Question 1

Assuming independent assortment at all loci, what is the probability that a cross between the following parents, AABbCc × AaBbCc, will produce an AaBbCc offspring?

Accepted Answer

The probability of Aa from AA x Aa is 1/2, Bb from Bb x Bb is 1/2, and Cc from Cc x Cc is 1/2. Multiplying these probabilities gives 1/2 * 1/2 * 1/2 = 1/8.

Question 2

In a cross of individuals who both possess this genotype, AaBbCc, what proportion of the offspring is predicted to have the genotype, AABBCC?

Accepted Answer

Each gene pair (Aa, Bb, Cc) segregates independently, and the probability of obtaining the homozygous dominant genotype (AA, BB, CC) for each pair is 1/4. Therefore, the combined probability for AABBCC is (1/4)^3 = 1/64.

Question 3

Pea plants produce either purple or white flowers with purple showing complete dominance. A gardener was given plants with purple flowers. Which of the following types of crosses would best allow her to determine the genotype of her plant in one generation?

Accepted Answer

A testcross involves crossing the plant in question with a homozygous recessive plant (white flowers in this case) to determine the genotype of the unknown. If any offspring show the recessive phenotype (white flowers), the unknown parent must be heterozygous. This method directly addresses the question of genotype in one generation.

Question 4

Two true-breeding stocks of pea plants are crossed. One parent has red, axial flowers, and the other has white, terminal flowers; all F1 individuals have red, axial flowers. Among the F2 offspring, what is the probability of producing plants with white axial flowers?

Accepted Answer

The traits follow Mendelian genetics, with red color (R) being dominant over white (r) and axial flowers (A) being dominant over terminal (a). The cross is RrAa x RrAa for the F1 generation. White axial flowers would be rrA_, which occurs with a probability of 1/4 (for rr) * 3/4 (for A_) = 3/16.

Question 5

When Mendel crossed yellow-seeded and green-seeded pea plants, all the offspring were yellow-seeded. Which of the following genotypic ratios was expected when Mendel crossed the F1 yellow-seeded plants with green-seeded plants?

Accepted Answer

When Mendel crossed F1 yellow-seeded plants (heterozygous for the seed color trait) with green-seeded plants (homozygous recessive), the expected genotypic ratio was 1:1, representing 1 heterozygous yellow-seeded to 1 homozygous green-seeded offspring.

Question 6

Which of the following phenotypes is an example of polygenic inheritance?

Accepted Answer

Skin pigmentation in humans is an example of polygenic inheritance, where multiple genes contribute to a single trait, resulting in a range of possible phenotypes.

Question 7

Black fur in mice (b) is dominant to brown fur (b). Short tails (S) are dominant to long tails (s). What fraction of the progeny of crosses BbSs × BBss will be expected to have black fur and long tails?

Accepted Answer

The cross BbSs × BBss results in offspring with genotypes: BBSs, BBss, BbSs, Bbss. All offspring will have black fur (BB or Bb). For tails, 1/2 will have long tails (ss). Therefore, 1/2 of the progeny will have black fur and long tails.

Question 8

A brown dog crossed with a yellow dog produced 12 brown offspring. A second brown individual was obtained. When the yellow dog was crossed with this animal, six brown and six yellow offspring were born. Which of the following best explains the results of the second cross?

Accepted Answer

The results of the second cross, where half of the offspring are brown and half are yellow, suggest that the yellow trait is recessive and the brown trait is dominant. This is because the yellow parent, being homozygous recessive, when crossed with a heterozygous brown (carrying one allele for yellow and one for brown), produces a 1:1 ratio of brown to yellow offspring.

Question 9

If the parent or offspring genotypes or phenotypes are known, the pattern of inheritance can be predicted. Two organisms, with genotypes BbDD and BBDd, are mated. Assuming independent assortment of the two loci, which of the following genotypic ratios in offspring would occur?

Accepted Answer

The correct genotypic ratio for the offspring of the cross between BbDD and BBDd, assuming independent assortment, is 1 BBDD: 1 BbDD: 1 BBDd: 1 BbDd. This is because each parent can produce two types of gametes with respect to the two genes in question, and when these gametes combine, they produce offspring with these four different genotypes.

Question 10

In a particular plant, green seed color is dominant to blue. If two plants with green seeds were crossed and resulted in 302 green and 98 blue seed plants, what was the most probable genotype of each parent?

