the simplest form: S^2 + 2Ss + s^2 = 1. You’ve been given s^2. You need to find S^2 + 2Ss. You should work it out both ways as a check. You should get the same answer either way. - blcr11
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**Posts:**672**Joined:**Fri Mar 30, 2007 4:23 am
## Re: PLEASE HELP! Genetics HW: Hardy-Weinburg QuestionThank you...that does help a lot! I think it's the different variables that gets a little confusing especially since this is not something I have had a background in.
I guess my only other is applying to the application of the numbers back to the question. What will the number I get represent? The percentage more resistant to malaria? or the percentage that would suffer from malaria but not sickle-cell anemia? - laghmani786
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**Posts:**6**Joined:**Tue May 06, 2008 12:18 pm
## Re: PLEASE HELP! Genetics HW: Hardy-Weinburg QuestionS^2 + 2Ss is the proportion of genomes (individuals) in the population that have at least one S allele. Since S is presumed to be a simple dominant allele, anyone who has the S allele should be resistant to malaria.
- blcr11
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**Posts:**672**Joined:**Fri Mar 30, 2007 4:23 am
I may have misled you - not intentionally. The problem asks for the proportion who are resistant but don't have sickle cell disease. That will be 2Ss, the proportion of heterozygotes. Sorry, I read it as only resistance to malaria. Both SS and Ss individuals will be resistant to malaria, but SS individuals also have sickle cell disease while Ss individuals do not. Sorry if I threw you off. You are going to have to do at least part of it the long way so long as you need to know 2Ss.
- blcr11
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**Posts:**672**Joined:**Fri Mar 30, 2007 4:23 am
Umm ok...I think that should still be alright. So it would probably easier to do the whole thing the "long way" ?
- laghmani786
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**Posts:**6**Joined:**Tue May 06, 2008 12:18 pm
## Re: PLEASE HELP! Genetics HW: Hardy-Weinburg QuestionYou can easily calculate s (from s^2) and from S+s=1 get S, from which you can calculate S^2. The question is, how do you want to calculate 2Ss. You can plug in the values you have for S and s and calculate it directly, or you can find it from the other relation: S^2 + 2Ss + s^ = 1 without having to multiply 2xSxs. So long as you need to know the proportion of heterozygotes, you at least have to calculate s and S. If all you needed was the proportion of genotypes with at least one S allele, then you really didn't have to do much of anything because if S^2 + 2Ss + s^2 = 1 and you know that s^2 = 0.19 then S^2 + 2Ss = 1 - 0.19 and you'd be done. Unfortunately, I think you need to know 2Ss and not just the sum as I'd thought the first time I looked at the problem.
- blcr11
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**Posts:**672**Joined:**Fri Mar 30, 2007 4:23 am
That is true this is for a Human Variation class and grrr this is the only section I am having a lot of trouble with...or the only question just because there is a lot to account for...whereas the rest are pretty easy. And I will just take your word about the fly paper...this is frustrating as it is!
- laghmani786
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**Posts:**6**Joined:**Tue May 06, 2008 12:18 pm
## Re:
But I think Genetics is the most interesting subject in biology.. Or it just me who think that is true ? But why the number is not so good.. I mean 19%. It's hard for us to count the result manually - F4T32008
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**Posts:**25**Joined:**Tue May 06, 2008 4:58 pm
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