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Ok so this is the first question.
1. In a randomly mating population of mice, 6 out of every  100 mice born have white fur, a recessive trait. Calculate the heterozygous frequency for the population.
So what I did was
6/100= .006
square root is .244
1-.244=.756
my answer is 75%



2. I have no idea how to even start so if some one could walk me through this I would greatly appreciate it.
A dominant allele, T, codes for the ability to taste the compound phenylthiocarbamide (PTC). People who are homozygous for the recessive allele, t, are unable to taste PTC. In a genetics class of 170 students, and 23 cannot taste the PTC. Calculate the expected frequencies of the T allele in the student population


Thanks for any help!!

1. According to hardy-weinberg principle p2+ 2pq+ q2 = 1
Here frequency of recessive allele,(q=recessive , p=dominant), q= square root of (6/100) = .244
So frequency of dominant allele p is (1-.244)= .756
Heterozygous condition means 2pq, So the frequency is 2*(.756)*(.244)= 0.368 or 36.8%

2. Taste blindness of PTC or Phenyl Thiocarbamide is a genetic trait.
People with TT & Tt can taste its sour taste. But people with tt can't.
Here (23/170)*100 % or 13.52% can't taste it. So frequency of t = square root of (13.52/100) = .135
Hence frequency of T is (1-.135)= .864

Avradeep got it right in the first one the answer is 36,8%

However the second problem has a flaw. The answer is that the T alele ( marked as p for dominant) has the frequency of 0,63 or 63%.So p=0,63.

The number of students N is
N=170
the number of resessive homozygotes is 23
the percentage of recessive homozygotes in your population is thus:
Y=23/170=0,135%.
To get the frequency of the recessive allele q according to HV equlibrium you need to square root your Y, in your case 0,135 and you will get 0,367. So your resessive q allele frequency is q=0,367.
To get the opposite frequency of the dominant allele wich you lack is simple. The sum of frequencys must build up to 1. So p + q = 1. You have q so to get p you need to subtract your q from 1 to get: p = 1 - q = 1 - 0,367 = 0,633.
So the freq. of the dominant allele p is 0,63

Hi Mephesto.. I don't understand why are saying that the percentage is .135% in case of the recessive allele.
For example if I say that 50 students out of 200 can't solve a question..what will be the percentage of students who can't solve it? Of course it is (50/200)*100%=25%
So please justify you are saying that the percentage of homozygous recessive allele Y=23/170=.135%
This method you have used is against the rules of maths. I am expecting a prompt reply.

Hello there

My apology. I was refereing to genotype frequency but I dubed it percentege not knowing that it can create confusion with that % symbol added.

It doesn't change the fact that your equation wrong. The square root of 13,5/100 is not 0,135.
 It's 0,367...

Yes..You are right ..I had done wrong calculations... Sorry for wrong answer...  :-$

Oh there is no need to be sorry...You explained the Hardy-Weinberg pretty well with your formulas and surely the original poster got the answer he was looking for. I am glad that there are people on this forum that posses knowledge regarding population genetics.
I would love to know your opinion about a topic I started on the Environmental and Conservation Biology subforum:
Can you help me choose the right population genetics and statistical method  (https://biology-forums.com/index.php?topic=332494.0)

I am working with relative allele frequencys there also. If you could spare some of your time and look in to it I would be very grateful. Cheers!