HELP with Hardy-Weinberg (NBME Form 3)

[02115]

---

Members don't see this ad.

I cannot for the life of me figure out how to apply the Hardy-Weinberg in this case or maybe my math is just so bad:

In a clinical study 100 subjects are examined for a particular genetic alteration hypothesized to be related to cancer. The resulst are shown below:

GENOTYPE (# Controls)
AA (30)
AB (25)
BB (45)

If A is one allele and B is the other, which is the frequency of the B allele?

A) 25/200
B) 45/200
C) 50/200
D) 90/200
E) 115/200

So if I understand H-W correctly, psquare = .3, qsquare = .45 and 2pq = .25??

I think I am completely going up the wrong tree with this approach as my rudimentary math skills can't seem to figure this out! Any help you could provide would be greatly appreciated!

 

[keeping-it-real]

I cannot for the life of me figure out how to apply the Hardy-Weinberg in this case or maybe my math is just so bad:

In a clinical study 100 subjects are examined for a particular genetic alteration hypothesized to be related to cancer. The resulst are shown below:

GENOTYPE (# Controls)
AA (30)
AB (25)
BB (45)

If A is one allele and B is the other, which is the frequency of the B allele?

A) 25/200
B) 45/200
C) 50/200
D) 90/200
E) 115/200

So if I understand H-W correctly, psquare = .3, qsquare = .45 and 2pq = .25??

I think I am completely going up the wrong tree with this approach as my rudimentary math skills can't seem to figure this out! Any help you could provide would be greatly appreciated!

i threw hardy-weinberg out the window a while ago. i just use the 'rational person' approach 😉

the question basic is looking for how many B alleles out of all posible alleles.

AA (30) = 0 B alleles; 60 A alleles
AB (25) = 25 B alleles; 25 A alleles
BB (45) = 90 B allels; 0 A alleles

B alleles = 90 + 25 = 115

Total alleles = 60 + 25 + 25 + 90 = 200

answer = 115/200... at least that's what i think it should be. anyone else agree/disagree?

 

[FutureERDoc]

That's right-- I get 115/200 also. It is not really a H-W problem, just a matter of calculating how many B alleles out of the total alleles given. Since there are 100 people, each with 2 alleles, you are calculating a frequency out of 200 (2 alelles each x 100 people). Then you simply have to add up how many B alleles are in the group. The heterozygotes (AB people) have only 1 B allele so 1 allele x 25 people = 25 B alleles total for that group. The homozygote B people have 2 B alleles each, so 2 alleles x 45 people = 90. Then you add all the B alleles from these groups together 90+25 = 115. So the answer is 115/200.

 

[lord_jeebus]

anyone else agree/disagree?

agree

 

[02115]

I can't believe I am such an idiot-- I spent like an hour trying to figure this out using H-W, and to no avail. Thank you so much!!!