Accepted Answer

The ratio of green to blue seed plants is close to 3:1, which is typical for a monohybrid cross involving two heterozygous parents (Gg × Gg). This suggests that each parent carries both the dominant (G) and recessive (g) alleles.

Question 11

Having polydactyly (extra digits on hands and feet) is a dominant trait. A man has polydactyly. His wife and oldest daughter do not have polydactyly. The couple's second child has polydactyly. What is the probability that their next (third) child will have extra digits?

Accepted Answer

Since polydactyly is a dominant trait, the man must have at least one allele for polydactyly. His wife, not having polydactyly, must have two recessive alleles. Therefore, each child has a 1/2 chance of inheriting the dominant allele from the father.

Question 12

Which of the following calculations requires the use of the addition rule of probability?

Accepted Answer

The addition rule of probability is used when calculating the probability of either of two mutually exclusive events happening. In choice D, calculating the probability of a child having either sickle-cell anemia or cystic fibrosis involves adding the probability of one event happening to the probability of the other event happening, assuming the events are mutually exclusive.

Question 13

Phenylketonuria is an inherited disease caused by a recessive allele. If a woman and her husband are both carriers, what is the probability that their first child will be a girl without phenylketonuria?

Accepted Answer

Both parents are carriers (Pp), so the probability of a child without phenylketonuria (PP or Pp) is 3/4. The probability of having a girl is 1/2. Therefore, the probability of having a girl without phenylketonuria is 3/4 * 1/2 = 3/8.

Question 14

Ten plants of the same genotype are planted in a large flower garden. Later that year, some of the ten plants have blue flowers and others have pink flowers. This can be best explained by which of the following statements?

Accepted Answer

Environmental factors such as soil pH can affect the phenotype of plants, including the color of their flowers. In hydrangeas, for example, the color of the flowers can change from pink to blue or vice versa depending on the soil pH. This explains why genetically identical plants can exhibit different flower colors.

Question 15

Which of the following inheritance patterns describes the ability of a single allele to have multiple phenotypic effects?

Accepted Answer

Pleiotropy refers to the phenomenon where a single gene influences multiple phenotypic traits, which is the definition provided in the question.

Question 16

In cattle, roan coat color (mixed red and white hairs) occurs in the heterozygous (CRCW) offspring of red (CRCR) and white (CWCW) homozygotes. Which of the following crosses would produce offspring in the ratio of 1 red:2 roan:1 white?

Accepted Answer

Crossing two roan cattle (CRCW × CRCW) would produce offspring in the ratio of 1 red (CRCR):2 roan (CRCW):1 white (CWCW), as this is a classic example of a 1:2:1 genotypic ratio in a simple Mendelian inheritance involving codominance.

Question 17

Which of the following statements best describes the addition rule of probability?

Accepted Answer

The addition rule of probability is used to find the probability that either one of two independent events will occur. It is often expressed as P(A or B) = P(A) + P(B) - P(A and B), where P(A and B) is subtracted to correct for the fact that it was counted twice if events A and B are not mutually exclusive.

Question 18

Which of the following scenarios describes an example of epistasis?

Accepted Answer

Epistasis occurs when the effect of one gene is dependent on the presence of one or more 'modifier genes', i.e., the genetic background. In choice B, the genotype (ee) prevents any fur color from developing, regardless of the alleles present at other loci that normally control fur color, exemplifying epistasis where one gene's expression masks the effect of another gene.

Question 19

What fraction of the offspring from a cross between two animals with the genotype AaBbCc are expected to be homozygous recessive for the three traits?

Accepted Answer

Each trait (A, B, C) has a 1/4 chance of being homozygous recessive (aa, bb, cc) when crossing Aa x Aa, Bb x Bb, and Cc x Cc. The probability of all three traits being homozygous recessive simultaneously is (1/4) * (1/4) * (1/4) = 1/64.

Question 20

In pea plants, the tall phenotype is dominant to the dwarf phenotype. If a heterozygous pea plant is crossed with a homozygous tall pea plant, what is the probability that the offspring will be dwarf in size?

Accepted Answer

The dwarf phenotype can only appear if both alleles are recessive. Since one parent is homozygous dominant (tall), all offspring will receive at least one dominant allele, making it impossible for any offspring to be dwarf.

Quiz 14: Mendelian Genetics