Question 1

Radish flowers may be red, purple, or white. A cross between a red-flowered plant and a white-flowered plant yields all-purple offspring. The part of the radish we eat may be oval or long, with long being the dominant trait. If true-breeding red long radishes are crossed with true-breeding white oval radishes, the F1 will be expected to be which of the following?

Accepted Answer

When a red-flowered plant is crossed with a white-flowered plant and yields all-purple offspring, it indicates that the purple flower color is dominant over both red and white. Therefore, the offspring of the cross between true-breeding red long radishes and true-breeding white oval radishes (F1) will have purple flowers. With respect to the radish shape, long is the dominant trait. Thus, the F1 will have long radishes. Therefore, the best choice is D) purple and long.

Question 2

In rabbits, the homozygous CC is normal, Cc results in deformed legs, and cc results in very short legs. The genotype BB produces black fur, Bb brown fur, and bb white fur. If a cross is made between brown rabbits with deformed legs and white rabbits with deformed legs, what percentage of the offspring would be expected to have deformed legs and white fur?

Accepted Answer

The cross between brown rabbits with deformed legs (BbCc) and white rabbits with deformed legs (bbCc) would produce offspring with the following genotypes: BbCc (brown fur, deformed legs), Bbcc (brown fur, very short legs), bbCc (white fur, deformed legs), and bbcc (white fur, very short legs). The percentage of offspring with deformed legs and white fur would be 25% (bbCc).

Question 3

Suppose two AaBbCc individuals are mated. Assuming that the genes are NOT linked, what fraction of the offspring are expected to be homozygous recessive for the three traits?

Accepted Answer

For each gene (A, B, C), the chance of an offspring being homozygous recessive (aa, bb, cc) is 1/4, since each parent can contribute a recessive allele with a probability of 1/2. Since the genes are not linked and assort independently, the probability of an offspring being homozygous recessive for all three traits is (1/4) * (1/4) * (1/4) = 1/64.

Question 4

Given the parents AABBCc × AabbCc, assume simple dominance for each trait and independent assortment. What proportion of the progeny will be expected to phenotypically resemble the first parent with genotype AABBCc?

Accepted Answer

The first parent has the genotype AABBCc, which means it has dominant alleles for A and B, and a recessive allele for C. Therefore, all of its offspring will inherit at least one dominant A and one dominant B allele, and only 1/4 of them will inherit a recessive c allele from both parents. 
The second parent has the genotype AabbCc, which means it has dominant alleles for A and C, and recessive alleles for B. Therefore, half of its offspring will inherit a dominant A allele from this parent, and half will inherit a dominant C allele from this parent. 
By independent assortment, the probability of inheriting a dominant A allele from the first parent and a dominant C allele from the second parent is 1/2 x 1/2 = 1/4. 
Therefore, the proportion of progeny expected to phenotypically resemble the first parent with genotype AABBCc is 1/4.

Question 5

The fact that all seven of the pea plant traits studied by Mendel obeyed the principle of independent assortment most probably indicates which of the following?

Accepted Answer

If all seven of the pea plant traits studied by Mendel obeyed the principle of independent assortment, it means that the genes controlling these traits behaved as if they were located on different chromosomes. This is because independent assortment describes the random assortment of homologous chromosomes during meiosis, which results in the independent inheritance of different traits. If all the genes controlling the traits were located on the same chromosome, they would be linked and inherited together, violating the principle of independent assortment. Therefore, answer choice D is the best choice.

Question 6

The individual with genotype AaBbCCDdEE can make many kinds of gametes. Which of the following is the major reason?

Accepted Answer

The individual with genotype AaBbCCDdEE can make many kinds of gametes due to different possible assortments of chromosomes into gametes during meiosis, which can lead to various combinations of alleles.

Question 7

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. The genes for flower color and location assort independently. If 1000 F2 offspring resulted from a dihybrid cross, approximately how many of them would you expect to have red, terminal flowers?

Accepted Answer

Given that the genes for flower color and location assort independently and considering the Mendelian inheritance, we can determine the expected phenotypic ratio for a dihybrid cross, which is 9:3:3:1. In this scenario, red is dominant over white, and axial is dominant over terminal. The question asks for red, terminal flowers, which falls under the 3/16 category (since red is dominant and terminal is the recessive trait being expressed). Therefore, out of 1000 F2 offspring, 3/16 would be expected to have red, terminal flowers. Calculating this gives 1000 * (3/16) = 187.5, which rounds to approximately 190.

Question 8

When Thomas Hunt Morgan crossed his red-eyed F1 generation flies to each other, the F2 generation included both red- and white-eyed flies. Remarkably, all the white-eyed flies were male. What was the explanation for this result?

Accepted Answer

In fruit flies, eye color is determined by a gene on the X chromosome. Since males only have one X chromosome, they inherit the eye color gene from their mother. If the mother has the red-eye allele, then the male offspring will have red eyes. However, if the mother has the white-eye allele, then the male offspring will have white eyes since they do not inherit a second copy of the X chromosome to mask the effect of the white-eye allele. Females, on the other hand, have two X chromosomes and inherit one from each parent. This means that if they inherit the white-eye allele on one X chromosome, they can still have red eyes if they inherit the red-eye allele on the other X chromosome. Thus, both red- and white-eyed female offspring are possible.

Question 9

Gene S controls the sharpness of spines in a type of cactus. Cactuses with the dominant allele, S, have sharp spines, whereas homozygous recessive ss cactuses have dull spines. At the same time, a second gene, N, determines whether or not cactuses have spines. Homozygous recessive nn cactuses have no spines at all. If doubly heterozygous SsNn cactuses were allowed to self-pollinate, the F2 would segregate in which of the following ratios?

