Mendelian Genetics

Question 1

In pea plants, spherical seeds (S) are dominant to dented seeds (s). In a genetic cross of two plants that are heterozygous for the seed shape trait, what fraction of the offspring should have spherical seeds?

A. None
B. 1/4
C. 1/2
D. 3/4
E. All

Answer 1

D. 3/4

One fourth of the offspring will be homozygous dominant (SS), one half will be heterozygous (Ss), and one fourth will be homozygous recessive (ss).

Question 2

A phenotypic ratio of 3:1 in the offspring of a mating of two organisms heterozygous for a single trait is expected when:

A. the alleles segregate during meiosis.
B. each allele contains two mutations.
C. the alleles are identical.
D. the alleles are incompletely dominant.
E. only recessive traits are scored.

Answer 2

A. the alleles segregate during meiosis.

Mendel first proposed that alleles segregate from one another during the formation of gametes.

Question 3

In Mendel’s “Experiment 1,” true-breeding pea plants with spherical seeds were crossed with true-breeding plants with dented seeds. (Spherical seeds are the dominant characteristic.) Mendel collected the seeds from this cross, grew F1-generation plants, let them self-pollinate to form a second generation, and analyzed the seeds of the resulting F2 generation. The results that he obtained, and that you would predict for this experiment are:

A. 1/2 the F1 and 3/4 of the F2 generation seeds were spherical.
B. 1/2 the F1 and 1/4 of the F2 generation seeds were dented.
C. All of the F1 and F2 generation seeds were spherical.
D. 3/4 of the F1 and 9/16 of the F2 generation seeds were spherical.
E. All the F1 and 3/4 of the F2 generation seeds were spherical.

Answer 3

E. All the F1 and 3/4 of the F2 generation seeds were spherical.

All of the F1 plants were true hybrids with a phenotype of Ss. The recessive trait reappears in the F2 generation

Question 4

A genetic cross between two F1-hybrid pea plants for spherical seeds will yield what percent spherical-seeded plants in the F2 generation? (Recall, spherical-shaped seeds are dominant over dented seeds.)

A. 100%
B. 75%
C. 50%
D. 25%
E. 0%

Answer 4

B. 75%

Only 25% of F2 plants will have the recessive phenotype.

Question 5

A genetic cross between two F1-hybrid pea plants having yellow seeds will yield what percent green-seeded plants in the F2 generation? Yellow seeds are dominant to green.

A. 0%
B. 25%
C. 50%
D. 75%
E. 100%

Answer 5

B. 25%

Among the F2 plants of a Yy x Yy cross, 25% will be yy with the recessive, green-seeded phenotype.

Question 6

When true-breeding tall stem pea plants are crossed with true-breeding short stem pea plants, all of the _________ plants, and 3/4 of the __________ plants had tall stems. Therefore, tall stems are dominant.

A. F1, F2.
B. G1, G2.
C. parental, F2.
D. F2, parental.
E. P1, P2

Answer 6

A. F1, F2.

The F1 plants are all Tt hybrids. The recessive trait (tt) reappears in the F2 generation in about 25% of the plants.

Question 7

To identify the genotype of yellow-seeded pea plants as either homozygous dominant (YY) or heterozygous (Yy), you could do a test cross with plants of genotype _______.

A. y
B. Y
C. yy
D. YY
E. Yy

Answer 7

C. yy

A cross with the homozygous recessive (yy) is a test cross. If the parent of unknown genotype is heterozygous (Yy), half of the offspring will have the recessive trait. The unknown genotype could also be determined by a cross with a known heterozygote (Yy).

Question 8

A test cross is used to determine if the genotype of a plant with the dominant phenotype is homozygous or heterozygous. If the unknown is homozygous, all of the offspring of the test cross have the __________ phenotype. If the unknown is heterozygous, half of the offspring will have the __________ phenotype.

A. dominant, incompletely dominant
B. recessive, dominant
C. dominant, epistatic
D. co-dominant, complimentary
E. dominant, recessive

Answer 8

E. dominant, recessive

The test cross was invented by Mendel to determine the genotype of plants displaying the dominant phenotype.

Question 9

In Mendel’s experiments, if the gene for tall (T) plants was incompletely dominant over the gene for short (t) plants, what would be the result of crossing two Tt plants?

A. 1/4 would be tall; 1/2 intermediate height; 1/4 short
B. 1/2 would be tall; 1/4 intermediate height; 1/4 short.
C. 1/4 would be tall; 1/4 intermediate height; 1/2 short.
D. All the offspring would be tall.
E. All the offspring would be intermediate.

Answer 9

A. 1/4 would be tall; 1/2 intermediate height; 1/4 short

The heterozygous offspring (Tt) would be of intermediate height.

Question 10

A genetic cross of inbred snapdragons with red flowers with inbred snapdragons with white flowers resulted in F1-hybrid offspring that all had pink flowers. When the F1 plants were self-pollinated, the resulting F2-generation plants had a phenotypic ratio of 1 red: 2 pink: 1 white. The most likely explanation is:

A. pink flower color is epistatic to red flower color.
B. pink flowers are the result of a blending of the red and white genotypes.
C. flower color is due to 2 or more complementary genes.
D. heterozygous plants have a different phenotype than either inbred parent because of incomplete dominance of the dominant allele.
E. flower color inheritance in snapdragons does not behave as a Mendelian trait.

Answer 10

D. heterozygous plants have a different phenotype than either inbred parent because of incomplete dominance of the dominant allele.

The features of crosses involving incomplete dominance are intermediate phenotype of heterozygous individuals, and parental phenotypes reappear in F2 when heterozygotes are crossed.

Question 11

Human blood type is determined by codominant alleles. There are three different alleles, known as IA, IB, and i. The IA and IB alleles are co-dominant, and the i allele is recessive.

The possible human phenotypes for blood group are type A, type B, type AB, and type O. Type A and B individuals can be either homozygous (IAIA or IBIB, respectively), or heterozygous (IAi or IBi, respectively).

A woman with type A blood and a man with type B blood could potentially have offspring with which of the following blood types?

A. type A
B. type B
C. type AB
D. type O
E. all of the above

Answer 11

E. all of the above

But if the man was type O rather than type B, offspring of type B and type AB would not be possible.

Question 12

Manx cats are heterozygous for a dominant mutation that results in no tails (or very short tails), large hind legs, and a distinctive gait. The mating of two Manx cats yields two Manx kittens for each normal, long-tailed kitten, rather than three-to-one as would be predicted from Mendelian genetics. Therefore, the mutation causing the Manx cat phenotype is likely a(n) __________ allele.
Courtesy of PETNET in Australia

A. pleiotropic
B. co-dominant
C. epistatic
D. lethal
E. sex-linked

Answer 12

D. lethal

The predicted segregation pattern in the F2 generation is 1/4 normal (homozygous), 1/2 Manx phenotype (heterozygous), an 1/4 embryonic lethal (homozygous for the Manx allele).

Question 13

What are the possible blood types of the offspring of a cross between individuals that are type AB and type O? (Hint: blood type O is recessive)

A. AB or O
B. A, B, or O
C. A or B
D. A, B, AB, or O
E. A, B, or AB

Answer 13

C. A or B

A cross between individuals that are of genotype IAIBx ii can yield offspring that are either IAi or IBi. Their blood type will be A or B.

Question 14

A pea plant is heterozygous for both seed shape and seed color. S is the allele for the dominant, spherical shape characteristic; s is the allele for the recessive, dented shape characteristic. Y is the allele for the dominant, yellow color characteristic; y is the allele for the recessive, green color characteristic. What will be the distribution of these two alleles in this plant’s gametes?

A. 50% of gametes are Sy; 50% of gametes are sY
B. 25% of gametes are SY; 25% of gametes are Sy;
25% of gametes are sY; 25% of gametes are sy.
C. 50% of gametes are sy; 50% of gametes are SY
D. 100% of the gametes are SsYy
E. 50% of gametes are SsYy; 50% of gametes are SSYY.

Answer 14

B. 25% of gametes are SY; 25% of gametes are Sy;
25% of gametes are sY; 25% of gametes are sy.

Alleles of different genes are assorted independently of each other during the formation of gametes.

Question 15

A phenotype ratio of 9:3:3:1 in the offspring of a mating of two organisms heterozygous for two traits is expected when:

A. the genes reside on the same chromosome
B. each gene contains two mutations
C. the gene pairs assort independently during meiosis
D. only recessive traits are scored
E. none of the above

Answer 15

C. the gene pairs assort independently during meiosis

The parental organisms have the same phenotype, but their offspring have 4 different phenotypes. The 9: 3: 3: 1 ratio can only occur if the two traits assort independently during meiosis.

Question 16

Which of the following genetic crosses would be predicted to give a phenotypic ratio of 9:3:3:1?

A. SSYY x ssyy
B. SsYY x SSYy
C. SsYy x SsYy
D. SSyy x ssYY
E. ssYY x ssyy

Answer 16

C. SsYy x SsYy

The dihybrid cross was invented by Mendel to discover the independent assortment of alleles during gamete formation.

Question 17

The gametes of a plant of genotype SsYy should have the genotypes:

A. Ss and Yy
B. SY and sy
C. SY, Sy, sY, and sy
D. Ss, Yy, SY and sy
E. SS, ss, YY, and yy

Answer 17

C. SY, Sy, sY, and sy

Each gamete will receive either on S or s allele, and either a Y or y allele. All of S or s, and Y or y are possible.

Question 18

Which of the following genotypes would you not expect to find among the offspring of a SsYy x ssyy test cross:

A. ssyy
B. SsYy
C. Ssyy
D. ssYy
E. SsYY

Answer 18

E. SsYY

Offspring could not be homozygous for the dominant yellow seed color (YY), because one recessive y allele must be inhereited from the ssyy parent.

Question 19

The expected phenotypic ratio of the progeny of a SsYy x ssyy test cross is:

A. 9:3:3:1
B. 3:1
C. 1:1:1:1
D. 1:2:1
E. 3:1:1:3

Answer 19

C. 1:1:1:1.

SsYy, ssYy, Ssyy, ssyy are predicted to occur in a ratio of 1:1:1:1.

Question 20

In a dihybrid cross, AaBb x AaBb, what fraction of the offspring will be homozygous for both recessive traits?

A. 1/16
B. 1/8
C. 3/16
D. 1/4
E. 3/4

Answer 20

A. 1/16

1/4 of the gametes of each parent will be ab. The fraction of the offspring homozygous for both recessive traits will be 1/4 times 1/4, or 1/16.

Question 21

Following a SsYy x SsYy cross, what fraction of the offspring are predicted to have a genotype that is heterozygous for both characteristics?

A. 1/16
B. 2/16
C. 3/16
D. 4/16
E. 9/16

Answer 21

4/16

Four out of 16 possible combinations of alleles yield the SsYy genotype. They are:
SY and sy,
sY and Sy,
Sy and sY,
SY and sy.

Question 22

In a dihybrid cross, SsYy x SsYy, what fraction of the offspring will be homozygous for both traits?

A. 1/16
B. 1/8
C. 3/16
D. 1/4
E. 3/4

Answer 22

D. 1/4

There are four different genotypes that are homozygous for both traits. These are SSYY, SSyy, ssYY, and ssyy.

Question 23

If Mendel’s crosses between tall, spherical-seeded plants and short, dented-seeded plants had produced many more than 1/16 short, dented-seeded plants in the F2 generation, he might have concluded that:

A. the dented seed and short traits are unlinked.
B. he would not have concluded any of the above.
C. all traits in peas assort independently of each other.
D. the spherical seed and tall traits are linked.
E. all traits in peas are linked.

Answer 23

D. the spherical seed and tall traits are linked.

Linked traits do not segregate independently during gamete formation. Mendel was probably very fortunate to have selected unlinked traits to study. Had he observed linked or partially-linked traits, it would have been more difficult to arrive at his theory that alleles of different genes assort independently during gamete formation.

Question 24

In Mendel’s experiments, the spherical seed character (SS) is completely dominant over the dented seed character (ss). If the characters for height were incompletely dominant, such that TT are tall, Tt are intermediate and tt are short, what would be the phenotypes resulting from crossing a spherical-seeded, short (SStt) plant to a dented-seeded, tall (ssTT) plant?

A.
All the progeny would be spherical-seeded and tall.
B.
1/2 would be spherical-seeded and intermediate height; 1/2 would be spherical-seeded and tall.
C.
All the progeny would be spherical-seeded and short.
D.
You cannot predict the outcome.
E.
All the progeny would be spherical-seeded and intermediate height.

Answer 24

E.
All the progeny would be spherical-seeded and intermediate height.
All of the offspring would have the genotype of SsTt. Mendel selected traits to study that did not display partial dominance. Could you predict the result of an SsTt x SsTt cross if one or both traits dispayed partial dominance? There would be 9 different phenotypes if both traits showed partial dominance.

Question 25

Two unlinked loci effect mouse hair color. CC or Cc mice are agouti. Mice with genotype cc are albino because all pigment production and deposition of pigment in hair is blocked. At the second locus, the B allele (black agouti coat) is dominant to the b allele (brown agouti coat). A mouse with a black agouti coat is mated with an albino mouse of genotype bbcc. Half of the offspring are albino, one quarter are black agouti, and one quarter are brown agouti. What is the genotype of the black agouti parent?

A. BBCC
B. BbCc
C. bbCC
D. BbCC
E. BBcc

Answer 25

B. BbCc

This is the only possibility for this combination of offspring.

Question 26

Two unlinked loci effect mouse hair color. AA or Aa mice are agouti. Mice with genotype aa are albino because all pigment production is blocked, regardless of the phenotype at the second locus. At the second locus, the B allele (agouti coat) is dominant to the b allele (black coat). What would be the result of a cross between two agouti mice of genotype AaBb?

A. 4 agouti: 4 black: 8 albino
B. 9 agouti: 3 black: 3 albino: 1 grey
C. 9 agouti: 3 black: 4 albino
D. 8 agouti: 4 black: 4 albino

Answer 26

C. 9 agouti: 3 black: 4 albino

This is a variant of the normal 9:3:3:1 ratio for a normal dihybrid cross without epistasis.

Question 27

In a cross between a white-eyed female fruit fly and red-eyed male, what percent of the female offspring will have white eyes? (White eyes are X-linked, recessive)

A. 100%
B. 25%
C. 50%
D. 75%
E. 0%

Answer 27

E. 0%

All of the females are red-eyed and heterozygous. All of the males are white-eyed and hemizygous.

Question 28

A female Drosophila of unknown genotype was crossed with a white-eyed male fly, of genotype (w = white eye allele is recessive, w+= red-eye allele is dominant.) Half of the male and half of the female offspring were red-eyed, and half of the male and half of the female offspring were white-eyed. What was the genotype of the female fly?

A. xw+ Y
B. xw+ xw+
C. xw xw
D. xw Y
E. xw+ xw

Answer 28

E. xw+ xw

Female must be heterozygous because she can transmit either the red eye or white eye allele to her sons and daughters.

Question 29

In a cross between a pure bred, red-eyed female fruit fly and a white-eyed male, what percent of the male offspring will have white eyes? (white eyes are X-linked, recessive)

A. 100%
B. 75%
C. 50%
D. 25%
E. 0%

Answer 29

E. 0%

All of the males and all of the females are red-eyed.

Question 30

What is the genotype of a red-eyed, yellow-bodied female fruit fly who is homozygous for the eye color allele?
Red eyes (w+) and tan bodies (y+) are the dominant alleles. (Both traits are X chromosome linked).

A. xw+ Y
B. xw+ y xw+y
C. xw y+ xw y+
E. xw Y+
D. xw+ y xw+y+

Answer 30

B. xw+ y xw+y

The female is homozygous for dominant eye color (given), and homozygous for recessive body color allele.

Question 31

A white-eyed female fruit fly is crossed with a red-eyed male. Red eyes are dominant, and X-linked. What are the expected phenotypes of the offspring?

A. All of the females will have red eyes; half of the males will have red eyes, and half of the males will have white eyes.
B. All of the females and all of the males will have white eyes.
C. All of the females will have red eyes; all of the males will have white eyes.
D. All of the females and all of the males will have red eyes.
E. All of the females will have white eyes; half of the males will have red eyes, and half of the males will have white eyes.

Answer 31

C. All of the females will have red eyes; all of the males will have white eyes.

All of the females are red-eyed and heterozygous. All of the males are white-eyed and hemizygous.

Question 32

Hemophilia in humans is due to an X-chromosome mutation. What will be the results of mating between a normal (non-carrier) female and a hemophilac male?

A. half of daughters are normal and half of sons are hemophilic.
B. all sons are normal and all daughters are carriers.
C. half of sons are normal and half are hemophilic; all daughters are carriers.
D. all daughters are normal and all sons are carriers.
E. half of daughters are hemophilic and half of daughters are carriers; all sons are normal.

Answer 32

B. all sons are normal and all daughters are carriers.

Daughters inherit a normal allele from their mother and the hemophilia allele from their father. Sons inherit the normal allele from their mother.

Question 33

A human female “carrier” who is heterozygous for the recessive, sex-linked trait causing red-green color blindness (or alternatively, hemophilia), marries a normal male. What proportion of their male progeny will have red-green color blindness (or alternatively, will be hemophiliac)?

A. 100%
B. 75%
C. 50%
D. 25%
E. 0%

Answer 33

C. 50%

Half the sons would be expected to inherit the allele from their mother and be afflicted because they are hemizygous. Half the daughters would be carriers like their mothers.

Question 34

Women have sex chromosomes of XX, and men have sex chromosomes of XY.

Which of a man’s grandparents could not be the source of any of the genes on his Y-chromosome?

A. Father’s Mother.
B. Mother’s Father.
C. Father’s Father.
D. Mother’s Mother, Mother’s Father, and Father’s Mother.
E. Mother’s Mother.

Answer 34

D. Mother’s Mother, Mother’s Father, and Father’s Mother.

The Y chromosome is inherited solely from father to son in each generation.

Question 35

Women have sex chromosomes of XX, and men have sex chromosomes of XY.

Which of a women’s grandparents could not be the source of any of the genes on either of her X-chromosomes?

A. Mother’s Father.
B. Father’s Mother.
C. Mother’s Mother.
D. Father’s Father.
E. Mother’s Mother and Mother’s Father.

Answer 35

D. Father’s Father.

The father’s father contributes only the Y chromosome to his sons, and subsequently to his grandsons.

Question 36

A human female “carrier” who is heterozygous for the recessive, sex-linked trait red color blindness, marries a normal male.

What proportion of their female progeny will show the trait?

A. All
B. 1/2
C. 1/4
D. 0
E. 3/4

Answer 36

D. 0

Half the sons would be expected to inherit the color blindness allele from their mother. Half the daughters would be carriers like their mothers.

Question 37

The alleles for eye color and for body color are on the X chromosome of Drosophila, but not on the Y. Red eye color (w+) is dominant to white eye color (w), and tan body color (y+) is dominant to yellow body color (y).

What is the genotype of a yellow-bodied, red-eyed female who is homozygous for eye color?

A. xw+ y xw y+
B. xw y xw+ y
C. xw y xw y
D. xw+ y xw+ y
E. xw y+ xw y+

Answer 37

D. xw+ y xw+ y

The female with red eyes and a yellow body must be homozygous recessive for the yellow body color allele. She is homozygous dominant for the eye color allele.

Question 38

The alleles for eye color and for body color are on the X chromosome of Drosophila, but not on the Y. Red eye color (w+) is dominant to white eye color (w), and tan body color (y+ ) is dominant to yellow body color (y).
What is the genotype of a tan-bodied, white-eyed male?

A. Xw+y+ Y
B. Xwy+ Y
C. Xwy Y
D. Xwy Xwy
E. Xw+y Y

Answer 38

B. Xwy+ Y

The male is hemizygous for both the recessive (white) eye color allele and dominant (tan) body color allele.
His genotype is Xwy+ Y.

Question 39

What offspring would you expect from a cross between the female Drosophila described in problem 1 (red eyes and a yellow body, homozygous recessive for the yellow body color allele and homozygous dominant for the eye color allele) and the male described in problem 2 (hemizygous for both the recessive (white) eye color allele and dominant (tan) body color allele?)

A reminder that the alleles for eye color and for body color are on the X chromosome of Drosophila, but not on the Y. Red eye color (w+) is dominant to white eye color (w), and tan body color (y+) is dominant to yellow body color (y).

A. Daughters would be yellow-bodied, red-eyed; the sons would be tan-bodied, white-eyed
B. Daughters would be tan-bodied, red-eyed; the sons would be yellow-bodied, white-eyed
C. Daughters would be tan-bodied, red-eyed; the sons would be yellow-bodied, red-eyed
D. Daughters would be yellow-bodied, white-eyed; the sons would be tan-bodied, white-eyed
E. Daughters would be yellow-bodied, red-eyed; the sons would be tan-bodied, red-eyed

Answer 39

C. Daughters would be tan-bodied, red-eyed; the sons would be yellow-bodied, red-eyed

Question 40

If we mated the F1 female and male flies from the cross obtained in problem 3, what male phenotype in the F2 generation would be evidence that crossing over had occured during gamete formation?

Daughters were tan-bodied, red-eyed, heterozygous for both eye and body color. The sons were yellow-bodied, red-eyed hemizygous.

A. White eyes and Tanned body
B. Red eyes and Tanned body
C. Red eyes and Yellow body
D. White eyes and Yellow body and Red eyes and Tanned body
E. White eyes and Yellow body

Answer 40

D. White eyes and yellow bodies and red eyes and tan bodies

Either white eyes and yellow bodies or red eyes and tan bodies are combinations of alleles on the X chromosome that did not occur in either original parent. This is evidence for crossing over of the X-chromosome in the F1 females.