Genetics A Conceptual Approach 5th Edition by Benjamin A. Pierce Test bank

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Genetics A Conceptual Approach 5th Edition by Benjamin A. Pierce Test bank

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Test Bank for
Chapter 3: Basic Principles of Heredity

Multiple Choice Questions

1. Why was the pea plant an ideal plant for Mendel to use?

a. Generation time that is several years
b. Simple traits that are easy to identify
c. Low numbers of offspring produced
d. Expensive and time-consuming to grow
e. All of the above

Answer: b
Section 3.1
Comprehension

2. Genes come in different versions called

a. alleles.
b. loci.
c. genotypes.
d. chromosomes.
e. genomes.

Answer: a
Section 3.1
Comprehension

3. Which of the following statements is true?

a. The genotype is the physical appearance of a trait.
b. Alleles, genes, and loci are different names for the same thing.
c. The phenotype of a dominant allele is never seen in the F1 progeny of a monohybrid cross.
d. A testcross can be used to determine whether an individual is homozygous or heterozygous.
e. All of these statements are true.

Answer: d
Section 3.2
Comprehension

4. Gregor Mendel carried out a cross between two pea plants by taking pollen from a plant that was homozygous for round seeds and dusting the pollen onto the stigma of a plant homozygous for wrinkled seeds. Which of the following would be the reciprocal cross that Mendel had carried out for this experiment?

a. Homozygous round stigma pollinated with homozygous wrinkled
b. Homozygous round stigma pollinated with heterozygous wrinkled
c. Heterozygous round stigma pollinated with homozygous wrinkled
d. Homozygous wrinkled stigma pollinated with homozygous round
e. Homozygous wrinkled stigma pollinated with homozygous wrinkled

Answer: a
Section 3.2
Comprehension

5. Which of the following statements is true?

a. The probability of a woman giving birth to three girls in a row is 1/8.
b. The chi-square test is used to determine if observed outcomes are consistent with expected outcomes.
c. The probability of two or more independent events occurring together is calculated by multiplying their independent probabilities.
d. Branched diagrams are used for determining probabilities of various phenotypes or genotypes for genetic crosses involving more than one gene pair.
e. All of these statements are true.

Answer: e
Section 3.2
Comprehension

6. In Labrador retrievers, black coat color is dominant to brown. Suppose that a black Lab is mated with a brown one and the offspring are 4 black puppies and 1 brown. What can you conclude about the genotype of the black parent?

a. The genotype must be BB.
b. The genotype must be bb.
c. The genotype must be Bb.
d. The genotype could be either BB or Bb.
e. The genotype cannot be determined from these data.

Answer: c
Section 3.2
Comprehension

7. In Mendels peas, purple flower color is dominant to white. From which of the following descriptions can you not infer the genotype completely?

a. Purple
b. White
c. Pure-breeding purple
d. Heterozygous
e. More than one of the above

Answer: a
Section 3.2
Comprehension

8. Which of the following was not one of Mendels conclusions based on his monohybrid crosses?

a. Genes are carried on chromosomes.
b. Alleles exist in pairs.
c. Alleles segregate equally into gametes.
d. Alleles behave as particles during inheritance.
e. One allele can mask the expression of the other allele.

Answer: a
Section 3.2
Comprehension

9. In Mendels peas, yellow seeds are dominant to green. A pure-breeding yellow plant is crossed with a pure-breeding green plant. All of the offspring are yellow. If one of these yellow offspring is crossed with a green plant, what will be the expected proportion of plants with green seeds in the next generation?

a. 0%
b. 25%
c. 50%
d. 75%
e. 100%

Answer: c
Section 3.2
Comprehension

10. In a cross between pure-breeding tall plants with pure-breeding short plants, all of the F1 are tall. When these plants are allowed to fertilize themselves, the F2 plants occur in a ratio of 3 tall:1 short. Which of the following is not a valid conclusion from these results?

a. The allele that produces the tall condition is dominant to the allele that produces the short condition.
b. The difference between tall and short stature in these lines is controlled by a single gene pair.
c. During production of gametes in F1 plants, the tall and short alleles segregate from each other equally into the gametes.
d. The tall and short traits assort independently of each other in this cross.
e. Fertilization occurs randomly between gametes carrying the tall and short alleles.

