1. Mendel and the Gene Idea

2. Pea plants were particularly well suited for use in Mendel’s breeding experiments for all of the following reasons except that
peas show easily observed variations in a number of characters, such as pea shape and flower color.
it is possible to control matings between different pea plants.
it is possible to obtain large numbers of progeny from any given cross.
peas have an unusually long generation time.
many of the observable characters that vary in pea plants are controlled by single genes.
Answer: D 2

3. Pea plants were particularly well suited for use in Mendel’s breeding experiments for all of the following reasons except that
peas show easily observed variations in a number of characters, such as pea shape and flower color.
it is possible to control matings between different pea plants.
it is possible to obtain large numbers of progeny from any given cross.
peas have an unusually long generation time.
many of the observable characters that vary in pea plants are controlled by single genes. 3

4. A pea plant is heterozygous at the independent loci for flower color (white versus purple) and seed color (yellow versus green). What types of gametes can it produce?
two gamete types: white/white and purple/purple
two gamete types: white/yellow and purple/green
four gamete types: white/yellow, white/green, purple/yellow, and purple/green
four gamete types: white/purple, yellow/green, white/white, and purple/purple
one gamete type: white/purple/yellow/green
Answer: C
The purpose of this question is to help students figure out gamete types—a step they often rush past. Students need to realize that gametes are haploid and that each gamete contains one, and only one, allele for each of the traits being studied. They also need to know that one allele of one gene will be present with one allele of the other gene. In this case, we are emphasizing that two alleles and two genes are present, and students would use a Punnett square to pair up two alleles (each from a different gene). The presence of the second pea in the question stem is a distracter. Answer A is incorrect because there are four gamete types, and it shows gametes with two alleles for flower color and no alleles for seed color. Answer B is incorrect because it shows only two possible gamete types. Answer C is correct because it shows all four possible gamete types. Answer D is incorrect because it shows gametes that are diploid for one trait and containing an allele for only one locus: white and purple is not possible, and so on. Answer E is incorrect because it shows only one gamete type and a gamete diploid for both loci.

5. A pea plant is heterozygous at the independent loci for flower color (white versus purple) and seed color (yellow versus green). What types of gametes can it produce?
two gamete types: white/white and purple/purple
two gamete types: white/yellow and purple/green
four gamete types: white/yellow, white/green, purple/yellow, and purple/green
four gamete types: white/purple, yellow/green, white/white, and purple/purple
one gamete type: white/purple/yellow/green
Answer: C
The purpose of this question is to help students figure out gamete types—a step they often rush past. Students need to realize that gametes are haploid and that each gamete contains one, and only one, allele for each of the traits being studied. They also need to know that one allele of one gene will be present with one allele of the other gene. In this case, we are emphasizing that two alleles and two genes are present, and students would use a Punnett square to pair up two alleles (each from a different gene). The presence of the second pea in the question stem is a distracter. Answer A is incorrect because there are four gamete types, and it shows gametes with two alleles for flower color and no alleles for seed color. Answer B is incorrect because it shows only two possible gamete types. Answer C is correct because it shows all four possible gamete types. Answer D is incorrect because it shows gametes that are diploid for one trait and containing an allele for only one locus: white and purple is not possible, and so on. Answer E is incorrect because it shows only one gamete type and a gamete diploid for both loci.

6. A cross between homozygous purple-flowered and homozygous white-flowered pea plants results in offspring with purple flowers. This demonstrates
the blending model of genetics.
true breeding.
dominance.
a dihybrid cross.
the mistakes made by Mendel.
Answer: C 6

7. A cross between homozygous purple-flowered and homozygous white-flowered pea plants results in offspring with purple flowers. This demonstrates
the blending model of genetics.
true breeding.
dominance.
a dihybrid cross.
the mistakes made by Mendel. 7

8. Imagine a genetic counselor working with a couple who have just had a child who is suffering from Tay-Sachs disease. Neither parent has Tay-Sachs, nor does anyone in their families. Which of the following statements should this counselor make to this couple?
“Because no one in either of your families has Tay-Sachs, you are not likely to have another baby with Tay-Sachs. You can safely have another child.”
“Because you have had one child with Tay-Sachs, you must each carry the allele. Any child you have has a 50% chance of having the disease.”
“Because you have had one child with Tay-Sachs, you must each carry the allele. Any child you have has a 25% chance of having the disease.”
“Because you have had one child with Tay-Sachs, you must both carry the allele. However, since the chance of having an affected child is 25%, you may safely have thee more children without worrying about having another child with Tay-Sachs.”
“You must both be tested to see who is a carrier of the Tay-Sachs allele.”
Answer: C
The intent of the question is to help students understand that Tay-Sachs is a genetic disease caused by a deleterious recessive allele. It also should help students realize that, with independent events, prior events have no effect on subsequent events. The parents are both heterozygotes and students can construct a Punnett square. Answer A is likely incorrect because the fact that the couple had one child with Tay-Sachs indicates that each parent has an allele for Tay-Sachs (the one exception is if a new mutation, a rare event, occurred during the meiotic events leading up to the production of one of the gametes that formed the first child). In answer B, the first sentence is correct, but this answer is incorrect because each child has a 25% probability of having the disease (it would be 50% if the allele were dominant). Answer C is correct. Answer D is incorrect because each conception must be viewed as an independent event. If this rationale were true, every family that had a girl would next have a boy—and we’ve all seen examples of families with two girls or two boys. Answer E is incorrect because Tay-Sachs is recessive and a person must have two copies of the allele to have the disease. 8

