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| 1 | |
The tendency of genes to be transmitted together exemplifies __________ and a unit of genetic map distance is given by __________. |
| | A) | chromosomal interference; chi square |
| | B) | mosaic; centimorgan |
| | C) | genetic linkage; chi square |
| | D) | recombination; centimorgan |
| | E) | genetic linkage; centimorgan |
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| 2 | |
A visible site of crossing over is known as a __________ and recombination between genes can be moldulated by __________. |
| | A) | centromere; chromosomal interference |
| | B) | centromere; map units |
| | C) | chiasma; chromosomal interference |
| | D) | kinetochore; chromosomal interference |
| | E) | chiasma; chi square |
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| 3 | |
An underlying assumption about linkage can be given by __________ and a statistical test of linkage is called __________. |
| | A) | chromosomal interference; chi square |
| | B) | chromosomal interference; null hypothesis |
| | C) | null hypothesis; chromosomal interference |
| | D) | null hypothesis; chi square |
| | E) | chi square; null hypothesis |
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| 4 | |
Which of the following processes can generate recombinant gametes? |
| | A) | Segregation of alleles in a heterozygote. |
| | B) | Crossing over between two linked heterozygous loci. |
| | C) | Independent assortment of two unlinked heterozygous loci. |
| | D) | b and c |
| | E) | a, b and c |
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| 5 | |
In tetrad analysis, second-division segregations result from:
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| | A) | single crossovers between linked genes. |
| | B) | double crossovers between linked genes. |
| | C) | single crossovers between a gene and a centromere. |
| | D) | independent assortment of unlinked genes. |
| | E) | nondisjunction of homologs. |
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| 6 | |
The Q and P genes are linked and 20 map units apart. In the cross Qp/qP x qp/qp what fraction of the progeny will be QP/qp?
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| | A) | 5% |
| | B) | 10% |
| | C) | 25% |
| | D) | 40% |
| | E) | 45% |
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| 7 | |
If the map distance between genes A and B is 15 map units and the map distance between genes B and C is 40 map units, what is the map distance between genes A and C?
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| | A) | 25 map units |
| | B) | 55 map units |
| | C) | Either 25 map units or 55 map units, depending on the order of the genes. |
| | D) | 45 map units |
| | E) | The map distance between A and C can not be predicted from these data |
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| 8 | |
Suppose the L and M genes are on the same chromosome but separated by 65 map units. What fraction of the progeny from the cross LM/lm x lm/lm would be Lm/lm?
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| | A) | 10% |
| | B) | 25% |
| | C) | 32.5% |
| | D) | 50% |
| | E) | 75% |
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| 9 | |
In Drosophila, singed bristles (sn) and cut wings (ct) are both caused by recessive, X-linked alleles. The wild type alleles (sn+ and ct+) are responsible for straight bristles and intact wings, respectively. A female homozygous for sn and ct+ is crossed to a sn+ct male. The F1 flies are interbred. The F2 males are distributed as follows:
| genotype | number | | sn ct | 15 | | sn ct+ | 34 | | sn+ ct | 33 | | sn+ct+ | 18 |
What is the map distance between sn and ct?
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| | A) | 15 m.u. |
| | B) | 18 m.u. |
| | C) | 33 m.u. |
| | D) | 49 m.u. |
| | E) | 67 m.u. |
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| 10 | |
The table below contains critical chi square values:
| accept null hypothesis | | | reject null hypothesis |
| | degrees of freedom | 0.50 | 0.10 | 0.05 | 0.01 | 0.01 |
| | 1 | 0.45 | 2.71 | 3.84 | 6.64 | 10.83 | | 2 | 1.39 | 4.61 | 5.99 | 9.21 | 13.82 | | 3 | 2.37 | 6.25 | 7.81 | 11.35 | 16.27 | | 4 | 3.36 | 7.78 | 9.49 | 13.28 | 18.47 |
In Drosophila, singed bristles (sn) and carnation eyes (car) are both caused by recessive, X-linked alleles. The wild-type alleles (sn+ and car+) are responsible for straight bristles and red eyes, respectively. A sn car female is mated to a sn+ car+male and the F1 progeny are interbred. The F2 are distributed as follows:
| genotype | number | | sn car | 57 | | sn car+ | 43 | | sn+ car | 43 | | sn+car+ | 57 |
The p value from this test is __________, which means that the sn and car genes are __________.
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| | A) | 0.5>p>0.1; linked |
| | B) | 0.5>p>0.1; unlinked |
| | C) | 0.1>p>0.05; linked |
| | D) | 0.1>p>0.05; unlinked |
| | E) | 0.05>p>0.01; linked |
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| 11 | |
The pairwise map distances for four linked genes are as follows: A-B = 28 m.u., B-C = 15 m.u., C-D = 25 m.u., B-D = 10 m.u., A-D = 38 m.u., A-C = 13 m.u. What is the order of these four genes?
