# Using the day 7 data. what is the frequency of the cg allele (p)?

a. Use the observed genotype frequencies from the day 7 data to calculate the frequencies of the CG allele (p) and CY allele (q). (Remember that the frequency of an allele in a gene pool is the number of copies of that allele divided by the total number of copies of all alleles at that locus.

b. Next use Hardy Weinberg equation (p^2+2pg+q^2=1) to calculate expected frequencies of genotypes CGCG, CGCY, CYCY for a population in Hardy Weinberg equilibrium.

c. Calculate observed frequencies of genotypes CGCG, CGCY, CYCY at day 7. (Observed frequency of a genotype in a gene pool is the number of individuals with that genotype divided by total number of individuals). Compare these frequencies to the expected frequencies calculated in step 2. Is the seedling population in Hardy Weinberg equilibrium at day 7, or is evolution occurring? Explain reasoning and identify which genotypes appear to be selected for or against.

d. Calculate the observed frequencies of genotypes CGCG, CGCY, and CYCY at day 21. Compare frequencies to expected frequencies calculated in step b and observed frequencies at day 7. Is seedling population in Hardy Weinberg equilibrium at day 21 or is evolution occurring? Explain reasoning and identify which genotypes appear to be selected for or against.

e. Homozygous CYCY individuals cannot produce chlorophyll. The ability to photosynthesize becomes more critical as seedlings age and begin to exhaust the supply of food stored in seed from which they emerged. Develop a hypothesis that explains data for days 7 and 21. Based on this hypothesis predict how frequencies of CG and CY alleles will change beyond day 21.

a) Each individual has two alleles, so the total number of alleles at day 7 is 216 × 2 = 432. To calculate the frequency of the CG allele, here each of the 49 individuals of genotype CGCG has two CG alleles, and each of the 111 individuals of genotype CGCY has one CG allele. The 56 individuals of genotype CYCY have zero CG alleles. Thus, the frequency of the CG allele (p) is
p=(2×49)+(1×111)+(0×56)432=0.48

You can use the same procedure that you used to calculate p to calculate q. However, an easier way to calculate q is to remember that p + q = 1. Since you know that p = 0.484, you can calculate that q = 1 - p = 0.516.

b)  The expected frequency of genotype CGCG is p 2 = 0.484 × 0.484 = 0.234.
The expected frequency of genotype CGCY is 2pq = 2 × 0.484 × 0.516 = 0.499.
The expected frequency of genotype CYCY is q 2 = 0.516 × 0.516 = 0.266

c)  The observed frequency of genotype CGCG is 49÷216 = 0.227.
The observed frequency of genotype CGCY is 111÷216 = 0.514.
The observed frequency of genotype CYCY is 56÷216 = 0.259

The seedling population is in equilibrium at Day 7. The expected and observed genotypic frequencies for each seedling are similar. At Day 7, there is no particular allele that is being selected for or against.

d)  The observed frequency of genotype CGCG is 47÷173 = 0.272.

The observed frequency of genotype CGCY is 106÷173 = 0.613.

The observed frequency of genotype CYCY is 20÷173 = 0.116.

The data suggests that the seedling population is evolving at Day 21. The allele frequencies have changed from Day 7 to Day 21. The C Y C Y genotype is being selected against and the C G C Y is being selected for.

e)   The C Y C Y do not produce chlorophyll. As the plant begins to grow and develop, it cannot photosynthesize due to the lack of chlorophyll pigment. The plant is therefore unable to produce glucose for energy. Initially the plant is relying on stored sugar for growth and develop, but between Day 7 and Day 21 the plant begins to suffer from a lack of energy and as a result, the number of surviving plants each day decreases over time. Beyond Day 21, one would expect to see further increase of the CG allele (selected for), and a decrease in the CY allele (selected against)

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