1. Mendel and the Gene Idea11
Mendel and the Gene Idea
Questions prepared by
Brad Stith, University of Colorado Denver
Janet Lanza, University of Arkansas at Little Rock
Louise Paquin, McDaniel College

2. two gamete types: white/white and purple/purple
Imagine crossing a pea heterozygous at the loci for flower color (white versus purple) and seed color (yellow versus green) with a second pea homozygous for flower color (white) and seed color (yellow). What types of gametes will the first pea 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.

3. two gamete types: white/white and purple/purple
Imagine crossing a pea heterozygous at the loci for flower color (white versus purple) and seed color (yellow versus green) with a second pea homozygous for flower color (white) and seed color (yellow). What types of gametes will the first pea 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 3

4. it is possible to control matings between different pea plants.
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 4

5. it is possible to control matings between different pea plants.
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. 5

6. the blending model of genetics. true breeding. dominance.
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. the blending model of genetics. true breeding. dominance.
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. One normal allele produces as much melanin as two normal alleles.
Albinism in humans occurs when both alleles at a locus produce defective enzymes in the biochemical pathway leading to melanin. Given that heterozygotes are normally pigmented, which of the following statements is/are correct?
One normal allele produces as much melanin as two normal alleles.
Each defective allele produces a little bit of melanin.
Two normal alleles are needed for normal melanin production.
The two alleles are codominant.
The amount of sunlight will not affect skin color of heterozygotes.
Answer: A
This question is an attempt to help students understand how enzyme action can produce phenotypes. If a person with two defective alleles at a locus in the melanin production pathway is albino, no melanin is produced. However, a person with only one defective allele is normally pigmented, meaning that the person has a normal amount of melanin. Answer A is correct—one functional allele provides a normal amount of pigment. Answer B is wrong because albinos produce no melanin. Answer C is wrong because it contradicts the statement that heterozygotes are normally pigmented; if this statement were true, heterozygotes would have an intermediate amount of pigment. Answer D is wrong because a heterozygote in a codominance situation will have a different phenotype from both heterozygotes. Answer E is wrong because if heterozygotes have a normal amount of melanin, they will tan (produce more melanin when exposed to UV rays). This question simplifies the inheritance of albinism and its phenotypes. 10

11. One normal allele produces as much melanin as two normal alleles.
Albinism in humans occurs when both alleles at a locus produce defective enzymes in the biochemical pathway leading to melanin. Given that heterozygotes are normally pigmented, which of the following statements is/are correct?
One normal allele produces as much melanin as two normal alleles.
Each defective allele produces a little bit of melanin.
Two normal alleles are needed for normal melanin production.
The two alleles are codominant.
The amount of sunlight will not affect skin color of heterozygotes. 11

12. 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. Thinking about enzyme action, 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 12

13. 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. Thinking about enzyme action, 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. 13

14. Dark hair alleles are equally common in all parts of Europe.
In humans, alleles for dark hair are genetically dominant, while alleles for light hair are recessive. Which of the following statements is/are most likely to be correct?
Dark hair alleles are more common than light hair alleles in all areas of Europe.
Dark hair alleles are more common than light hair alleles in southern Europe but not in northern Europe.
Dark hair alleles are equally common in all parts of Europe.
Dark hair is dominant to light hair in southern Europe but recessive to light hair in northern Europe.
Dark hair is dominant to light hair in northern Europe but recessive to light hair in southern Europe.
Answer: B
This question confronts a widespread misconception. Many people think (incorrectly) that the genetically dominant allele will be most frequent and that dominant alleles increase in frequency within populations over time. The Hardy-Weinberg principle shows that this is not the case; dominance and frequency are not related. Allele frequency of dominant alleles can be low (dark hair color in northern Europe) or high (dark hair color in southern Europe). Answer B is the only answer that reflects this pattern. Even though the chapter does not address population genetic ideas, it’s a good idea to try to break this misconception early. 14

