Chapter 8 Experimental Questions




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CHAPTER 8
Experimental Questions

E1. With regard to the analysis of chromosome structure, explain the experimental advantage that polytene chromosomes offer. Discuss why changes in chromosome structure are more easily detected in polytene chromosomes compared to ordinary (nonpolytene) chromosomes.



Answer: Polytene chromosomes can be viewed in greater detail under the microscope because they are much larger. This makes it much easier to detect very small changes in chromosome structure. Polytene chromosomes are produced from the sequential replication and alignment of chromosomes. As an example, suppose a toothpick has fine lines written on it. You would probably have trouble seeing the lines. However, if you took 1,000 toothpicks with the same lines and stacked them up in a parallel manner, the lines would be much easier to see. Similarly, small changes in chromosome structure are hard to see in a single chromosome but much easier to detect in a polytene chromosome.

E2. Describe how colchicine can be used to alter chromosome number.



Answer: Colchicine interferes with the spindle apparatus and thereby causes nondisjunction. At high concentrations, it can cause complete nondisjunction and produce polyploid cells.

E3 Female fruit flies homozygous for the X-linked white-eye allele are crossed to males with red eyes. On very rare occasions, an offspring is a male with red eyes. Assuming that these rare offspring are not due to a new mutation in one of the mother’s X chromosomes that converted the white-eye allele into a red-eye allele, explain how this red-eyed male arose.



Answer: The male offspring is the result of nondisjunction during oogenesis. The female produced an egg without any sex chromosomes. The male parent transmitted a single X chromosome carrying the red allele. This produces an X0 male offspring

E4. A cytogeneticist has collected tissue samples from members of the same butterfly species. Some of the butterflies were located in Canada, while others were found in Mexico. Upon karyotyping, the cytogeneticist discovered that chromosome 5 of the Canadian butterflies had a large inversion compared to the Mexican butterflies. The Canadian butterflies were inversion homozygotes, whereas the Mexican butterflies had two normal copies of chromosome 5.

A. Explain whether a mating between the Canadian and Mexican butterflies would produce phenotypically normal offspring.

B. Explain whether the offspring of a cross between Canadian and Mexican butterflies would be fertile.



Answer:

A. The F1 offspring would probably be phenotypically normal, because they would carry the correct number of genes.

B. The F1 offspring would have lowered fertility because they are inversion heterozygotes. Because this is a large inversion, crossing over is fairly likely in the inverted region. When this occurs, it will produce deletions and duplications that will probably be lethal in the resulting F2 offspring.

E5. While conducting field studies on a chain of islands, you decide to karyotype two phenotypically identical groups of turtles, which are found on different islands. The turtles on one island have 24 chromosomes, while the turtles on another island have 48 chromosomes. How would you explain this observation? How do you think the turtles with 48 chromosomes came into being? If you mated the two types of turtles together, would you expect their offspring to be phenotypically normal? Would you expect them to be fertile? Explain.



Answer: The turtles are two distinct species that appear phenotypically identical. The turtles with 48 chromosomes are polyploid relatives (i.e., tetraploids) of the species with 24 chromosomes. In animals, it is somewhat hard to imagine how this could occur because animals cannot self-fertilize, so there had to be two animals (i.e., one male and one female) that became tetraploids. It is easy to imagine how one animal could become a tetraploid; complete nondisjunction could occur during the first cell division of a fertilized egg, thereby creating a tetraploid cell that continued to develop into a tetraploid animal. This would have to happen independently (i.e., in two individuals of opposite sex) to create a tetraploid species. If you mated a tetraploid turtle with a diploid turtle, the offspring would be triploid and probably phenotypically normal. However, the triploid offspring would be sterile because they would make highly aneuploid gametes.

E6. It is an exciting time to be a plant breeder because so many options are available for the development of new types of agriculturally useful plants. Let’s suppose you wish to develop a seedless tomato that could grow in a very hot climate and is resistant to a viral pathogen that commonly infects tomato plants. At your disposal, you have a seed-bearing tomato strain that is heat resistant and produces great-tasting tomatoes. You also have a wild strain of tomato plants (which have lousy-tasting tomatoes) that is resistant to the viral pathogen. Suggest a series of steps that you might follow to produce a great-tasting, seedless tomato that is resistant to heat and the viral pathogen.



Answer: First, you would cross the two strains together. It is difficult to predict the phenotype of the offspring. Nevertheless, you would keep crossing offspring to each other and backcrossing them to the parental strains until you obtained a great-tasting tomato strain that was resistant to heat and the viral pathogen. You could then make this strain tetraploid by treatment with colchicine. If you crossed the tetraploid strain with your great-tasting diploid strain that was resistant to heat and the viral pathogen, you may get a triploid that had these characteristics. This triploid would probably be seedless.

E7. What is a G band? Discuss how G bands are useful in the analysis of chromosome structure.



Answer: A G band is a dark band on a chromosome that has been stained with Giemsa. The pattern of G bands on chromosomes can be used to distinguish chromosomes from each other even if their sizes and centromere locations are very similar. Banding patterns are also used to detect changes in chromosome structure.

E8. A female fruit fly contains one normal X chromosome and one X chromosome with a deficiency. The deficiency is in the middle of the X chromosome and is about 10% of the entire length of the X chromosome. If you stained and observed the chromosomes of this female fly in salivary gland cells, draw what the polytene arm of the X chromosome would look like. Explain your drawing.



Answer: A polytene chromosome is formed when a chromosome replicates many times, and the chromatids lie side by side, as shown in Figure 8.18. The homologous chromosomes also lie side by side. Therefore, if there is a deletion, there will be a loop. The loop is the segment that is not deleted from one of the two homologs.

E9. Describe two different experimental strategies to create an allotetraploid from two different diploid species of plants.



Answer: You could cross the two species together to first create an allodiploid. The allodiploid could be treated with colchicine to make a segment of the plant an allotetraploid. As described in Figure 8.27, a cutting of the allotetraploid segment of the plant could be rooted to create a new allotetraploid plant. Alternatively, one could first use colchicine to make two different tetraploids (autopolyploidy), which would make diploid gametes. These two autopolyploids could be crossed to each other to make an allotetraploid.


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