9. Imagine a genetic counselor working with a couple who have just had a child who is suffering from Tay-Sachs disease. Neither parent has Tay-Sachs, nor does anyone in their families. Which of the following statements should this counselor make to this couple?
“Because no one in either of your families has Tay-Sachs, you are not likely to have another baby with Tay-Sachs. You can safely have another child.”
“Because you have had one child with Tay-Sachs, you must each carry the allele. Any child you have has a 50% chance of having the disease.”
“Because you have had one child with Tay-Sachs, you must each carry the allele. Any child you have has a 25% chance of having the disease.”
“Because you have had one child with Tay-Sachs, you must both carry the allele. However, since the chance of having an affected child is 25%, you may safely have thee more children without worrying about having another child with Tay-Sachs.”
“You must both be tested to see who is a carrier of the Tay-Sachs allele.” 9

10. Imagine a locus with four different alleles for fur color in an animal
Imagine a locus with four different alleles for fur color in an animal. The alleles are named Da, Db, Dc, and Dd. If you crossed two heterozygotes, DaDb and DcDd, what genotype proportions would you expect in the offspring?
25% DaDc, 25% DaDd, 25% DbDc, 25% DbDd
50% DaDb, 50% DcDd
25% DaDa, 25% DbDb, 25% DcDc, 25% DdDdDcDd
50% DaDc, 50% DbDd
25% DaDb, 25% DcDd, 25% DcDc, 25% DdDd
Answer: A
This question is designed to reinforce the idea that diploid organisms have two, but only two, copies of alleles for each locus. The existence of four alleles will confuse some students but is not an uncommon situation. To answer this question, students need to figure out gametes produced by the parents (Da or Db for the first parent and Dc or Dd for the second parent). Then using a Punnett square, they will get 25% each of four genotypes, DaDc, DaDd, DbDc, and DbDd. The correct answer is A. All offspring should have two alleles, and there cannot be any homozygotes from this set of parents. 10

11. Imagine a locus with four different alleles for fur color in an animal
Imagine a locus with four different alleles for fur color in an animal. The alleles are named Da, Db, Dc, and Dd. If you crossed two heterozygotes, DaDb and DcDd, what genotype proportions would you expect in the offspring?
25% DaDc, 25% DaDd, 25% DbDc, 25% DbDd
50% DaDb, 50% DcDd
25% DaDa, 25% DbDb, 25% DcDc, 25% DdDdDcDd
50% DaDc, 50% DbDd
25% DaDb, 25% DcDd, 25% DcDc, 25% DdDd 11

12. John, age 47, has just been diagnosed with Huntington’s disease, which is caused by a dominant allele. His daughter, age 25, now has a 2-year-old son. No one else in the family has the disease. What is the probability that the daughter will contract the disease? 0%
25%
50%
75%
100%
Answer: C
This question begins a series of three questions about Huntington’s disease and seeks to make sure that students understand that the daughter has a 50% probability of inheriting the normal allele (normal phenotype) and a 50% probability of inheriting the Huntington allele (Huntington phenotype). The question will also help students understand the implications of the fact that the Huntington allele is dominant. When the students construct a Punnett square, because no one in the family has the disease, they will note that the mother is probably homozygous recessive, the grandfather is heterozygous with the dominant Huntington’s allele, and the father of the boy is probably homozygous recessive. 12

13. John, age 47, has just been diagnosed with Huntington’s disease, which is caused by a dominant allele. His daughter, age 25, now has a 2-year-old son. No one else in the family has the disease. What is the probability that the daughter will contract the disease? 0%
25%
50%
75%
100% 13

14. An individual with the genotype AaBbEeHH is crossed with an individual who is aaBbEehh. What is the likelihood of having offspring with the genotype AabbEEHh?
1/8
1/16
1/32
1/64
That genotype would be impossible.
Answer: C

15. An individual with the genotype AaBbEeHH is crossed with an individual who is aaBbEehh. What is the likelihood of having offspring with the genotype AabbEEHh?
1/8
1/16
1/32
1/64
That genotype would be impossible.

