Plant Chapter 14 Concept Check

Explanation
According to the law of independent assortment, genes for different traits segregate independently during the formation of gametes. In this case, the plants are heterozygous for flower position (Aa) and stem length (Tt). When these plants self-pollinate, there are four possible combinations of alleles in the offspring: AT, At, aT, and at. Among these combinations, the offspring with the genotype aatt would have terminal flowers and be dwarf. Since there are four possible combinations and each has an equal chance of occurring, 25% of the offspring (or 25 out of 100) would be predicted to have terminal flowers and be dwarf.

Explanation
A pea plant heterozygous for seed color, seed shape, and pod shape (YyRrIi) can produce eight different gametes, which are YRI, YRi, YrI, Yri, yRI, rRi, yrI, and yri. To predict the offspring of a self-pollination of this "trihybrid," a Punnett square would need 8 rows and 8 columns. This Punnett square would have spaces for the 64 possible unions of gametes in the offspring.

Explanation
In a cross between CC and Cc, the dominant allele C will always be expressed in the offspring. Therefore, half of the offspring will be homozygous dominant (CC) and the other half will be heterozygous (Cc). Since the recessive allele c is masked by the dominant allele, there will be no homozygous recessive offspring (cc) in this cross.

Explanation
The given genotypes of all possible offspring from the cross are BBDD, BbDD, BBDd, and BbDd. Since the genes assort independently, the chance of each genotype occurring is equal. Therefore, the probability of each genotype occurring is 1/4 or 25%.

Explanation
The fraction of the offspring predicted to be homozygous recessive for at least two of the three characters can be calculated by adding up the frequencies of the genotypes that fulfill this condition. In this case, the genotypes that fulfill the condition are ppyyIi, ppYyii, Ppyyii, ppYYii, and ppyyii. The frequencies of these genotypes are 1/16, 2/16, 1/16, 1/16, and 1/16 respectively. Adding up these frequencies gives a total of 6/16, which simplifies to 3/8. Therefore, the fraction of the offspring predicted to be homozygous recessive for at least two of the three characters is 3/8.

Explanation
Incomplete dominance describes the relationship between two alleles of a single gene, where neither allele is completely dominant over the other. In incomplete dominance, a heterozygous individual displays an intermediate phenotype that is a blend of the two homozygous phenotypes. On the other hand, epistasis relates to the genetic relationship between two genes. It occurs when the expression of one gene masks or modifies the expression of another gene. Epistasis can result in the suppression or alteration of a trait controlled by another gene.

Explanation
When a man with type AB blood marries a woman with type O blood, their children can inherit either type A or type B blood from the father, and they can only inherit type O blood from the mother. Therefore, it is expected that half of their children will have type A blood and the other half will have type B blood.

Explanation
The simplest explanation for the inheritance of these colors in chickens is incomplete dominance. This means that neither the gray allele nor the black allele is completely dominant over the other, resulting in a blending of the two colors in the offspring. The phenotype of the heterozygotes, which have one gray allele and one black allele, is gray. Therefore, in a cross between a gray rooster and a black hen, we would expect to see gray and black offspring in equal numbers.

Explanation
Beth and Tom each have a sibling with cystic fibrosis, indicating that they are both carriers of the recessive allele for the disease. Since neither Beth nor Tom has cystic fibrosis, they must each have a 2/3 chance of being carriers. If they are both carriers, there is a 1/9 chance that they will have a child with cystic fibrosis. If a test reveals that Tom is a carrier but Beth is not, there is a 1/4 chance that they will produce a child with cystic fibrosis.

Explanation
Joan's genotype is Dd because the allele for polydactyly is dominant to the allele for five digits per appendage. Because Joan's father does not have polydactyly, his genotype must be dd, therefore Joan, who does have the trait, must be heterozygous.