In the year 2500, five male space colonists and five female space colonists (all unrelated to each other) settle on an uninhabited earthlike planet in the andromeda galaxy. the colonists and their offspring randomly mate for generations. all ten of the original colonists had free earlobes, and two were heterozygous for that trait. the allele for free earlobes is dominant to the allele for attached earlobes. 67) which of these is closest to the allele frequency in the founding population?

a) 0.1 a, 0.9 a
b) 0.2 a, 0.8 a
c) 0.5 a, 0.5 a
d) 0.8 a, 0.2 a
e) 0.4 a, 0.6 a

A) 0.1 a, 0.9 A

Explanation:

According to the given information, all the individuals in the founder population had free earlobes. Since the allele for the free earlobes (A) is dominant over the one for the attached earlobes (a), the genotype of each of the homozygous dominant founder individual would be "AA". Humans are diploid and one individual has two alleles for each gene in its genome. Total number of alleles for the earlobe in the founder population = 10 x 2 = 20.

Frequency of dominant allele, A in the founder population = total number of homozygous dominant individuals x 2 + the total number of heterozygous genotype / total alleles for the earlobe in the population. There were 8 homozygous dominant and 2 heterozygous dominant individuals in the population.  

So, frequency of dominant allele = 8 x 2 + 2 / 20 = 16 + 2 / 20 = 18/20 = 0.9

Since p+q=1; frequency of recessive allele "a" = 1-p = 1-0.9 = 0.1

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not sure but try c two haploid cells

explanation: