Question 1

What would be TRUE if linkage of two genes was complete?
A) All possible combinations of alleles in the offspring would be produced.
B) The genes would assort independently.
C) All possible combinations of alleles in the gametes would be produced.
D) Crossing over between the two genes would rarely occur.

Accepted Answer

Complete linkage between two genes means they are located very close together on the same chromosome, making it unlikely for crossing over to occur between them during meiosis. This results in the parental allele combinations being inherited together more frequently than not, as opposed to producing a wide variety of recombinant gametes.

Question 2

Albinism results from a recessive allele. Which of the following describes the expected offspring from a normally pigmented male (who had an albino father) and an albino wife?
A) All normal
B) 75% albino; 25% normal
C) 75% normal; 25% albino
D) All albino
E) 50% normal; 50% albino

Accepted Answer

The normally pigmented male must be heterozygous (carrying one normal and one albino allele) because he had an albino father, indicating he carries the recessive albino allele. An albino individual must have two copies of the recessive allele. When a heterozygous individual mates with a homozygous recessive individual, the offspring have a 50% chance of being heterozygous (and thus normally pigmented) and a 50% chance of being homozygous recessive (and thus albino).

Question 3

In crossing a homozygous recessive with a heterozygote, what is the chance of getting a homozygous recessive phenotype in the F1 generation?
A) 100%
B) 25%
C) 0%
D) 50%
E) 75%

Accepted Answer

In a cross between a homozygous recessive individual and a heterozygote, there is a 50% chance of producing offspring with a homozygous recessive phenotype, as one allele from the heterozygote parent can either be dominant or recessive, and the homozygous recessive parent can only pass on the recessive allele.

Question 4

If you cross pea plants that are both heterozygous for purple flowers and yellow seeds ( PpYy), and you examine 800 offspring for flower and seed color, about how many do you expect to have white flowers and green seeds?
A) 450
B) All of them (800)
C) 50
D) 150
E) None of them

Accepted Answer

The cross of two heterozygous pea plants (PpYy x PpYy) involves two traits, each with two alleles: P (purple flowers) is dominant over p (white flowers), and Y (yellow seeds) is dominant over y (green seeds). To find the offspring with white flowers and green seeds (ppy), we use a Punnett square or the forked-line method. Each trait segregates independently according to Mendel's law of independent assortment. The probability of pp (white flowers) is 1/4, and the probability of yy (green seeds) is also 1/4. Multiplying these probabilities gives 1/16 for the combination of pp and yy in any given offspring. Therefore, out of 800 offspring, 800 * (1/16) = 50 are expected to have white flowers and green seeds.

Question 5

What is the genotype of a dominant individual if, when selfed, some of its offspring show the recessive phenotype?
A) Either DD or Dd
B) DD
C) dd
D) Dd

Accepted Answer

For an individual showing a dominant phenotype to produce offspring with a recessive phenotype when selfed, it must be heterozygous (Dd). This is because two copies of the recessive allele (dd) are required for the recessive phenotype to be expressed, indicating that the parent must carry one recessive allele to pass it on.

Question 6

When genes are linked, a few of the recombinant genotypes still occur in the offspring because of
A) linkage.
B) independent assortment.
C) crossing over.
D) natural selection.
E) mutation.

Accepted Answer

Crossing over during meiosis can result in recombinant genotypes even when genes are linked, as it allows for the exchange of genetic material between homologous chromosomes.

Question 7

Crossing smooth-seeded pea plants with wrinkled-seeded pea plants resulted in progeny that all had smooth seeds. This indicates that the wrinkled-seed trait is
A) codominant and dominant.
B) recessive only.
C) codominant and recessive.
D) dominant only.
E) codominant only.

Accepted Answer

The wrinkled-seed trait is recessive because it does not appear in the progeny when crossed with the dominant smooth-seed trait, indicating that the presence of the dominant allele for smooth seeds masks the expression of the recessive wrinkled-seed allele.

Question 8

31)	Imagine that a newly discovered species of fish normally has three eyes ( E), but the rare mutation
(e)	causes ?diopthalmic? fish with two eyes to be born. In addition, eye color in these fish is inherited similarly to humans, with black eyes (B) being dominant to blue (b). Assuming these two genes are carried on separate chromosomes, what must the genotypes of two black -eyed,
three-eyed parents be if they have a blue-eyed, diopthalmic son?
A) eeBB and EEbb
B) Both EeBb
C) Both eebb
D) EEBB and eebb

Accepted Answer

Given that the parents are both three-eyed and black-eyed, and they produce a blue-eyed, diopthalmic son, the parents must be heterozygous for both traits (EeBb). This allows for the combination of recessive alleles (eebb) in their son, resulting in two eyes and blue eye color.

