Chapter 18 | Mendel’s Experiments and Heredity

  1. Figure 18.6 In pea plants, round peas (R) are dominant to wrinkled peas (r). You do a test cross between a pea plant with wrinkled peas (genotype rr) and a plant of unknown genotype that has round peas. You end up with three plants, all which have round peas. From this data, can you tell if the round pea parent plant is homozygous dominant or heterozygous? If the round pea parent plant is heterozygous, what is the probability that a random sample of 3 progeny peas will all be round?
    In a test cross, a parent with a dominant phenotype but unknown genotype is crossed with a recessive parent. If the parent with the unknown phenotype is homozygous dominant, all of the resulting offspring will have at least one dominant allele. If the parent with the unknown phenotype is heterozygous, fifty percent of the offspring will inherit a recessive allele from both parents and will have the recessive phenotype.
    Figure 18.6 A test cross can be performed to determine whether an organism expressing a dominant trait is a homozygote or a heterozygote.
  2. Figure 18.7 What are the genotypes of the individuals labeled 1, 2 and 3?
    This is a pedigree of a family that carries the recessive disorder alkaptonuria. In the second generation, an unaffected mother and an affected father have three children. One child has the disorder, so the genotype of the mother must be upper case A lower case a, and the genotype of the father is lower case a lower casea. One unaffected child goes on to have two children, one affected and one unaffected. Because her husband was not affected, she and her husband must both be heterozygous. The genotype of their unaffected child is unknown, and is designated upper A question mark. In the third generation, the other unaffected child had no offspring, and his genotype is therefore also unknown. The affected third-generation child goes on to have one child with the disorder. Her husband is unaffected and is labeled 3. The first generation father is affected and is labeled 1; The first generation mother is unaffected and is labeled 2 The Visual Connection question asks the genotype of the three numbered individuals.
    Figure 18.7 Alkaptonuria is a recessive genetic disorder in which two amino acids, phenylalanine and tyrosine, are not properly metabolized. Affected individuals may have darkened skin and brown urine, and may suffer joint damage and other complications. In this pedigree, individuals with the disorder are indicated in blue and have the genotype aa. Unaffected individuals are indicated in yellow and have the genotype AA or Aa. Note that it is often possible to determine a person’s genotype from the genotype of their offspring. For example, if neither parent has the disorder but their child does, they must be heterozygous. Two individuals on the pedigree have an unaffected phenotype but unknown genotype. Because they do not have the disorder, they must have at least one normal allele, so their genotype gets the “A?” designation.
  3. Figure 18.13 What ratio of offspring would result from a cross between a white-eyed male and a female that is heterozygous for red eye color?
    This illustration shows a Punnett square analysis of fruit fly eye color, which is a sex-linked trait. A red-eyed male fruit fly with the genotype X superscript w baseline, Y, is crossed with a white-eyed female fruit fly with the genotype X superscript w, X superscript w baseline. All of the female offspring acquire a dominant upper case W allele from the father and a recessive lower case w allele from the mother, and are therefore heterozygous dominant with red eye color. All of the male offspring acquire a recessive w allele from the mother and a Y chromosome from the father and are therefore hemizygous recessive with white eye color.
    Figure 18.13 Punnett square analysis is used to determine the ratio of offspring from a cross between a red-eyed male fruit fly and a white-eyed female fruit fly.
  4. Figure 18.17 In pea plants, purple flowers (P) are dominant to white flowers (p) and yellow peas (Y) are dominant to green peas (y). What are the possible genotypes and phenotypes for a cross between PpYY and ppYy pea plants? How many squares do you need to do a Punnett square analysis of this cross?
    This illustration shows a dihybrid cross between pea plants. In the upper case P generation, a plant that has the homozygous dominant phenotype of round, yellow peas is crossed with a plant with the homozygous recessive phenotype of wrinkled, green peas. The resulting F subscript 1 baseline offspring have a heterozygous genotype and round, yellow peas. Self-pollination of the F subscript 1 baseline generation results in F subscript 2 baseline offspring with a phenotypic ratio of 9 colon 3 colon 3 colon 1 for yellow round, green round, yellow wrinkled and green wrinkled peas, respectively.
    Figure 18.17 This dihybrid cross of pea plants involves the genes for seed color and texture.
  5. Figure 18.23 In a test cross for two characteristics such as the one shown here, can the predicted frequency of recombinant offspring be 60 percent? Why or why not?
    The illustration shows the possible inheritance patterns of linked and unlinked genes. The example used includes fruit fly body color and wing length. Fruit flies may have a dominant gray color lowercase b superscript +, or a recessive black color lowercase b. They may have dominant long wings lowercase v g superscript +, or recessive short wings, lowercase v g. Three hypothetical inheritance patterns for a test cross between a heterozygous and a recessive fruit fly are shown, based on gene placement. The actual experimental results published by Thomas Hunt Morgan in 1912 are also shown. In the first hypothetical inheritance pattern in part a, the genes for the two characteristics are on different chromosomes. Independent assortment occurs so that the ratio of genotypes in the offspring is 1 b + b v g + v g colon 1. b b v g v g colon 1. b + b v g v g colon 1. v g v g v g + v g, and 50% of the offspring are nonparental types. In the second hypothetical inheritance pattern in part b, the genes are close together on the same chromosome so that no crossover occurs between them. The ratio of genotypes is 1 b + b v g + v g colon 1. b b v g v g, and none of the offspring are recombinant. In the third hypothetical inheritance pattern in part c, the genes are far apart on the same chromosome so that crossing over occurs 100% of the time. The ratio of genotypes is the same as for genes on two different chromosomes, and 50% of the offspring are recombinant, nonparental types. Part d shows that the number of offspring that Thomas Hunt Morgan actually observed was 965 colon 944 colon 206 colon 185, which is b + b v g + v g colon b b v g v g colon b + b v g v g colon, b b v g + v g. Seventeen percent of the offspring were recombinant, indicating that the genes are on the same chromosome and crossing over occurs between them some of the time.
    Figure 18.23 Inheritance patterns of unlinked and linked genes are shown. In (a), two genes are located on different chromosomes so independent assortment occurs during meiosis. The offspring have an equal chance of being the parental type (inheriting the same combination of traits as the parents) or a nonparental type (inheriting a different combination of traits than the parents). In (b), two genes are very close together on the same chromosome so that no crossing over occurs between them. The genes are therefore always inherited together and all of the offspring are the parental type. In (c), two genes are far apart on the chromosome such that crossing over occurs during every meiotic event. The recombination frequency will be the same as if the genes were on separate chromosomes. (d) The actual recombination frequency of fruit fly wing length and body color that Thomas Morgan observed in 1912 was 17 percent. A crossover frequency between 0 percent and 50 percent indicates that the genes are on the same chromosome and crossover occurs some of the time.
  6. Figure 18.24 Which of the following statements is true?
    1. Recombination of the body color and red/ cinnabar eye alleles will occur more frequently than recombination of the alleles for wing length and aristae length.
    2. Recombination of the body color and aristae length alleles will occur more frequently than recombination of red/brown eye alleles and the aristae length alleles.
    3. Recombination of the gray/black body color and long/short aristae alleles will not occur.
    4. Recombination of the red/brown eye and long/ short aristae alleles will occur more frequently than recombination of the alleles for wing length and body color.
      The illustration shows a Drosophila genetic map. The gene for aristae length occurs at 0 centimorgans, or lower case c upper case M. The gene for body color occurs at 48.5 lower c upper M. The gene for red versus cinnabar eye color occurs at 57.5 lower c upper M. The gene for wing length occurs at 65.5 lower c upper M, and the gene for red versus brown eye color occurs at 104.5 lower c upper M. One lower c upper M is equivalent to a recombination frequency of 0.01.
      Figure 18.24 This genetic map orders Drosophila genes on the basis of recombination frequency.
  7. Figure 18.26 Which of the following statements about nondisjunction is true?
    1. Nondisjunction only results in gametes with n+1 or n–1 chromosomes.
    2. Nondisjunction occurring during meiosis II results in 50 percent normal gametes.
    3. Nondisjunction during meiosis I results in 50 percent normal gametes.
    4. Nondisjunction always results in four different kinds of gametes.
      This illustration shows nondisjunction that occurs during meiosis I. Nondisjunction during meiosis I occurs when a homologous pair fails to separate, and results in two gametes with n + 1 chromosomes, and two gametes with n − 1 chromosomes. Nondisjunction during meiosis II would occur when sister chromatids fail to separate, and results in one gamete with n + 1 chromosomes, one gamete with n − 1 chromosomes, and two normal gametes.
      Figure 18.26 Nondisjunction occurs when homologous chromosomes or sister chromatids fail to separate during meiosis, resulting in an abnormal chromosome number. Nondisjunction may occur during meiosis I or meiosis II.

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