Answer: d
Section 3.2
Comprehension

11. In poodles, black fur is dominant to white fur. A black poodle is crossed with a white poodle. In a litter of four, all of the puppies are black. What is the best conclusion?

a. The black poodle is definitely homozygous.
b. The black poodle is probably homozygous.
c. The black poodle is definitely heterozygous.
d. The black poodle is probably heterozygous.
e. The genotype of the black poodle cannot be inferred with this information.

Answer: b
Section 3.2
Comprehension

12. Honeybees have a haplo-diploid sex determination system where females develop from a fertilized egg (they are diploid, having one allele from the female queen and one allele from the male), and males develop from unfertilized eggs (they are haploid, having only one allele from the queen). Assuming that the queen is heterozygous for a particular gene, what is the probability that a female will inherit the recessive allele from her mother? What is the probability that a male will inherit a recessive allele from his mother?

a. The probability that a daughter will inherit a recessive allele from her mother is 50%; the probability that a son will inherit a recessive allele from his mother is 50%.
b. The probability that a daughter will inherit a recessive allele from her mother is 50%; the probability that a son will inherit a recessive allele from his mother is 100%.
c. The probability that a daughter will inherit a recessive allele from her mother is 100%; the probability that a son will inherit a recessive allele from his mother is 50%.
d. The probability that a daughter will inherit a recessive allele from her mother is 100%; the probability that a son will inherit a recessive allele from his mother is 100%.
e. The probability that a daughter will inherit a recessive allele from her mother is 0%; the probability that a son will inherit a recessive allele from his mother is 100%.

Answer: a
Section 3.2
Comprehension

13. While doing field work in Madagascar, you discover a new dragonfly species that has either red (R) or clear (r) wings. Initial crosses indicate that R is dominant to r. You perform three crosses using three different sets of red-winged parents with unknown genotype and observe the following data:

Cross Phenotypes
1 72 red-winged, 24 clear-winged
2 12 red-winged
3 96 red-winged

Which cross is likely to have at least one parent with the genotype RR?

a. Cross 1
b. Cross 2
c. Cross 3
d. Crosses 1 and 2
e. Crosses 2 and 3

Answer: e
Section 3.2
Comprehension

14. A couple has six daughters and is expecting a seventh child. What is the probability that this child will be a boy?

a. 1/2
b. 1/4
c. 1/16
d. 1/64
e. 1/128

Answer: a
Section 3.2
Comprehension

15. If an organism of genotype Aa is used for a test cross, what is the genotype of the other individual used in the cross?

a. AA
b. Aa
c. Aa
d. The genotype cannot be known
e. Either a or b

Answer: c
Section 3.2
Application

16. Which of the following crosses would produce a 1:1 ratio of phenotypes in the next generation?

a. AA AA
b. AA aa
c. Aa Aa
d. Aa aa
e. aa aa

Answer: d
Section 3.2
Application

17. Which of the following crosses would produce a 3:1 ratio of phenotypes in the next generation?

a. AA AA
b. AA aa
c. Aa Aa
d. Aa aa
e. aa aa

Answer: c
Section 3.2
Application

18. Freckles are caused by a dominant allele. A man has freckles but one of his parents does not have freckles. What is his genotype?

a. Homozygous dominant
b. Homozygous recessive
c. Heterozygous
d. Heterologous
e. Homologous

Answer: c
Section 3.2
Application

19. Freckles are caused by a dominant allele. A man has freckles but one of his parents does not have freckles. The man has fathered a child with a woman that does not have freckles. What is the probability that their child has freckles?

a. 1/4
b. 1/3
c. 1/2
d. 2/3
e. 3/4

Answer: c
Section 3.2
Application

20. In animals, the inability to make the pigment melanin results in albinism, a recessive condition. Two unaffected parents, who have decided to have three children, have a first child that has albinism (genotype aa). What is the probability that the second and third children will also have albinism?

a. 1/4
b. 1/2
c. 1/16
d. 9/16
e. 1 (100%)

Answer: c
Section 3.2
Application

21. The ability to curl ones tongue into a U-shape is a genetic trait. Curlers always have at least one curler parent but noncurlers can have one or both parents who are curlers. Using C and c to symbolize the alleles that control this trait, what is the genotype of a noncurler?

a. CC
b. Cc
c. cc
d. Any of the above could be correct.

Answer: c
Section 3.2
Application

22. Two gene loci, A and B, assort independently, and alleles A and B are dominant over alleles a and b. What is the probability of producing an AB gamete from an AaBb individual?

a. 1/4
b. 1/2
c. 1/16
d. 9/16
e. 1 (100%)

Answer: a
Section 3.3
Comprehension

23. Two gene loci, A and B, assort independently, and alleles A and B are dominant over alleles a and b. What is the probability of producing an AB gamete from an AABb individual?

a. 1/4
b. 1/2
c. 1/16
d. 9/16
e. 1 (100%)

Answer: b
Section 3.3
Comprehension

24. Two gene loci, A and B, assort independently, and alleles A and B are dominant over alleles a and b. What is the probability of producing an AABB zygote from a cross AaBb AaBb?

a. 1/4
b. 1/2
c. 1/16
d. 9/16
e. 1 (100%)

Answer: c
Section 3.3
Comprehension

25. Two gene loci, A and B, assort independently, and alleles A and B are dominant over alleles a and b. What is the probability of producing an AaBb zygote from a cross AaBb AABB?

a. 1/4
b. 1/2
c. 1/16
d. 9/16
e. 1 (100%)

Answer: a
Section 3.3
Comprehension

26. Two gene loci, A and B, assort independently, and alleles A and B are dominant over alleles a and b. What is the probability of producing an AB phenotype from a cross AaBb AaBb?

a. 1/4
b. 1/2
c. 1/16
d. 9/16
e. 1 (100%)

Answer: d
Section 3.3
Comprehension

27. Two gene loci, A and B, assort independently, and alleles A and B are dominant over alleles a and b. What is the probability of producing an AB phenotype from a cross aabb AABB?

a. 1/4
b. 1/2
c. 1/16
d. 9/16
e. 1 (100%)

Answer: e
Section 3.3
Comprehension

28. In a cross between AaBbCc and AaBbcc, what proportion of the offspring would be expected to be A_bbcc? (A_ means AA or Aa.)

a. 3/256
b. 3/32
c. 3/16
d. 3/8
e. 3/4

Answer: b
Section 3.3
Comprehension

29. In a cross between AABbCcDD and AaBbccdd, what proportion of the offspring would be expected to be A_B_C_D_? (A_ means AA or Aa.)

a. 3/256
b. 3/32
c. 3/16
d. 3/8
e. 3/4

Answer: d
Section 3.3
Comprehension

30. In a cross between AaBbCcDdEe and AaBbccDdee, what proportion of the offspring would be expected to be A_bbC_ddE_? (A_ means AA or Aa.)

a. 3/256
b. 3/32
c. 3/16
d. 3/8
e. 3/4

Answer: a
Section 3.3
Comprehension

31. Round seeds (R) is dominant to wrinkled seeds (r), and yellow seeds (Y) is dominant to green seeds (y). A true-breeding plant with round and yellow seeds is crossed to a true-breeding plant with wrinkled and green seeds. What is the genotype of the F1 progeny?

a. RRYY
b. RrYY
c. RRYy
d. RrYy
e. rryy

Answer: d
Section 3.3
Application

32. Round seeds (R) is dominant to wrinkled seeds (r), and yellow seeds (Y) is dominant to green seeds (y). A true-breeding pea plant with round and yellow seeds is crossed to a true-breeding plant with wrinkled and green seeds. The F1 progeny are allowed to self-fertilize. What is the probability of obtaining a round, yellow seed in the F2?

a. 3/4
b. 1/16
c. 9/16
d. 3/16
e. 1/2

Answer: c
Section 3.3
Application

33. Round seeds (R) is dominant to wrinkled seeds (r), and yellow seeds (Y) is dominant to green seeds (y). A true-breeding pea plant with round and yellow seeds is crossed to a true-breeding plant with wrinkled and green seeds. The F1 progeny are allowed to self-fertilize. What is the probability of obtaining a wrinkled, green seed in the F2?

a. 3/4
b. 1/16
c. 9/16
d. 3/16
e. 1/2

Answer: b
Section 3.3
Application

34. Round seeds (R) is dominant to wrinkled seeds (r), and yellow seeds (Y) is dominant to green seeds (y). A true-breeding pea plant with round and yellow seeds is crossed to a true-breeding plant with wrinkled and green seeds. The F1 progeny are allowed to self-fertilize. What is the probability of obtaining a yellow seed in the F2?

a. 3/4
b. 1/16
c. 9/16
d. 3/16
e. 1/2

Answer: a
Section 3.3
Application

35. Round seeds (R) is dominant to wrinkled seeds (r), and yellow seeds (Y) is dominant to green seeds (y). A plant of unknown genotype is testcrossed to a true-breeding plant with wrinkled and green seeds. The offspring produced were 53 round and yellow, 49 round and green, 44 wrinkled and yellow, 51 wrinkled and green. What is the likely genotype of the parent in question?

a. RRYY
b. RrYY
c. RRYy
d. RrYy
e. rryy

Answer: d
Section 3.3
Application

36. In dogs, black coat color (B) is dominant over brown (b), and solid coat color (S) is dominant over white spotted coat (s). A cross between a black, solid female and a black, solid male produces only puppies with black, solid coats. This same female was then mated with a brown, spotted male. Have of the offspring from this cross were black and solid, and half of the offspring were black and spotted. What is the genotype of the female?

a. BBSS
b. BbSS
c. BBSs
d. BbSs
e. bbss

Answer: c
Section 3.3
Application

37. In dogs, black coat color (B) is dominant over brown (b), and solid coat color (S) is dominant over white spotted coat (s). A cross between a black, solid female and a black, solid male produces only puppies with black, solid coats. This same female was then mated with a brown, spotted male. Have of the offspring from this cross were black and solid, and half of the offspring were black and spotted. Which of the following could be the genotype of the black, solid male?

a. BBSs
b. BBss
c. BbSS
d. BbSs
e. Bbss

Answer: c
Section 3.3
Application

38. In dogs, black coat color (B) is dominant over brown (b), and solid coat color (S) is dominant over white spotted coat (s). A cross between a black, solid female and a black, solid male produces only puppies with black, solid coats. This same female was then mated with a brown, spotted male. Have of the offspring from this cross were black and solid, and half of the offspring were black and spotted. What is the genotype of the brown, spotted male?

a. BBSS
b. BbSS
c. BBSs
d. BbSs
e. bbss

Answer: e
Section 3.3
Application

39. A space capsule crashes to earth with an alien life form aboard. Two creatures emerge from the capsule, one with green skin and one with yellow skin. The yellow creature soon gives birth to offspring fathered by the green creature, producing 12 green and 8 yellow offspring. Green skin in these diploid creatures is dominant to yellow skin. You are curious to find out if the number of offspring significantly different from expected Mendelian ratios, so you perform a chi-square test. What is the chi-square value for this cross?

a. 0.2
b. 0.4
c. 0.8
d. 1.2
e. 1.6

Answer: c
Section 3.4
Application

40. A chi-square test was performed and indicated that the observed numbers of offspring were significantly different from the expected. Which of the following P-values would support this conclusion?

a. 0.995
b. 0.536
c. 0.024
d. 0.752
e. 0.159

Answer: c
Section 3.4
Application

Short-Answer Questions

41. How did Suttons chromosome theory of inheritance link Mendels work with a more mechanistic understanding of heredity?

Answer: Sutton documented the fact that each homologous pair of chromosomes consists of one maternal chromosome and one paternal chromosome. Showing that these pairs segregate independently into gametes in meiosis, he concluded that this process is the biological basis for Mendels principles of heredity.
Section 3.2
Comprehension

42. What conclusions did Mendel make from his monohybrid crosses?

Answer:
(1) Progeny inherit genetic factors from both parents.
(2) Each individual possesses two factors (alleles) that control the appearance of each phenotypic trait.
(3) The two alleles in each individual separate (segregate) during gametogenesis and are randomly distributed with equal probability of being distributed into the gametes.
(4) From a cross between two true-breeding (homozygous) parents expressing different phenotypes for a given trait, traits that appeared unchanged in the F1 heterozygous offspring were dominant, and traits that disappeared in F1 heterozygous offspring were recessive.
Section 3.2
Comprehension

43. What is the difference between a backcross and a testcross?

Answer: A testcross is used to determine genotypes of individuals with a dominant phenotype that may be heterozygous or homozygous for a dominant allele. The unknown genotypes are revealed by crossing the dominant individual to a tester that is known to be homozygous for the recessive allele in question.

A backcross is the mating of F1 progeny back to one of their parents. A backcross can also be a testcross if the original parent is homozygous for the recessive allele. Backcrosses are typically used to introgress an allele of interest back into the genetic background of one of the original parents.
Section 3.2
Comprehension

44. Using the diploid cell shown below (at interphase), illustrate/describe Mendels principles of segregation and independent assortment.

A b

a B

Answer:
(1) Principle of segregation: The separation of paired homologs distributes the alleles contained on each homolog to different gametes.

A b

a B

(2) Principle of independent assortment: Homologs are randomly assorted along the metaphase plate, and are subsequently distributed (in complete sets) to gametes.

A b

a B

OR

A B

a b

Section 3.2
Application

45. Albinism is a recessive condition resulting from the inability to produce the dark pigment melanin in skin and hair. A man and woman with normal skin pigmentation want to have two children. The man has one albino parent; the woman has parents with normal pigmentation, but an albino brother.

a. What is the probability that at least one child will be albino?
b. What is the probability of both children being normal?

Answer: If we use the allele symbols A (normal) and a (albino), the mans genotype must be Aa since he is normal but one of his parents is aa. The woman has an albino brother, which means both her parents must be carriers (Aa). However, the woman (who is not albino) could have either an AA or Aa genotype. In the womans case, the aa (albino) genotype must be excluded as a possibility. Therefore, the probability of the woman being AA is 1/3, and the probability that she is Aa is 2/3.

First, we should calculate the probability of the couple having an albino child each time a child is born. If the woman is Aa, then the mating is Aa Aa, and P(albino) = 1/4. However, there is only a 2/3 chance that she is Aa. So overall for this mating, P(albino) = P(man is Aa) P(woman is Aa) 1/4 = 1 2/3 1/4 = 2/12 = 1/6. Conversely, the probability that a child will be normal P(normal) = 1 P(albino) = 5/6.

If the couple plans to have 2 children, there are 4 possible outcomes, which are given in the table below along with each probability. The overall probability of each outcome is calculated using the multiplication rule.

Child 1 Child 2 Probability
Normal (5/6) Normal (5/6) 25/36
Normal (5/6) Albino (1/6) 5/36
Albino (1/6) Normal (5/6) 5/36
Albino (1/6) Albino (1/6) 1/36

a. The probability that at least one child will be albino corresponds to the last three outcomes of the table. Since it can happen in any of three different ways, the three probabilities should be added to get the final probability. P(at least one albino) = 11/36 = 0.306.

b. The probability of both children being normal is given in the table. P(both are normal) = 25/36 = 0.694.
Section 3.2
Application

46. In peas, tall (T) is dominant to short (t). A homozygous tall plant is crossed with a short plant. The F1 are self-fertilized to produce the F2. Both tall and short plants appear in the F2. If the tall F2 are self-fertilized, what types of offspring and proportions will be produced?

Answer: 1/3 of the tall F2 plants are TT, and 2/3 are Tt. The TT plants when selfed will produce all tall plants. The Tt plants when selfed will produce 3 tall:1 short. If all F2 plants produce an equal number of offspring, then 1/3 of the offspring will be from TT plants and will be tall. Three-quarters of the plants from Tt will be tall, but since only 2/3 of the F2 plants are Tt, this represents 2/3 3/4 = 6/12 = 1/2 of the total. Therefore, 1/3 + 1/2 = 5/6 will be tall. The remaining 1/6 of the offspring will be short.
Section 3.2
Application

47. In peas, tall (T) is dominant to short (t). A homozygous tall plant is crossed with a short plant. The F1 are self-fertilized to produce the F2. Both tall and short plants appear in the F2. If the short F2 are self-fertilized, what types of offspring and proportions will be produced?

Answer: Selfing the short F2 progeny (tt) will only yield short progeny.
Section 3.2
Application

48. In deer mice, red eyes (r) is recessive to normal black eyes (R). Two mice with black eyes are crossed. They produce two offspring, one with red eyes and one with black eyes. Give the genotypes of parents and offspring of this cross.