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| | A) | ABCD |
| | B) | ACBD |
| | C) | ABDC |
| | D) | BADC |
| | E) | CADB |
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| 12 | |
The measured distance between genes D and E in a two point test cross is 50 map units. What does this mean in physical terms?
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| | A) | D and E are on different pairs of chromosomes. |
| | B) | D and E are linked and exactly 50 map units apart. |
| | C) | D and E are linked and at least 50 map units apart. |
| | D) | either a or b |
| | E) | either a or c |
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| 13 | |
The map of a chromosome interval is X 5 Y 10 Z. From the cross Xyz/xYZ x xyz/xyz, how many double crossovers would be expected out of 1000 progeny?
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| | A) | 5 |
| | B) | 10 |
| | C) | 20 |
| | D) | 40 |
| | E) | 80 |
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| 14 | |
In humans, the genes for red-green color blindness (R=normal, r=color-blind) and hemophilia A (H=normal, h=hemophilia) are both X-linked and only 3 map units apart. Suppose a woman has four sons, and two are colorblind and have hemophilia and two are normal. What is the probable genotype of the woman?
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| | A) | HR/hr |
| | B) | Hr/hr |
| | C) | hr/hr |
| | D) | Hr/hR |
| | E) | HR/Hr |
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| 15 | |
Both red/green color blindness (R=normal, r=colorblind) and Duchenne-type muscular dystrophy (d=normal, d=muscular dystrophy) are X-linked recessive traits that map close to each other. A woman has a father who is red/green color blind. Her mother's family has a history of Duchenne's muscular dystrophy. This woman is apparently healthy with neither color blindness nor muscular dystrophy. She marries a healthy man and they have four sons and two daughters. Half the sons are healthy but color blind, the other half have normal color vision, but have Duchenne's muscular dystrophy. The daughters are both normal. What is the genotype of the woman?
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| | A) | RD/rd |
| | B) | rd/rd |
| | C) | Rd/rD |
| | D) | RD/RD |
| | E) | cannot determine |
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| 16 | |
Suppose a three-point testcross was conducted involving the genes A, B, and C. If the most abundant classes are CaB and cAb and the rarest classes are cab and CAB, which gene is in the middle?
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| | A) | A |
| | B) | B |
| | C) | C |
| | D) | the genes are linked, but order cannot be determined |
| | E) | the genes are not linked |
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| 17 | |
Determine which gene is in the middle for each of the three experiments in the table below.
| | Experiment | | Gametes produced from heterozygote | A | B | C | | + + + | 303 | 1 | 41 | | + + c | 61 | 3 | 232 | | + b + | 11 | 29 | 82 | | + b c | 3 | 75 | 201 | | a + + | 1 | 77 | 194 | | a + c | 19 | 31 | 77 | | a b + | 72 | 4 | 235 | | a b c | 253 | 1 | 46 |
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| | A) | a, b, c |
| | B) | a, c, c |
| | C) | a, a, a |
| | D) | a, a, c |
| | E) | b, c, a |
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| 18 | |
In Drosophila, the genes A, B, and C are linked and in the order given. The distances are A-B = 5 m.u. and B-C = 15 m.u. The interval exhibits 20% interference. How many double crossovers would be recovered in a three-point cross involving A, B, and C out of 500 progeny?
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| | A) | 3 |
| | B) | 6 |
| | C) | 30 |
| | D) | 80 |
| | E) | 100 |
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| 19 | |
In tetrad analysis, NPD asci result from:
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| | A) | independent assortment of unlinked genes |
| | B) | double crossovers between linked genes |
| | C) | single crossovers between linked genes |
| | D) | single crossovers between a gene and a centromere |
| | E) | a or b |
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| 20 | |
A wild-type yeast strain is crossed with another strain that is auxotrophic for both arginine and uracil (needs both arginine and uracil added to medium to grow). The resulting heterozygous diploid cells are allowed to undergo meiosis and the products of meiosis are analyzed. There are three classes of unordered tetrads observed:
| Class one: | 2 spores that fail to grow without arginine and uracil | | (125 tetrads) | 2 wild-type spores
| | Class two: | 2 spores that fail to grow without arginine, but can grow without uracil | | (5 tetrads) | 2 spores that fail to grow without uracil, but can grow without arginine
| | Class three: | 1 spore that fails to grow without arginine and uracil | | (70 tetrads) | 1 wild-type spore | | | 1 spore that fails to grow without arginine, but can grow without uracil | | | 1 spore that fails to grow without uracil, but can grow without arginine |
What is the map distance between the arg and ura genes?
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| | A) | 40 m.u. |
| | B) | 20 m.u. |
| | C) | 2 m.u. |
| | D) | 36 m.u. |
| | E) | unlinked |
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| 21 | |
A nic+ ad haploid strain of Nerurospora is crossed with a nic ad+ haploid strain. A total of 1000 ordered tetrads are isolated and categorized as follows:
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | | nic+ad | nic+ad+ | nic+ad+ | nic+ad | nic+ad | nic+ad+ | nic+ad+ | | nic+ad | nic+ad+ | nic+ad+ | nic+ad | nic+ad | nic+ad+ | nic+ad+ | | nic+ad | nic+ad+ | nic+ad | nic ad | nic ad+ | nic ad | nic ad | | nic+ad | nic+ad+ | nic+ad | nic ad | nic ad+ | nic ad | nic ad | | nic ad+ | nic ad | nic ad+ | nic+ad+ | nic+ad | nic+ad+ | nic+ad | | nic ad+ | nic ad | nic ad+ | nic+ad+ | nic+ad | nic+ad+ | nic+ad | | nic ad+ | nic ad | nic ad | nic ad+ | nic ad+ | nic ad | nic ad+ | | nic ad+ | nic ad | nic ad | nic ad+ | nic ad+ | nic ad | nic ad+ | | 800 | 1 | 95 | 4 | 94 | 1 | 5 |
Which statement below is true?
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| | A) | nic and ad are unlinked. |
| | B) | nic and ad are on the same chromosome, on opposite sides of the centromere. |
| | C) | nic and ad are on the same chromosome, on the same side of the centromere, with nic closer to the centromere than ad. |
| | D) | nic and ad are on the same chromosome, on the same side of the centromere, with ad closer to the centromere than nic. |
| | E) | none of the above. |
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| 22 | |
A recessive mutation in the X-linked yellow gene of Drosophila confers a yellow body color to the flies, whereas a recessive mutation in another X-linked gene, singed, gives singed hairs. The singed gene (sn) maps closer to the centromere than the yellow (y) gene. In an animal of genotype y sn/ y+sn+, what type of spot(s) might you see as a result of mitotic recombination if a crossover occurs between the sn gene and the centromere?
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| | A) | a yellow/singed spot and normal tissue |
| | B) | twin spot |
| | C) | one yellow spot and normal tissue |
| | D) | all normal tissue |
| | E) | a or d |
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| 23 | |
A trihybrid (3 genes heterozygous) produces the following gametes:
| ABC | 46 | | Abc | 63 | | ABc | 4 | | aBc | 381 | | abc | 38 | | aBC | 71 | | abC | 2 | | AbC | 395 | | total | 1000 |
What is the order of and map distance between the three genes?
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| | A) | A←9→B←14→C |
| | B) | B←9→A←14→C |
| | C) | A←0.6→B←14→C |
| | D) | B←14→A←9→C |
| | E) | B←8.4→A←13.4→C |
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| 24 | |
A nematode homozygous for two genes, dpy (dumpy) and bli (blister) is crossed to an unc (uncoordinated) nematode. All three traits are recessive and map to the same chromosome. The F1 nematodes from the cross above are then testcrossed and the F2 phenotypes are listed below. Draw a linkage map showing the order and distance between the genes.
| Phenotype | n | | normal (wild type) | 40 | | blister | 7 | | uncoordinated | 561 | | dumpy, blister, uncoordinated | 36 | | dumpy | 97 | | uncoordinated, dumpy | 4 | | blister, uncoordinated | 102 | | dumpy, blister | 573 | | Total | 1420 |
What is the order of and map distance between the three genes?
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| | A) | unc←6→dpy←15→bli |
| | B) | unc←15→dpy←6→bli |
| | C) | dpy←1→bli←6→unc |
| | D) | bli←15→dpy←6→unc |
| | E) | a or d |
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| 25 | |
A true-breeding wild-type female fly is crossed to a true-breeding ebony male with vestigial wings and cardinal eyes. All traits are autosomal recessive. The resulting heterozygotes are testcrossed and the frequencies of the phenotypes are shown in the table below. Draw a linkage map of the three genes.
| eye color | body color | wings | number | | cardinal | ebony | vestigial | 245 | | red | tan | normal | 241 | | cardinal | ebony | normal | 235 | | red | tan | vestigial | 225 | | cardinal | tan | vestigial | 10 | | red | ebony | normal | 14 | | cardinal | tan | normal | 14 | | red | ebony | vestigial | 16 | | Total | | | 1000 |
What is the order of and map distance between the three genes?
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| | A) | c←5.4→e←48.4→v |
| | B) | c←48.4→e←5.4→v |
| | C) | c←3→e←46→v |
| | D) | c←46→e←3→v |
| | E) | c←5.4→e; v is unlinked to c and v |
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