1. 11
Mendel and the Gene Idea

2. Imagine crossing a pea heterozygous at the loci for flower color (Pp) and seed color (Yy) with a second pea homozygous for flower color (pp) and seed color (yy). What genotypes of gametes will the first pea produce?
two gamete types: pp and Pp
two gamete types: pY and Py
four gamete types: pY, py, PY, and Py
four gamete types: Pp, Yy, pp, PP
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.
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3. Imagine crossing a pea heterozygous at the loci for flower color (Pp) and seed color (Yy) with a second pea homozygous for flower color (pp) and seed color (yy). What genotypes of gametes will the first pea produce?
two gamete types: pp and Pp
two gamete types: pY and Py
four gamete types: pY, py, PY, and Py
four gamete types: Pp, Yy, pp, PP
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4. 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
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5. 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.
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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
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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.
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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 three 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.
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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 three 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.”
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10. 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 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.
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11. 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 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.
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12. In humans, alleles for dark hair are genetically dominant, while alleles for light hair are recessive. Dark hair is also more prevalent in southern Europe than in northern Europe. Which of the following statements about hair color alleles is 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.
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13. In humans, alleles for dark hair are genetically dominant, while alleles for light hair are recessive. Dark hair is also more prevalent in southern Europe than in northern Europe. Which of the following statements about hair color alleles is 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.
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14. Imagine a locus with four different alleles for fur color in an animal, 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.
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15. Imagine a locus with four different alleles for fur color in an animal, 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
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16. Envision a family in which a man, age 47, has just been diagnosed with Huntington’s disease, which is caused by a dominant allele (and the man is a heterozygote). His daughter, age 25, 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.
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17. Envision a family in which a man, age 47, has just been diagnosed with Huntington’s disease, which is caused by a dominant allele (and the man is a heterozygote). His daughter, age 25, 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%
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18. Review the family described in the previous question
Review the family described in the previous question. What is the probability that the 2-year-old son 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%.
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19. Review the family described in the previous question
Review the family described in the previous question. What is the probability that the 2-year-old son will contract the disease? 0%
25%
50%
75%
100%
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20. 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
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21. 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
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22. 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.
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23. Which of the following observations supported Mendel’s hypothesis that inheritance is “particulate” rather than due to blending?
There are two distinct flower colors in pea plants.
White-flowered plants are true-breeding.
Crossing true-breeding purple-flowered and white-flowered plants produced all purple-flowered plants.
Crossing two purple-flowered heterozygotes produced purple-flowered and white-flowered plants.
Answer: D. This is a fundamental and often underappreciated observation that does not even require looking at ratios of offspring. The white allele can remain intact, although hidden, in the purple heterozygotes.
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24. Which of the following observations supported Mendel’s hypothesis that inheritance is “particulate” rather than due to blending?
There are two distinct flower colors in pea plants.
White-flowered plants are true-breeding.
Crossing true-breeding purple-flowered and white-flowered plants produced all purple-flowered plants.
Crossing two purple-flowered heterozygotes produced purple-flowered and white-flowered plants.
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25. Manx cats have characteristic stubby tails due to being heterozygous at a single locus. Homozygotes for the Manx allele die before birth with severe spinal deformities. What is the expected phenotypic ratio of live offspring of two Manx cats?
3 normal:1 Manx
3 Manx:1 normal
2 normal:1 Manx
2 Manx:1 normal
Answer: D. This is a good question to set up lethal genes and their characteristic 2:1 ratio. It is also a good segue for a history lesson on Cuenot, Castle, and Little’s discovery of lethal genes.
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26. Manx cats have characteristic stubby tails due to being heterozygous at a single locus. Homozygotes for the Manx allele die before birth with severe spinal deformities. What is the expected phenotypic ratio of live offspring of two Manx cats?
3 normal:1 Manx
3 Manx:1 normal
2 normal:1 Manx
2 Manx:1 normal
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27. The allele for inflated pods (I) is dominant to that for constricted pods (i), and the allele for green pods (G) is dominant to that for yellow pods (g). A plant of unknown genotype with inflated green pods is crossed with a plant with constricted yellow pods. Among the offspring are 49 plants with inflated green pods, and 53 plants with constricted green pods. What was the previously unknown genotype?
IiGg
IiGG
IIGg
IIGG
Answer: B. This is a straightforward application of the laws of segregation and independent assortment. It is good for showing that that loci truly are independent in terms of genotype (some can be homozygous, while others are heterozygous).
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28. The allele for inflated pods (I) is dominant to that for constricted pods (i), and the allele for green pods (G) is dominant to that for yellow pods (g). A plant of unknown genotype with inflated green pods is crossed with a plant with constricted yellow pods. Among the offspring are 49 plants with inflated green pods, and 53 plants with constricted green pods. What was the previously unknown genotype?
IiGg
IiGG
IIGg
IIGG
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29. Imagine a cross of two triple heterozygous pea plants with purple flowers and yellow round seeds (genotype PpYyRr). If you were to create a Punnett square for this cross (not recommended!) what would be its dimensions? Recall that all three loci assort independently.
3 × 3
4 × 4
6 × 6
8 × 8
Answer: D. This is a good question to connect meiosis with the process of solving genetics problems. Students should be able to reason that independent assortment will create 8 different kinds of gametes from each parent. A very good follow-up exercise is to have students actually draw the 4 different arrangements at metaphase I and write the 8 different gamete genotypes.
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30. Imagine a cross of two triple heterozygous pea plants with purple flowers and yellow round seeds (genotype PpYyRr). If you were to create a Punnett square for this cross (not recommended!) what would be its dimensions? Recall that all three loci assort independently.
3 × 3
4 × 4
6 × 6
8 × 8
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31. Imagine the same cross of two triple heterozygous pea plants with purple flowers and yellow round seeds (genotype PpYyRr). Using the rules of probability (and not a Punnett square!), determine what proportion of offspring will have purple flowers and green wrinkled seeds.
¼ × ¼ × ¼ = 1/64
¾ + ¼ + ¼ = 5/4
¾ × ¼ × ¼ = 3/64
½ × ½ × ½ = 1/8
Answer: C. This is a good question for reinforcing the multiplication rule. Also, in tandem with the preceding question, students can see that the denominator is the same as the total number of squares in the 8 8 Punnett. They can “see” how using the rules of probability is much faster than the Punnett square and gives the same answer.
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32. Imagine the same cross of two triple heterozygous pea plants with purple flowers and yellow round seeds (genotype PpYyRr). Using the rules of probability (and not a Punnett square!), determine what proportion of offspring will have purple flowers and green wrinkled seeds.
¼ × ¼ × ¼ = 1/64
¾ + ¼ + ¼ = 5/4
¾ × ¼ × ¼ = 3/64
½ × ½ × ½ = 1/8
© 2016 Pearson Education, Inc.