Accepted Answer

The correct segregation ratio for the F2 generation from a doubly heterozygous SsNn self-pollination is 9 sharp-spined:3 dull-spined:4 spineless. This follows from applying the dihybrid cross principle, where each gene segregates independently, resulting in a 9:3:3:1 ratio for the phenotypes, with the modification that the nn genotype for the N gene results in spinelessness, overriding the S gene's effect.

Question 10

Feather color in budgies is determined by two different genes, Y and B, one for pigment on the outside and one for the inside of the feather. YYBB, YyBB, or YYBb is green; yyBB or yyBb is blue; YYbb or Yybb is yellow; and yybb is white. A blue budgie is crossed with a white budgie. Which of the following results is NOT possible?

Accepted Answer

The answer of Feather color in budgies is determined by...

Question 11

Gene S controls the sharpness of spines in a type of cactus. Cactuses with the dominant allele, S, have sharp spines, whereas homozygous recessive ss cactuses have dull spines. At the same time, a second gene, N, determines whether or not cactuses have spines. Homozygous recessive nn cactuses have no spines at all. A cross between a true-breeding sharp-spined cactus and a spineless cactus would produce ________.

Accepted Answer

The true-breeding sharp-spined cactus must have the genotype SSNN (homozygous dominant for sharp spines and having spines), and the spineless cactus must be nn for the N gene (homozygous recessive for no spines). Since the spineless cactus can only contribute the n allele and the sharp-spined cactus can only contribute the S and N alleles, all progeny will have at least one S and one N allele, resulting in all sharp-spined progeny.

Question 12

Assuming independent assortment for all gene pairs, what is the probability that the following parents, AABbCc × AaBbCc, will produce an AaBbCc offspring?

Accepted Answer

First, we need to determine the possible alleles from each parent:

AABbCc 
Possible alleles: ABc, Abc

AaBbCc 
Possible alleles: ABc, Abc, aBc, abc

To get an AaBbCc offspring, we need to get at least one A from the first parent, one B from either parent, and one C from either parent.

Probability of getting an A from the first parent: 
1 (AA is homozygous dominant)

Probability of getting a B from either parent: 
3/4 (Bb or bB)

Probability of getting a C from either parent: 
3/4 (Cc or cC)

Multiplying these probabilities together gives: 
1 × (3/4) × (3/4) = 9/16

Therefore, the probability of producing an AaBbCc offspring is 9/16, which is answer choice C.

Question 13

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. The genes for flower color and location assort independently. Among the F2 offspring of a dihybrid cross, what is the probability of plants with white axial flowers?

Accepted Answer

The probability of white axial flowers can be calculated using the principles of independent assortment. Since all F1 individuals have red, axial flowers, red and axial are dominant traits. For F2 offspring to have white axial flowers, they must inherit a recessive allele for flower color (white) and a dominant allele for flower position (axial). The probability of inheriting the recessive white allele is 1/4 (since the genotype must be ww), and the probability of inheriting the dominant axial allele is 3/4 (since the genotype can be AA or Aa). Multiplying these probabilities gives 1/4 * 3/4 = 3/16.

Question 14

In humans, male-pattern baldness may be assumed to be controlled by an autosomal gene that occurs in two allelic forms. Allele B determines nonbaldness, and allele b determines pattern baldness. In males, because of the presence of testosterone, allele b is dominant over B. If a man and woman both with genotype Bb have a son, what is the chance that he will eventually be bald?

Accepted Answer

The answer of In humans, male-pattern baldness may be assumed...

Question 15

How many unique gametes could be produced through independent assortment by an individual with the genotype AaBbCCDdEE?

Accepted Answer

The answer of How many unique gametes could be produced...

Question 16

Mendel's observation of the segregation of alleles in gamete formation has its basis in which of the following phases of cell division?

Accepted Answer

Mendel's observation of the segregation of alleles in gamete formation is based on anaphase I of meiosis, where homologous chromosomes separate and move toward opposite poles of the cell, resulting in the separation of alleles.

Question 17

Which of the following statements about independent assortment and segregation is correct?

Accepted Answer

The law of independent assortment describes how different genes separate from one another during gamete formation, and it requires describing two or more genes relative to one another. The law of segregation describes how alleles of the same gene separate from each other during gamete formation and does not require multiple genes or generations to be described. The observation of prophase I helps to account for the law of independent assortment, while anaphase of meiosis I helps to account for the law of segregation.

Question 18

What do we mean when we use the terms "monohybrid cross" and "dihybrid cross"?

Accepted Answer

A monohybrid cross involves organisms that are heterozygous for one character being studied, whereas a dihybrid cross involves organisms that are heterozygous for two characters being studied. The terms do not refer to the number of parents, the number of offspring, or the number of generations involved in the cross. The ratios of progeny produced by each type of cross can vary depending on the specific traits being studied.

Question 19

In the cross AaBbCc × AaBbCc, what is the probability of producing the genotype AABBCC?

Accepted Answer

The answer of In the cross AaBbCc × AaBbCc, what...

Question 20

Mendel crossed yellow-seeded and green-seeded pea plants and then allowed the offspring to self-pollinate to produce an F2 generation. The results were as follows: 6022 yellow and 2001 green (8023 total). The allele for green seeds has what relationship to the allele for yellow seeds?

Accepted Answer

The allele for green seeds is recessive to the allele for yellow seeds, as in the F2 generation 1/4th of the offspring with the genotype for green seeds (yy) are expressed phenotypically, while the other 3/4th have the dominant yellow-seeded phenotype.

Quiz 14: Mendel and the Gene