Answer: For red eyes (recessive) to be expressed in the progeny (i.e., to segregate in the progeny), both parents must be heterozygotes (Rr). Rr Rr 1 (RR), 2 (Rr), 1 (rr) = 3:1 (black eyes:red eyes). Note that even with only two progeny produced, you have determined the genotypes of the parents because you observe segregation of the recessive allele (red eyes).
Section 3.2
Application

49. Sex in mammals is determined by the X and Y sex chromosomes: males are XY and females are XX. How do you explain the 5050 sex ratio in mammalian progeny?

Answer: Sex ratio segregates like any other trait. By forming a Punnett square, you see that an XX mother and XY father form offspring in a 5050 ratio of males and females.

Section 3.2
Application

50. You have learned that purple flowers are dominant to white in Mendels peas. When walking the grounds of Mendels monastery, you come across a stray purple pea plant. You suspect that it is descended from Mendels experimental plants, but you have no idea of its exact heritage. Propose two ways that you could determine the plants genotype with respect to the flower color. Assume that you have any other pea plants that you might want to use in your analysis. Provide expected results and interpretations of possible results for your experiments. Which of the two ways would be easier and why?

Answer: The easiest way would be to allow the plant to fertilize itself. If it produces only purple offspring, it must be PP. If it produces 3 purple:1 white, it must be Pp. This is easy because you dont have to do any manipulations.

Another way would be to do a testcross of the unknown purple plant with a white tester (pp). If the plant is PP, then the offspring of this testcross would be all purple. If the unknown plant is Pp, then the offspring will be 1 purple:1 white. This is harder because you have to set up the crosses.
Section 3.2
Application

51. A man has either an AaBB or AABb genotype with equal probability. Assume these genes assort independently. What is the overall probability that the man will produce an Ab gamete?

Answer: Only the AABb genotype can produce Ab gametes, and the expected probability of Ab gametes produced will be 1/2. Because the parent has an equal probability (50%) of having either genotype (AaBB or AABb), the final probability of the parent producing an Ab gamete is (1/2) (1/2) = 1/4 = 0.25 = 25%.
Section 3.3
Comprehension

52. A man has either an AaBB or AABb genotype with equal probability. Assume these genes assort independently. What is the probability of the man producing an AB gamete?

Answer: For genotype AaBB, the probability of generating an AB gamete is 1/2. For genotype AABb, the probability of generating an AB gamete is 1/2. Again, because both genotypes have equal probabilities, the final probability of the parent generating an AB gamete is (1/2) (1/2) + (1/2) (1/2) = 2/4 = 0.5 = 50%.
Section 3.3
Comprehension

53. Compare and contrast Mendels principle of segregation and the principle of independent assortment.

Answer: The principle of segregation involves segregation (separation) of the alleles of a gene pair as the paired homologs on which they reside separate during anaphase I of meiosis. This principle can be demonstrated using a single pair of homologous chromosomes.

The principle of independent assortment involves the random assortment of alleles from different homologs into separate daughter cells during meiosis I. This principle can only be demonstrated using two or more pairs of homologous chromosomes.
Section 3.3
Comprehension

54. For a particular plant, red flowers (A) are dominant over yellow flowers (a). An initial cross was made between a plant that was true-breeding for red flowers, and another plant true-breeding for yellow flowers. F1 progeny, all having red flowers, were allowed to form seeds, which were then planted to generate F2 progeny. Pollen from all the resulting F2 plants was pooled and used to fertilize true-breeding yellow plants. What proportion of the progeny resulting from this cross would be expected to have yellow flowers?

Answer: The F2 progeny will have the genotypes AA:Aa:aa in a 1:2:1 ratio (out of 4 plants, there will be 1 AA plant, 2 Aa plants, and 1 aa plant). When pooled, 50% of the gametes produced by these plants will contain only a alleles. One hundred percent of the gametes produced by the true-breeding yellow plant will contain a alleles. This means that 50% of the F3 progeny will be homozygous recessive (aa) with yellow flowers.
Section 3.3
Application

55. The following cross produced 125 progeny: aaBbCcDd AaBbccDD. Assume all the genes assort independently and that the uppercase letters represent dominant alleles.

a. How many offspring are expected to express the phenotype ABCD?
b. How many offspring are expected to have the genotype aaBBccDd?
c. How many offspring are expected to have the genotype AaBbCcDd?

