Draw a Punnett square to represent the mating of a tall heterozygous pea plant w
ID: 191528 • Letter: D
Question
Draw a Punnett square to represent the mating of a tall heterozygous pea plant with a short pea plant. What phenotype ratios do you expect to find in their offspring? Instructions: Show your work and circle your final answers. 1. t Tt tt 2. You are growing 100 snapdragon plants in your garden. You have 68 red flowers, of which 42 are homozygous dominant and 26 are heterozygous. The other 32 flowers are white (homozygous recessive). What are the allele frequencies of R (red) and r (white) in the population? Note: do not assume Hardy-Weinberg equilibrium Suppose that a single gene controls the color of a species of beetle living on an island, such that green and g brown. You find 36 green beetles on the island and 64 brown beetles. Assuming Hardy-Weinberg equilibrium, what is the frequency of the brown allele in this population? 3. 4. How many heterozygotes are present in the island beetle population from Question 3? 5. In another garden, you have another 100 snapdragon plants. In this population, 20 are hom are heterozygous, and 60 are homozygous recessive for flower color. Is this p in Hardy-Weinberg equilibrium? How can you tell? (Hint: compare the actual genotype freq to the predicted genotype frequencies under HW.)Explanation / Answer
Answer:
1). Tt (tall heterozygous) x tt (short)
t
T
Tt (tall)
t
tt (short)
Tall : Short = 1 : 1
2).
Homozygous red = PP = 42 = 84 P alleles
Heterozygous red = Pp = 26 = 26 P alleles & 26 p Alleles
Homozygous white = pp = 32 = 64 p alleles
Total P alleles = 84+26= 110
Total p alleles = 26+64 = 90
Total alleles = 200
The frequency of allele P = R = 110/200= 0.55
The frequency of allele p = r = 90/200= 0.45
3).
The frequency of green beeteles = (gg) = 36/100 = 0.36
The frequency of green allele = g = 0.6
G+g= 1
Like p+q = 1
The frequency of brown allele = G = 1-g = 1-0.6 = 0.4
4). Heterozygous frequency = 2Gg= 2 * 0.4 * 0.6= 0.48 = 48%
5).
PP = homozygous dominant = 20/100 = 0.2
Pp = Heterozygous = 20/100 = 0.2
pp = Recessive = 60/100 = 0.6
Hardy-Weinberg equilibrium = p^2 + 2pq + q^2 = 1
0.2 + 0.2 + 0.6 = 1
So the population is in Hardy-Weinberg equilibrium.
t
T
Tt (tall)
t
tt (short)
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