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1. ### Hardy-Weinberg problem?

I have no idea how to do this problem. Ha. See if any of this looks right: p + q = 1 (p + q) * (p + q) = p^2 + 2pq + q^2 = 1 Okay. Let p = freq(T) Let q = freq(t) then through Hardy-Weinberg equilibrium calculation: q^2 = (102/320) Gene...

1 Answers · Science & Mathematics · 18/12/2017

2. ### How does coevolution lead to nontasters of PTC?

...like that, so the pressure on bitter taste alleles is reduced. PTC non-tasters do still taste many bitter compounds, just not certain ...

1 Answers · Science & Mathematics · 14/05/2017

3. ### Inheritance percentage question?

A. Since those who cannot taste PTC are homozygous tt, the frequency of the tt genotype (q^2) = 65/215 = 0...

1 Answers · Science & Mathematics · 16/05/2016

4. ### How to answer these questions (biology)?

The ability to taste or not to taste is genetic. Tasting is a dominant phenotype but seems to be variable in its expressivity: some tasters are "supertasters," and are assumed to be those individuals who...

1 Answers · Science & Mathematics · 06/04/2016

5. ### in humans, the trait for the ability to taste phenylthiocarbamide (PTC) is dominant over that for the inability to taste PTC.?

If you're crossing BbTt x BbTt, the proportion of bbT_ would be 3/16. When doing a dihybrid cross, 9/16 are dominant for both traits, 3/16 are dominant for one and recessive for the other, 3/16 are the other way around, and 1/16 is...

1 Answers · Science & Mathematics · 15/11/2015

6. ### Calculate the frequency of the genotypes using Hardy-Weinberg?

Since 37/125 could not taste PTC, fr(tt) must be 37/125 = 0.2960. Therefore, fr(t) = sqrt...

1 Answers · Science & Mathematics · 08/03/2015

7. ### Hardy-Weinberg problem help?

...0.4 * 0.6 * 1092 = 524 (approximately) (answer) C) How many people can taste PTC? 64% * 1092 = 699 (approx) (answer) Check: This should be the same as (p...

1 Answers · Science & Mathematics · 02/10/2014

8. ### biology help Hardy Weinberg question? Calculating allele frequencies?

Okay. You have 150 tasters and 215-150 = 65 nontasters. q^2 = 65/215 = .302 q = 0.55 freq(t) = q = 0.55 (answer) freq(T) = p = 1-q = 0.45 (answer) Check that: p^2 + 2pq + q^2 = 1 .45^2 + 2 * .45*.55 + .55^2 = 1 check 215 * (.45^2 + 2 * .45*.55) = 150 check 215 * (.55^2) = 65 check

1 Answers · Science & Mathematics · 26/09/2014

9. ### You are a PTC taster, your brother is not, your father is, your mother is not?

Since it is a dominant trait you must be heterozygous. So if P=taster, your mother is pp, father is Pp (because your brother is a non-taster so pp) and so you are Pp.

1 Answers · Science & Mathematics · 25/02/2014

10. ### Are most human traits one allele or multiple alleles?

... alleles, but of several genes. Tongue-rolling, PTC-tasting, and Rh antigen presence are traits determined...

1 Answers · Science & Mathematics · 18/01/2014