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AP BIOLOGY: Chapter Thirteen Review Answers 
1. The hybridization showed that breeding between species produced fertile offspring and that traits were masked through one generation and not blended as expected. The progeny were all different from one another, exhibiting different forms of a trait. Recognition of the significance of Koelreuter's theories was delayed because subsequent workers for almost a century did not quantify their results. 2. (1) He used previous researchers' success to direct his work, (2) he picked seven clearly distinguishable traits to examine closely, (3) peas are small, are easy to grow, and have short generation times, and (4) peas are self-fertile but can be easily cross-fertilized as well. 3. The garden pea was a good choice for research because (1) many people had already done research on it and many hybrids were available; (2) a large number of fine-breeding varieties were available; (3) pea plants are small, easy to grow, and have a short generation time, and (4) the sexual organs of the pea were enclosed in the flower. 4. He isolated single plants to allow them to self-fertilize; to cross-fertilize he cut the anthers off of one plant and applied it to the stigma of another. 5. All the F1 offspring were purple, which was dominant over the recessive white color. F1 crosses yielded a 3:1, purple:white (dominant:recessive) F2 ratio. 6. An allele is one expressible version of a gene. A homozygous individual has two of the same alleles, an individual who is heterozygous possess two different alleles (typically one dominant and one recessive). 7. An individual's phenotype is what is expressed by the combination of the two alleles (what one would see, such as a purple color). The genotype is the actual genetic make-up of the individual (for example, either a WW or Ww genotype would be expressed as the same purple phenotype). 8. To determine an unknown genotype, cross the plant with a homozygous recessive (white) plant. If the unknown plant was homozygous (WW) for the purple color, 100% of the offspring will also be purple. If the genotype of the unknown plant was, on the other hand, heterozygous (Ww) for purple color, the offspring of this cross would be half white (ww) and half purple (Ww). 9. The ratio of expected phenotypes in a dihybrid cross is 9:3:3:1. 1/16 should be homozygous recessive for both traits, and 1/16 should be homozygous dominant for both traits. 10. Epistasis is the effect of one gene on another such that expression becomes altered. Pleiotropy is an allele having more than one effect on an organism (i.e., height and color). Both interfere with interpreting inheritance because they interfere with typical Mendelian ratios. 11. Morgan noticed that some mutant flies had white eyes. In an effort to see if the eye color assorted in a Mendelian fashion, he noticed that eye color was strongly associated with sex, leading him to conclude that those chromosomes that designate sex must also designate eye color. 12. Stern noticed that if an exchange of genetic material occurred resulting in a change in eye color, "abnormal ends" of chromosomes also seemed to have exchanged, indicating the phenomenon of crossing over. Relative position of genes on a chromosome (a genetic map) can be determined by how frequently they do or do not cross over: genes farther apart on chromosomes cross over more frequently than those that are closer together. 13. Primary nondisjunction is the failure of chromosomes to separate from one another in meiosis. Down syndrome is usually caused by primary nondisjunction. Women who are more than 35 years old are more prone to produce these gametes because gametes in women are held in prophase I from birth to whenever the egg is released to develop further (ovulation), and the chance for damage increases if this holding pattern is lengthy. In men, sperm are produced continuously from puberty and are not stored for long periods. 14. The genotype of a person with Klinefelter syndrome is XXY, which makes him genetically male, since the presence of the Y chromosome determines maleness. The individual appears female. 15. Huntington's disease is dominant. Its frequency is maintained because the physiological effects are not evident until relatively late in life, generally after some offspring have already been produced. Answers to Mendelian Genetics Problems 1. Alleles segregate in meiosis, and the products of that segregation are contained within a pod. Each pea is a gamete. In this diagram, the segregation is incorrectly shown as being between pods, each pod shown as uniformly wrinkled or round. 2. The probability of getting two genes on the same chromosome is 1/223. 