# 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. Number of Seedlings Green Green-yellow Yellow Time (days) (C°C) (C C) 49 47 (CC Total 216 173 56 20 21 106

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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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