Question 9

In garden peas, the allele for tall plants is dominant over the allele for short plants. Imagine that a true-breeding tall plant is crossed with a short plant. Then one of their offspring is test crossed. Out of 20 offspring resulting from the test cross, about how many should be tall?
A) 0
B) 10
C) 15
D) 20
E) 5

Accepted Answer

A true-breeding tall plant has the genotype TT (where T is the dominant allele for tallness), and a short plant has the genotype tt (since shortness is recessive and can only be expressed in the homozygous condition). When a TT plant is crossed with a tt plant, all offspring (F1 generation) will be Tt (heterozygous but tall, due to the dominance of the T allele). A test cross involves crossing an individual with a homozygous recessive individual (tt) to determine the genotype of the first individual. In this case, crossing a Tt plant with a tt plant would result in approximately half of the offspring being tall (Tt) and half being short (tt). Therefore, out of 20 offspring, about 10 should be tall.

Question 10

In a test cross, what percentage of the offspring will have the same phenotype as the tested parent if the parent to be tested is homozygous?
A) 100%
B) 0%
C) 75%
D) 50%
E) 25%

Accepted Answer

In a test cross, if the parent to be tested is homozygous, all offspring will have the same phenotype as the tested parent, resulting in 100% of the offspring displaying that phenotype.

Question 11

Imagine you do a test cross between a purple -flowered pea plant having serrated leaves (both dominant traits) and a white-flowered pea plant having smooth edges. If the purple -flowered plant is heterozygous for both traits, the expected ratio in the offspring is 1 purple -serrated:1 purple-smooth:1 white-serrated:1 white-smooth. Instead, you see 7 purple-serrated:1 purple-smooth:1 white-serrated:7 white-smooth. What is the explanation of this ratio?
A) The genes are linked.
B) There is incomplete dominance.
C) The genes are sex-linked and inherited by incomplete dominance.
D) There has been a mutation during the cross.

Accepted Answer

The observed ratio suggests that the genes for flower color and leaf shape are linked, meaning they are located close together on the same chromosome and tend to be inherited together, which deviates from the expected 1:1:1:1 Mendelian ratio for independent assortment.

Question 12

Suppose you cross a pea plant with yellow, smooth seeds with one that has green, wrinkled seeds. You examine the seeds of 967 offspring and find that some have yellow, smooth seeds and some have yellow, wrinkled seeds, but no green seeds of either type show up. What is the genotype of the yellow-seeded parent? (Y = yellow, y = green; S = smooth, s = wrinkled)
A) YYSs
B) YYss
C) YYSS
D) YySS
E) YySs

Accepted Answer

The yellow-seeded parent must be homozygous for the yellow allele (YY) because no green seeds appeared among the offspring, indicating that the green allele (y) is recessive and was not contributed by the yellow-seeded parent. The presence of both smooth and wrinkled seeds among the offspring indicates that the yellow-seeded parent must have one allele for smooth seeds (S) and one for wrinkled seeds (s), making the genotype YYSs.

Question 13

All of the following combinations are possible in the gametes of an organism that is AaBb EXCEPT
A) aa.
B) AB.
C) ab.
D) aB.
E) Ab.

Accepted Answer

In an organism with the genotype AaBb, gametes can only contain one allele from each gene due to the process of segregation during meiosis. Therefore, a gamete cannot have both alleles (aa or BB) from the same gene, making option A (aa) impossible.

Question 14

Although the law of independent assortment is generally applicable, when two loci are located physically close to each other on the same chromosome, the phenotypes of the offspring sometimes do not fit the phenotypes predicted. This is due to
A) linkage.
B) inversions.
C) translation.
D) chromosome exchange.
E) chromatid abnormalities.

Accepted Answer

Linkage refers to the phenomenon where genes that are located close to each other on the same chromosome tend to be inherited together, because they are less likely to be separated by recombination during meiosis. This can result in offspring phenotypes that do not fit the predictions made by the law of independent assortment, which assumes genes are inherited independently of each other.

Question 15

A pea plant with red flowers is test crossed; half of the resulting progeny have red flowers, whereas the other half have white flowers. The genotype of the test -crossed parent was
A) rr.
B) either RR or Rr.
C) Rr.
D) RR.

Accepted Answer

The test cross (crossing with a homozygous recessive individual) results in half of the progeny showing the recessive phenotype (white flowers), indicating that the test-crossed parent was heterozygous (Rr) for the trait.

Question 16

If two parents are carriers of albinism (an autosomal recessive trait), what is the chance that their fourth child will have a homozygous dominant genotype?
A) 50%
B) 0%
C) 25%
D) 100%
E) 75%

Accepted Answer

Albinism is an autosomal recessive trait, meaning that an individual must inherit two recessive alleles (one from each parent) to express the condition. When both parents are carriers (heterozygous, having one dominant and one recessive allele), each child has a 25% chance of being homozygous dominant (not carrying the trait), a 50% chance of being a carrier like the parents (heterozygous), and a 25% chance of being homozygous recessive (having albinism). The birth order of the child does not affect these probabilities.

Question 17

Which of the following is TRUE of genes that are physically located close to each other on the same chromosome?
A) They are not linked.
B) They tend to not be inherited together.
C) They are linked and tend to be inherited together.
D) They will be expressed differently in males than in females.