3. Somewhere in your herd you have cows and bulls that are not homozygous for the dominant gene "polled." Since you have many cows and probably only one or some small number of bulls, it would make sense to concentrate on the bulls. If you have only homozygous "polled" bulls, you could never produce a horned offspring regardless of the genotype of the mother. The most expedient thing to do would be to keep track of the matings and the phenotype of the offspring resulting from these matings and render ineffective any bull found to produce horned offspring. 4. It would not be possible on the basis of the information presented to substantiate a claim of infidelity. You do not know if the woolly trait is the result of a single gene product, or even if the trait is dominant or recessive. Assuming for the moment that it was the effect of a single dominant allele W, the man could still be a heterozygote for the gene and, when mated to a recessive homozygous female, would expect to produce woolly headed offspring only one-half the time. 5. One-half of her offspring would be expected to be affected. 6. Albinism, a, is a recessive gene. If heterozygotes mated you would have the following:
Clearly one-fourth would be expected to be albinos. 7. The best thing to do would be to mate the racehorse to several mares homozygous for the recessive gene that causes the brittle bones. Half of the offspring would be expected to have brittle bones if the racehorse were a heterozygous carrier of the disease gene. Although you could never be 100% certain your horse was not a carrier, you could reduce the probability to a reasonable level. 8. Your mating of DDWw and Ddww individuals would look like the following:
Long-wing, red-eyed individuals would result from eight of the possible 16 combinations, and dumpy, white-eyed individuals would never be produced. 9. Breed the fly to one from the white-eyed stock. If half of the offspring are white eyed, then your fly is a heterozygote. 10. Both parents carry at least one of the recessive genes, even if only in the heterozygous condition (in which the trait would not be expressed in the parent). Since it is recessive, the trait is not manifest until they produce an offspring who is homozygous. 11. To solve this problem, let's first look at the second cross, where the individuals were crossed with the homozygous recessive sepia flies se/se. In one case, all the flies were red eyed: Unknown genotype Sepia
The only way to have all red-eyed flies when bred to homozygous sepia flies is to mate the sepia fly with a homozygous red-eyed fly. In the other case, half of the offspring were black eyed and the other half red eyed: Unknown genotype Sepia
The unknown genotype in this case must have been Se/se, since this is the only mating that will produce the proper ratio of sepia-eyed flies to red-eyed flies. Since the ratio of this unknown genotype and the one previously determined was 1:1, we must deduce the genotype of the original flies, which when mated, will produce a 1:1 ratio of Se/se to Se/Se flies. Unknown original 1 Original 2
You can see from this diagram that if one of the original flies was homozygous for red eyes and the other was a heterozygous individual, the proper ratio of heterozygous and homozygous offspring would be obtained. 12. a. It could have originated as a mutation in his germ cell line. b. Since their son Alexis was a hemophiliac, the disease almost certainly originated with Alexandra; Nicholas II would have contributed only a silent Y chromosome to Alexis's genome. There is a 50% chance that Anastasia was a carrier. 13. The ability to smell musk is an example of dominant inheritance. 14. 15. c is a rare allele, and Mabel's child can develop cystic fibrosis only if her husband also carries the allele: [probability that husband has c allele] = frequency of c = 1/20 16. You could expect 45 chromosomes (44 autosomes plus one extra copy of an X). 17. The woman's blood genotype and blood type is AO. 18. It expresses dominant inheritance. 19. The chances are 25%. 20. Let a = albino. The genotype of the father is Aa. 21.
22. It would be very difficult to establish negligence, especially if the child is a boy. If the child's maternal grandmother was a carrier, there's a 50:50 chance that the mother is a carrier. If the mother is a carrier, there's a 50:50 chance that a boy will express the disease. One would be suspicious about the causes of Duchenne muscular dystrophy in a girl. In the first place, symptomatic males may not live to reproductive age, and it would be rare for a symptomatic male to be virile. A symptomatic girl from a nonsymptomatic father could result from a genetic anomaly in the father, or from a mutation unrelated to radiation. The case for mutation on the job is clearly stronger for the girl, but very weak in either case. | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
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