- Figure 8.9 You cannot be sure if the plant is homozygous or heterozygous as the data set is too small: by random chance, all three plants might have acquired only the dominant gene even if the recessive one is present.
- Figure 8.10 The possible genotypes are PpYY, PpYy, ppYY, and ppYy. The former two genotypes would result in plants with purple flowers and yellow peas, while the latter two genotypes would result in plants with white flowers with yellow peas, for a 1:1 ratio of each phenotype. You only need a 2 × 2 Punnett square (four squares total) to do this analysis because two of the alleles are homozygous.
- Figure 8.16 Half of the female offspring would be heterozygous (XWXw) with red eyes, and half would be homozygous recessive (XwXw) with white eyes. Half of the male offspring would be hemizygous dominant (XWY) with red eyes, and half would be hemizygous recessive (XwY) with white eyes.
- The garden pea has flowers that close tightly during self-pollination. This helps to prevent accidental or unintentional fertilizations that could have diminished the accuracy of Mendel’s data.
- The Punnett square would be 2 × 2 and will have T and T along the top and T and t along the left side. Clockwise from the top left, the genotypes listed within the boxes will be Tt, Tt, tt, and tt. The phenotypic ratio will be 1 tall:1 dwarf.
- The Punnett square will be 2 × 2 and will have T and t along the top and T and t along the left side. Clockwise from the top left, the genotypes listed within the boxes will be TT, Tt, Tt, and tt. The genotypic ratio will be 1TT:2Tt:1tt.
- No, males can only express color blindness and cannot carry it because an individual needs two X chromosomes to be a carrier.
- Yes this child could have come from these parents. The child would have inherited an i allele from each parent and for this to happen the type A parent had to have genotype IAi and the type b parent had to have genotype IBi.
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