Suppose that allele A is dominant to allele a, that genotypes AA and Aa produce phenotype B, and that genotype aa produces
phenotype C
If the frequency of allele A in a certain population is 0.6 (or 60%) and the frequency of allele a is 0.4 (or 40%), estimate the
frequencies of genotypes AA, AA, and aa. Then, determine the frequencies of phenotypes B and C.
The frequency of genotype AA is about
The frequency of genotype Aa is about
The frequency of genotype aa is about
The frequency of phenotype B is about
The frequency of phenotype C is about
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Answers:

The frequency of genotype AA is about 0.36

The frequency of genotype Aa is about 0.48

The frequency of genotype aa is about 0.16

The frequency of phenotype B is about 0.84

The frequency of phenotype C is about 0.16

Explanation:

Allele A is 60% of the population, so the chance of getting one of them is 60% or 0.60. Getting two of them in a row is a probability of 0.60*0.60 = 0.36 = 36%

The allele a's frequency is 0.4. Getting A then a would have a frequency of 0.6*0.4 = 0.24. Since Aa is the same as aA, we basically need to double 0.24 to get 0.48 to account for both scenarios of this genotype.

Getting two 'a's in a row would yield 0.4*0.4 = 0.16

Phenotype B occurs when the genotype is AA or Aa

So the probability of this happening is 0.36+0.48 = 0.84

You just add the probabilities of genotypes AA and Aa as calculated in the section above.

The probability of phenotype C is the complement of phenotype B's probability, so 1-0.84= 0.16 are the chances of randomly getting phenotype C. Note that this is the same as the probability of getting genotype aa, as this genotype directly corresponds to the phenotype C.

Side note: it might help to make a dihybrid cross to visually keep track of all the frequencies.

acceleration = 18/3 = 6 m/s per second.

hope i

-lavira: )

a. 35°c

b. 25°c

c. 10°c

d. 40°c