33. The Rh blood factor in humans is perhaps the most important after the ABO system. The + phenotype is dominant to the – phenotype, and it is encoded on a separate autosome from the ABO locus (i.e., the two loci assort independently). A child of blood type A+ is born to a mother of blood type O. What can be concluded about the blood type of the biological father?
A+ or AB+
A+ only
AB+ only
any possible blood type
Answer: A. This question is good for working with multiple alleles and independent assortment simultaneously.
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34. The Rh blood factor in humans is perhaps the most important after the ABO system. The + phenotype is dominant to the – phenotype, and it is encoded on a separate autosome from the ABO locus (i.e., the two loci assort independently). A child of blood type A+ is born to a mother of blood type O. What can be concluded about the blood type of the biological father?
A+ or AB+
A+ only
AB+ only
any possible blood type
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35. Assume there are 50 people in the classroom
Assume there are 50 people in the classroom. Theoretically speaking, what is the maximum number of different alleles there could be at a hypothetical autosomal locus? 2 3 50
100
Answer: D. This question is a good follow-up to ABO blood type examples. It can be adapted to any size classroom.
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36. Assume there are 50 people in the classroom
Assume there are 50 people in the classroom. Theoretically speaking, what is the maximum number of different alleles there could be at a hypothetical autosomal locus? 2 3 50
100
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37. Consider the case of dominant epistasis in squash, where YY and Yy = yellow and yy = green. At a second, independently assorting locus, WW and ww = white, and ww allows for yellow or green. What is the expected phenotypic ratio among the progeny of the cross YyWw × YyWw? (Note these are white squash.)
1:1:1:1
9:3:3:1
9:4:3
12:3:1
Answer: D. This is a good question for understanding epistasis and using the multiplication rule.
¾ = white (WW and Ww)
¼ = ww, of which ¾ are yellow and ¼ are green.
12: 3: 1
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38. Consider the case of dominant epistasis in squash, where YY and Yy = yellow and yy = green. At a second, independently assorting locus, WW and ww = white, and ww allows for yellow or green. What is the expected phenotypic ratio among the progeny of the cross YyWw × YyWw? (Note these are white squash.)
1:1:1:1
9:3:3:1
9:4:3
12:3:1
© 2016 Pearson Education, Inc.

39. Regarding the simplified model for polygenic inheritance of human skin color presented in the text, since there are more than seven skin color phenotypes in humans, what modification(s) could make the model more complete? More than one may apply.
more loci
more alleles per locus
more individuals in the population
differing levels of effect per allele
differing levels of environmental exposure
Answers: A, B, and D
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40. Regarding the simplified model for polygenic inheritance of human skin color presented in the text, since there are more than seven skin color phenotypes in humans, what modification(s) could make the model more complete? More than one may apply.
more loci
more alleles per locus
more individuals in the population
differing levels of effect per allele
differing levels of environmental exposure
© 2016 Pearson Education, Inc.