Accepted Answer

Genes that are physically close to each other on the same chromosome are considered linked and tend to be inherited together because they are less likely to be separated by recombination during meiosis.

Question 18

Imagine that you self-pollinate heterozygous purple-flowered, long-pollened pea plants and get mostly purple-flowered, long-pollened plants and white-flowered, short-pollened plants, with a few purple-flowered, short-pollened or white-flowered, long-pollened plants. What is the most likely explanation?
A) The genes for flower color and pollen length are linked.
B) The gene for flower color is on a sex chromosome.
C) Long pollen is often lethal, so few plants with long pollen survive to adulthood.
D) The purple allele influences pollen to become shorter.
E) This is a normal random variation from the expected ratio for independent assortment.

Accepted Answer

The most likely explanation for the observed pattern of inheritance is that the genes for flower color and pollen length are linked, meaning they are located close together on the same chromosome and tend to be inherited together. This linkage disrupts the expected Mendelian ratio for independent assortment, leading to a predominance of parental phenotypes (purple-flowered, long-pollened and white-flowered, short-pollened) and fewer recombinant phenotypes (purple-flowered, short-pollened or white-flowered, long-pollened).

Question 19

In pea plants, tall plants are dominant over short, and green pods are dominant over yellow. If you crossed true-breeding tall, green-podded plants with true-breeding short, yellow-podded plants, and then crossed the F1 individuals among themselves, what proportion of the F2 offspring would you expect to be tall and yellow-podded?
A) 1/4
B) 1/16
C) 3/16
D) 3/4
E) 9/16

Accepted Answer

The F1 generation, being a cross between true-breeding tall, green-podded plants (TTGG) and true-breeding short, yellow-podded plants (ttgg), would all be heterozygous tall and green-podded (TtGg). Crossing these F1 individuals (TtGg x TtGg) would result in a 9:3:3:1 ratio for the F2 generation, according to Mendelian genetics. The tall, yellow-podded phenotype corresponds to the 3/16 portion of this ratio, as it combines the dominant trait from one gene (tall) with the recessive trait from the other (yellow pods).

Question 20

Yellow-seeded pea plants may be homozygous or heterozygous. To find out which, cross the plants with
A) heterozygous yellow-seeded plants.
B) true-breeding yellow-seeded plants.
C) heterozygous green-seeded plants.
D) true-breeding green-seeded plants.
E) the same genotype.

Accepted Answer

To determine if yellow-seeded pea plants are homozygous or heterozygous, crossing them with true-breeding green-seeded plants (D) is the most effective method. True-breeding green-seeded plants are homozygous recessive for the seed color trait. If any offspring from this cross are green-seeded, the original yellow-seeded plant must have been heterozygous, carrying one allele for yellow and one for green. If all offspring are yellow-seeded, the original plant is likely homozygous for the yellow seed color allele. This method utilizes the principle of a test cross, where an individual of unknown genotype is crossed with a homozygous recessive individual to reveal the unknown genotype based on the phenotypes of the offspring.

Quiz 11: Patterns of Inheritance

What would be TRUE if linkage of two genes was complete?

Albinism results from a recessive allele. Which of the following describes the expected offspring from a normally pigmented male (who had an albino father) and an albino wife?

In crossing a homozygous recessive with a heterozygote, what is the chance of getting a homozygous recessive phenotype in the F1 generation?

If you cross pea plants that are both heterozygous for purple flowers and yellow seeds ( PpYy) , and you examine 800 offspring for flower and seed color, about how many do you expect to have white flowers and green seeds?

What is the genotype of a dominant individual if, when selfed, some of its offspring show the recessive phenotype?

When genes are linked, a few of the recombinant genotypes still occur in the offspring because of

Crossing smooth-seeded pea plants with wrinkled-seeded pea plants resulted in progeny that all had smooth seeds. This indicates that the wrinkled-seed trait is

In garden peas, the allele for tall plants is dominant over the allele for short plants. Imagine that a true-breeding tall plant is crossed with a short plant. Then one of their offspring is test crossed. Out of 20 offspring resulting from the test cross, about how many should be tall?

In a test cross, what percentage of the offspring will have the same phenotype as the tested parent if the parent to be tested is homozygous?

All of the following combinations are possible in the gametes of an organism that is AaBb EXCEPT

Although the law of independent assortment is generally applicable, when two loci are located physically close to each other on the same chromosome, the phenotypes of the offspring sometimes do not fit the phenotypes predicted. This is due to

A pea plant with red flowers is test crossed; half of the resulting progeny have red flowers, whereas the other half have white flowers. The genotype of the test -crossed parent was

If two parents are carriers of albinism (an autosomal recessive trait) , what is the chance that their fourth child will have a homozygous dominant genotype?

Which of the following is TRUE of genes that are physically located close to each other on the same chromosome?

Yellow-seeded pea plants may be homozygous or heterozygous. To find out which, cross the plants with