Answer:
a. Figure out the likelihood of a dominant phenotype for each of the four independent monohybrid cross components (e.g., aa Aa = 1/2 Aa, 1/2 aa, etc.). Therefore, A_ = 1/2 (62 or 63); B_ = 3/4 (92 or 93); C_ = 1/2 (62 or 63); D_ = 1 (125). Thus, for phenotype ABCD: (1/2)(3/4)(1/2)(1) = 3/16 of the progeny, which represents 23 or 24 offspring.
b. Simplify this quadric hybrid cross by breaking it down into four independent monohybrid cross components (e.g., aa Aa = 1/2 Aa, 1/2 aa, etc.). This strategy can be used for any multihybrid cross involving genes that assort independently. Thus, for aaBBccDd: (1/2)(1/4)(1/2)(1/2) = 1/32. 1/32 (125) = 3.9 = 3 or 4 offspring.
c. For AaBbCcDd: (1/2)(1/2)(1/2)(1/2) = 1/16. 1/16 (125) = 7 or 8 offspring.
Section 3.3
Application

56. Let us assume that hairy toes (hh) and brittle ear wax (ww) are both recessive traits in humans. In families of three children where both parents are heterozygotes at both loci (they have smooth toes and sticky ear wax) what is the probability that the family will consist of one hairy-toed, brittle ear-waxed boy and two smooth-toed, sticky ear-waxed girls?

Answer: Three traits are segregating in these crossestoe hair, ear wax, and sex; we assume that these three loci all assort independently. The cross is Hh Ww XX by Hh Ww XY. The probability of getting one hh ww XY and two H_ W_ XX in any order is the product of the probabilities: the probability of getting the hairy-toed, brittle ear-waxed boy (hh ww XY) is 1/4 1/4 1/2 = 1/32 and the probability of getting a smooth-toed, sticky ear-waxed girl (H_ W_ XX) is 3/4 3/4 1/2 = 9/32. The probability of getting a second smooth-toed, sticky ear-waxed girl (H_ W_ XX) is the same: 9/32. There are three possible orders for the three children (boy first, second or third), thus the overall probability is 3 1/32 9/32 9/32 = 243/32,768.
Section 3.3
Challenge

57. Two pea plants with purple flowers are crossed. Among the offspring, 63 have purple flowers, and 17 have white flowers. With a chi-square test, compare the observed numbers with a 3:1 ratio and determine if the difference between observed and expected could be a result of chance.

Answer: Plugging the numbers into the chi-square equation: (63 60)2/60 + (17 20)2/20 = 0.15 + 0.45 = 0.6 (the chi-square statistic). Degrees of freedom = the number of classes (phenotypes) 1 = 2 1 = 1. Looking at the chi-square table for one degree of freedom and using the standard critical value of 5%, we do not reject the hypothesis and thus conclude that the differences between observed and expected progeny numbers are a result of chance. We can further conclude that both purple parents are heterozygous at the single gene locus controlling flower color (at least for purple and white).
Section 3.4
Comprehension

58. In a plant species, you notice that purple and yellow leaf colors, as well as hairy and smooth stems, segregate. You cross a plant with purple leaves and hairy stems to a plant with yellow leaves and hairy stems, and generate the progeny indicated below:

Class Phenotypes
1 68 yellow leaves, hairy stems
2 66 purple leaves, hairy stems
3 22 purple leaves, smooth stems
4 25 yellow leaves, smooth stems
Total: 181

a. Indicate the most likely genotypes for each parent. Yellow (P) is dominant to purple (p), and hairy (S) is dominant to smooth (s).
b. Propose a hypothesis to explain the progeny results. Based on this hypothesis, what ratios are expected for each of the four classes of progeny?
c. Using the chi-square method, test your hypothesis and indicate whether you accept or reject it.

Answer:
a. Using P for the leaf color locus and S for the stem type locus: PpSs ppSs.
b. A possible hypothesis is that each trait is controlled by independently assorting single gene pairs, and that one allele in each pair exhibits complete dominance over the other. Class 1 = 3/8, class 2 = 3/8, class 3 = 1/8, class 4 = 1/8.
c. (68 68)2/68 + (66 68)2/68 + (22 23)2/23 + (25 23)2/23 = 0.276. This number (the chi-square statistic) is small compared to the critical value; therefore you confidently accept your hypothesis.
Section 3.4
Application

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