16. ABO blood type in humans exhibits codominance and multiple alleles
ABO blood type in humans exhibits codominance and multiple alleles. What is the likelihood of a type A father and a type A mother having a type O child?
It is impossible.
25% if both parents are heterozygous
50% if both parent are heterozygous
25% if only the father is heterozygous
25% if only the mother is heterozygous
Answer: B

17. ABO blood type in humans exhibits codominance and multiple alleles
ABO blood type in humans exhibits codominance and multiple alleles. What is the likelihood of a type A father and a type A mother having a type O child?
It is impossible.
25% if both parents are heterozygous
50% if both parent are heterozygous
25% if only the father is heterozygous
25% if only the mother is heterozygous

18. If roan cattle (incomplete dominance) are allowed to breed, what ratio of phenotypes is expected in the offspring?
1:1 red:white
all roan
1:2:1 red:roan:white
3:1 red:white
1:1:1 red:roan:white
Answer: C

19. If roan cattle (incomplete dominance) are allowed to breed, what ratio of phenotypes is expected in the offspring?
1:1 red:white
all roan
1:2:1 red:roan:white
3:1 red:white
1:1:1 red:roan:white

20. The phenotype will probably be yellow but cannot be red.
Imagine that the last step in a biochemical pathway to the red skin pigment of an apple is catalyzed by enzyme X, which changes compound C to compound D. If an effective enzyme is present, compound D is formed and the apple skin is red. However, if the enzyme is not effective, only compound C is present and the skin is yellow. What can you accurately say about a heterozygote with one allele for an effective enzyme X and one allele for an ineffective enzyme X?
The phenotype will probably be yellow but cannot be red.
The phenotype will probably be red but cannot be yellow.
The phenotype will be a yellowish red.
The phenotype will be either yellow or red.
The phenotype will be either yellowish red or red.
Answer: E
This question again tries to connect enzyme action to phenotype. There is a yellow/red polymorphism in apple skin that appears to be controlled by one locus. Answers A and D are wrong because a heterozygote will produce some red pigmentation, so it cannot be yellow. Answers B and C are possible, but there is not enough information given to know which is more likely. Answer E is the best answer because yellowish red or red are possible—yellowish red if a heterozygote produces less pigment than the red homozygote, or red if a heterozygote produces as much pigment as a purple homozygote. Students with more sophistication may suggest that the phenotype would be a patchwork of yellow and red cells—red where the purple allele is “turned on” and yellow where the yellow allele is “turned on,” but this is not what happens in the apples. You can read about one gene involved in the pathway at 20

21. The phenotype will probably be yellow but cannot be red.
Imagine that the last step in a biochemical pathway to the red skin pigment of an apple is catalyzed by enzyme X, which changes compound C to compound D. If an effective enzyme is present, compound D is formed and the apple skin is red. However, if the enzyme is not effective, only compound C is present and the skin is yellow. What can you accurately say about a heterozygote with one allele for an effective enzyme X and one allele for an ineffective enzyme X?
The phenotype will probably be yellow but cannot be red.
The phenotype will probably be red but cannot be yellow.
The phenotype will be a yellowish red.
The phenotype will be either yellow or red.
The phenotype will be either yellowish red or red. 21

22. Examine this genetic pedigree
Examine this genetic pedigree. What mode of inheritance does this trait follow?
autosomal dominant
autosomal recessive
sex-linked recessive
mitochondrial
Answer: A

23. Examine this genetic pedigree
Examine this genetic pedigree. What mode of inheritance does this trait follow?
autosomal dominant
autosomal recessive
sex-linked recessive
mitochondrial

24. Examine this genetic pedigree
Examine this genetic pedigree. What mode of inheritance does this trait follow?
autosomal dominant
autosomal recessive
sex-linked recessive
mitochondrial
Answer: B

25. Examine this genetic pedigree
Examine this genetic pedigree. What mode of inheritance does this trait follow?
autosomal dominant
autosomal recessive
sex-linked recessive
mitochondrial

26. The following offspring were observed from many crossings of the same pea plants. What genotypes were the parents? 465 purple axial flowers 152 purple terminal flowers 140 white axial flowers 53 white terminal flowers
PpAa x PpAA
PpAa x ppAA
PPAA x ppaa
PpAa x PpAa
PPaa x ppAA
Answer: D

27. The following offspring were observed from many crossings of the same pea plants. What genotypes were the parents? purple axial flowers 152 purple terminal flowers white axial flowers 53 white terminal flowers
PpAa x PpAA
PpAa x ppAA
PPAA x ppaa
PpAa x PpAa
PPaa x ppAA