Choose the best answer.
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1 | |
Linked genes: |
| | A) | are located near each other on the same chromosome. |
| | B) | violate the law of independent assortment. |
| | C) | segregate together during meiosis. |
| | D) | All of the above. |
| | E) | None of the above. |
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2 | |
You cross a truebreeding normal corn plant with a true breeding plant with variegated seed coat and colorless endosperm. You suspect the genes are linked, so you cross one of the resulting normal offspring to a true breeding variegated, colorless plant. What are your expected results if the genes are completely linked? |
| | A) | 1 normal: 1 variegated: 1 colorless: 1 variegated and colorless |
| | B) | 9 normal: 3 variegated: 3 colorless: 1 variegated and colorless |
| | C) | 3 normal: 1 variegated and colorless |
| | D) | 3 variegated and colorless: 1 normal |
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3 | |
For the cross in the previous question, you get the following results. Calculate a chi square value using an expectation of independent assortment and determine if the genes are likely to be linked or not. Normal 350; variegated 63; colorless 57; variegated and colorless 330. |
| | A) | The chi square value is greater than the critical value; these genes are likely to be linked. |
| | B) | The chi square value is greater than the critical value; these genes are unlikely to be linked. |
| | C) | The chi square value is less than the critical value; these genes are likely to be linked. |
| | D) | The chi square value cannot help us predict linkage. |
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4 | |
From the results in the previous question, calculate a map distance between these two genes. |
| | A) | 0.15 map units |
| | B) | 0.85 map units |
| | C) | 1.5 map units |
| | D) | 15 map units |
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5 | |
How did Morgan demonstrate crossing over? |
| | A) | A fly that received only y+w+m+ and ywm chromosomes from her parents was able to transmit different combinations of alleles to her offspring. |
| | B) | A fly with a gray body and red eyes (wild type) was able to produce offspring with gray bodies and white eyes. |
| | C) | True breeding yellow, white, miniature flies always had yellow, white, miniature offspring. |
| | D) | None of these examples demonstrate crossing over. |
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6 | |
What was unique about the chromosomes studied by Creighton and McClintock? |
| | A) | They carried different alleles for two linked genes. |
| | B) | They were unable to cross over. |
| | C) | These chromosomes were unable to synapse during meiosis I. |
| | D) | Physical structures of the chromosomes allowed them to be distinguished from each other. |
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7 | |
Which of the following would be a reasonable way to demonstrate that twin spots are caused by mitotic recombination? |
| | A) | Collect a large number of flies with twin spots and verify that in all cases, both halves of the twin spot are the same size. |
| | B) | Collect a large number of petunias with twin spots and verify that they are able to pass this phenotype to their progeny. |
| | C) | Design cytologically distinct chromosomes that must recombine to cause the twin spot; analyze chromosomes from the spots with a microscope. |
| | D) | Try to clone new individuals from the cells in the twin spots and analyze their phenotype. |
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8 | |
You begin to study a novel plant species and discover that this diploid plant has 16 chromosomes. How many linkage groups would you expect to find? |
| | A) | 16 |
| | B) | 8 |
| | C) | 4 |
| | D) | It is impossible to tell from the information provided. |
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9 | |
You calculate map distances between genes A, B, and C based on all pairwise dihybrid crosses. When you perform the trihybrid cross to verify your results, you discover that, despite analysis of a very large number of progeny, you have only about half as many double crossover progeny as you expect. Which explanation is the most reasonable? |
| | A) | Your dihybrid cross data are an underestimate of the distance between the more widely separated genes. |
| | B) | You are seeing an example of interference. |
| | C) | You are seeing an example of random sampling error. |
| | D) | None of these. |
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10 | |
Consult Fig 5.8. In a dihybrid cross involving Bar eye and little fly genes, what frequency of recombinant progeny do you expect? |
| | A) | 2.1 |
| | B) | 11.1 |
| | C) | 57.0 |
| | D) | 68.1 |
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11 | |
Consult Fig 5.8. What percentage of double crossover progeny would you expect in a trihybrid cross involving aristaless, dumpy, and black? (Assume no interference.) |
| | A) | 0.05% |
| | B) | 5% |
| | C) | 13% |
| | D) | 48.5% |
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12 | |
You set up the following dihybrid mapping cross in fruit flies.
P e+e+ roro cq cq x ee ro+ro+ cq+cq+. After backcrossing F1 males to ee roro cq cq females, you get the following results: gray body, rough eyes, claret eyes 576; ebony body, smooth, red eyes 564. How can you explain this result? |
| | A) | The genes are not linked. |
| | B) | Crossing over does not occur in male Drosophila. |
| | C) | These genes are X linked. |
| | D) | A and B only. |
| | E) | None of the above. |
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13 | |
What is a convenient way to identify gene order in a trihybrid mapping cross? |
| | A) | Look for double crossover progeny and identify the gene that was flipped relative to parentals. |
| | B) | Look for the largest class of single crossovers and the two genes are the ones on the ends. |
| | C) | Look for the smallest class of single crossovers to identify the genes which are closest together. |
| | D) | A and B only. |
| | E) | All of the above. |
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14 | |
Why is tetrad analysis useful for studying genetics? |
| | A) | Cells within a tetrad are haploid so genotype is directly reflected in the phenotype. |
| | B) | Cells within a tetrad reflect the events of meiosis. |
| | C) | Analysis of simple systems can elucidate general rules that apply more broadly. |
| | D) | All of the above. |
| | E) | None of the above. |
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15 | |
Tetrad analysis automatically includes the centromere as an additional point in mapping crosses, because |
| | A) | crossovers that are not between a gene and the centromere do not affect the relationship of that gene to the chromosome during division. |
| | B) | crossovers that are between a gene and the centromere cannot be identified in this system. |
| | C) | when studying crossovers between genes, the position of the centromere is irrelevant. |
| | D) | All of the above. |
| | E) | None of the above. |
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16 | |
Formation of the following pattern in an ordered octad is due to what event during meiosis? Pattern: AAaaaaaAA |
| | A) | No crossing over and normal segregation in meiosis I. |
| | B) | Crossing over and normal segregation during meiosis I. |
| | C) | Crossing over and segregation during meiosis II. |
| | D) | No crossing over and segregation during meiosis II. |
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17 | |
In an unordered tetrad with the parental ditype: |
| | A) | all spores have the same genotype. |
| | B) | all spores have the genotype of one of the parental strains. |
| | C) | all four spores have different genotypes. |
| | D) | all spores are diploiD. In a parental ditype, the spores are of both parental types in a 2:2 ratio. |
| | E) | Correct! |
| | F) | This would be a tetratype. |
| | G) | The diploid zygote divides by meiosis to generate haploid spores. |
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18 | |
You set up a mapping cross involving two genes in Aspergillus and determine the following percentages of tetrad types: parental ditype 51%, nonparental ditype 49%. What can you say about your genes? |
| | A) | They are closely linked. |
| | B) | They are unlinked. |
| | C) | They are lethal. |
| | D) | They are not expressed in the phenotype. |
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19 | |
You set up a mapping cross involving your favorite gene (YFG) in Neurospora and determine the following octad types: 4:4 arrangement 83; 2:4:2 arrangement 7; 2:2:2:2 arrangement 10. What is the distance between your gene and the centromere? |
| | A) | 7 cm |
| | B) | 10 cm |
| | C) | 17 cm |
| | D) | 83 cm |
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20 | |
You set up a mapping cross involving two genes of interest in Saccharomyces and determine the following tetrad types: 8 nonparental ditype; 37 tetratype; 80 parental ditype. Calculate the most accurate map distance between these two genes. |
| | A) | 18.7mu |
| | B) | 32.3mu |
| | C) | 64.6mu |
| | D) | None of these is accurate. |
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21 | |
Every chromosome in a diploid cell comprises a linkage group. |
| | A) | True |
| | B) | False |
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22 | |
Crossing over involves the physical exchange of DNA between chromatids. |
| | A) | True |
| | B) | False |
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23 | |
If two genes are linked, a double heterozygote will only be able to produce two possible types of gametes. |
| | A) | True |
| | B) | False |
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24 | |
Crossing over only occurs within a bivalent. |
| | A) | True |
| | B) | False |
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25 | |
Parental phenotypes in the offspring are the most common in a mapping cross. |
| | A) | True |
| | B) | False |
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26 | |
Morgan was able to demonstrate linkage of genes to chromosomes because he studied X linked genes. |
| | A) | True |
| | B) | False |
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27 | |
Chi square analysis can help distinguish between two competing hypotheses. |
| | A) | True |
| | B) | False |
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28 | |
Creighton and McClintock's important discovery was that they were able to demonstrate linkage between a translocation and a neighboring gene. |
| | A) | True |
| | B) | False |
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29 | |
Twin spots are only possible in doubly heterozygous individuals. |
| | A) | True |
| | B) | False |
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30 | |
In linkage mapping, one map unit is equal to one % crossover frequency. |
| | A) | True |
| | B) | False |
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31 | |
Trihybrid mapping crosses are more accurate than dihybrid mapping crosses. |
| | A) | True |
| | B) | False |
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32 | |
You can predict the number of double crossovers in a region accurately if you have a detailed genetic map. |
| | A) | True |
| | B) | False |
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33 | |
Unordered tetrad analysis allows determination of the specific events of meiosis I and II. |
| | A) | True |
| | B) | False |
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34 | |
Unordered tetrad analysis is useful for determining recombination frequencies. |
| | A) | True |
| | B) | False |
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35 | |
Double crossovers can occur between multiple chromatids. |
| | A) | True |
| | B) | False |
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