man blood type is determined by three alleles IA, IB, and IO. The alleles IA and IB are co-dominant to each other, and both are dominant to IO. Within a large, randomly mating population (540,000 individuals), the frequencies for the blood type alleles are 0.3 for the IA allele; 0.6 for IO, and 0.1 for the IB allele.
a. Calculate the expected numbers of people in the population having each blood type A, B, AB, and O.
b. Determine the percentage of type B people that are heterozygotes (IBIO)
Flower diameter in sunflowers is a quantitative trait. A plant with 6-cm flowers, from a highly inbred strain, is crossed to a plant with 30-cm flowers, also from a highly inbred strain. The F1 have 18-cm flowers. F1 × F1 crosses yield F2 plants with flowers ranging from 6 to 30 cm in diameter, in approximately 4-cm intervals (6, 10, 14, 18, 22, 26, 30).
An 18-cm F1 plant is crossed to a 6-cm plant. What is the probability of an offspring with one additive allele, if all genes that influence this trait are unlinked?
In Drosophila, segmentation genes function in a sequential manner in the following order:
a. gap, segment-polarity, pair-rule.
b. transdeterminal, gap, pair-rule.
c. segmentational, helical, spherical.
d. pair-rule, transdeterminal, gap.
e. gap, pair-rule, segment-polarity.
a. Assuming the population is in H-W equilibrium, let p = the dominant purple allele (C), and q = the recessive allele (c). Applying the H-W equation, the frequency of the pink allele = q^2 = 153/1000 = 0.153; and q = (0.153)^1/2 = 0.39. Because p + q = 1, p = 1 - q = 0.61.
b. If the population is in H-W equilibrium, p^2 + 2pq + q^2 = 1.
For the heterozygotes, 2pq = 2(0.61)(0.39) = 0.476. Therefore, 47.6%, or (0.476)(1000) = 476 of all the purple flowered plants in this population are expected to be heterozygotes (Cc), and consequently, (p^2)(1000) = (0.61)^2 (1000) = 371 plants are expected to be
homozygous dominants (CC)