15. Dark hair alleles are equally common in all parts of Europe.
In humans, alleles for dark hair are genetically dominant, while alleles for light hair are recessive. Which of the following statements is/are most likely to be correct?
Dark hair alleles are more common than light hair alleles in all areas of Europe.
Dark hair alleles are more common than light hair alleles in southern Europe but not in northern Europe.
Dark hair alleles are equally common in all parts of Europe.
Dark hair is dominant to light hair in southern Europe but recessive to light hair in northern Europe.
Dark hair is dominant to light hair in northern Europe but recessive to light hair in southern Europe. 15

16. 25% DaDc, 25% DaDd, 25% DbDc, 25% DbDd 50% DaDb, 50% DcDd
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. 16

17. 25% DaDc, 25% DaDd, 25% DbDc, 25% DbDd 50% DaDb, 50% DcDd
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 17

18. Imagine a family with two parents who both maintain low fat levels through a combination of aerobic activity and weight training. Which of the following statements is/are most likely to apply to their two children?
The parents’ fat levels are irrelevant to the fat levels of the children.
One child is likely to have low fat levels but the other is more likely to have high fat levels because of independent assortment of genes.
The children may not have the same fat levels as their parents because genes independently assort during meiosis.
Answer: Several answers are possible because both genes (nature) and environment (nurture) interact to produce the phenotype. However, none of these three answers is “more right” than the others. Genes certainly affect a person’s basal metabolism (although specific genes are not yet identified). Exercise level will also affect a person’s weight. Furthermore, other factors (e.g., total food intake, choice of food types, and even rebellious attitudes toward their parents) can also affect fat levels. After discussion of this question, a faculty member might ask students to write one good answer. One possible, though obvious, answer would be, “Fat levels of the children will depend on the genes they inherit from their parents, their activity levels, and their eating habits.” 18

19. Envision a family in which the grandfather, age 47, has just been diagnosed with Huntington’s disease, which is caused by a dominant allele (and the father is a heterozygote). 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
Concept 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. 19

20. Envision a family in which the grandfather, age 47, has just been diagnosed with Huntington’s disease, which is caused by a dominant allele (and the father is a heterozygote). 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% 20

21. Review the family described in the previous question
Review the family described in the previous question. What is the probability that the baby will contract the disease? 0%
25%
50%
75%
100%
Answer: B
This is the second of three questions on Huntington’s disease. To choose the correct answer (B), students must realize that the probability of the daughter getting the Huntington allele is (as in the previous question) 50%. If she has the allele, the probability of her son getting the allele from her is also 50%. To calculate the probability of two independent events both occurring, one must multiply the probabilities. Thus, the grandson’s probability of inheriting the allele from his grandfather is (0.5)(0.5) = 0.25 = 25%. 21

22. Review the family described in the previous question
Review the family described in the previous question. What is the probability that the baby will contract the disease? 0%
25%
50%
75%
100% 22

23. Imagine that you are the daughter in the family described in the previous questions. You had been planning on having a second child. What kind of choices would you make about genetic testing, for yourself and for your child?
Answer: This last question in the Huntington’s disease sequence describes a situation that has applied to many families. Part of the tragedy of Huntington’s disease is that it is a gradual degeneration for which there is no cure. However, some of the tragedy of Huntington’s disease is that it is caused by a dominant allele (though the severity of the disease varies among people) and that the onset of the disease often occurs only after a person has had his or her children and even grandchildren. There is no right answer to this question, and the question can lead into a discussion of ethics (for example, some people advocate that minors should not be tested, and many people argue that the results of genetic testing should not affect insurance rates or employment opportunities). One possible option is to become pregnant, have the fetus tested, and abort an affected fetus. A good site for information is 23

24. When a disease is said to have a multifactorial basis, it means that
both genetic and environmental factors contribute to the disease.
it is caused by a gene with a large number of alleles.
it affects a large number of people.
it has many different symptoms.
it tends to skip a generation.
Answer: A 24

25. When a disease is said to have a multifactorial basis, it means that
both genetic and environmental factors contribute to the disease.
it is caused by a gene with a large number of alleles.
it affects a large number of people.
it has many different symptoms.
it tends to skip a generation. 25