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Anonymous
Anonymous asked in Science & MathematicsBiology · 9 years ago

64% of a population is able to taste a chemical called PTC. Tasters are either homozygous dominant or?

64% of a population is able to taste a chemical called PTC. Tasters are either homozygous dominant or heterozygous for the trait. For the parts that follow, show all formulas, calculations and statements:

a) Calculate the percentage of the population that is Unable to taste PTC ?

b) Calculate the percentage of the population that is heterozygous for the trait?

c) Calculate the percentage of the population that has the recessive allele ? (Hint: heterozygous + homozygous recessive.)

2 Answers

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  • Anonymous
    9 years ago
    Favorite Answer

    Using the Hardy-Weinberg equations:

    f(non tasters) = q^2 = 0.36 a), so q = 0.6 and p=0.4

    b) f(heterozygotes) = 2pq = 0.6*0.4*2 = 0.48

    c) 0.48 + 0.36 = 0.84 = 84%

  • 9 years ago

    64% = DD (must carry a dominant allele)

    so D = square root of 64% (0.64) is 0.8

    d + D = 1 d = recessive

    1 - 0.8 = 0.2 carry recessive allele (d)

    So dd (people that cant taste it as they must carry a dominant allele, and so no dominant allele = cant taste it) is 0.2 squared = 0.04 (4%) homozygous recessive. That is the answer to A.

    Heterozygous.

    You must use the equation:

    q^2 + p^2 + 2pq = 1

    2pq is for the heterozygous.

    You know that the d (recessive) is 0.2 and D (dominant) is 0.8

    So heterozygous is 2pq so, (0.2 x 0.8) x 2 = 0.32 = 32% (heterozygous). This is the answer to B.

    You know this is correct because it must add up to 1 (or 100%). 32% is heterozygous, 4% is homozygous recessive and 64% is homozygous dominant = 100%

    Wow this took LONG! Points please!

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