Fall 2004 Animal Science 213 Animal Genetics EXAM Points Exam must be completed in INK!

Transcription

1 Fall 2004 Animal Science 213 Animal Genetics EXAM Points Exam must be completed in INK! Name: KEY Multiple Choice (4 points each) 1. If an individual plant homozygous dominant for the single gene tall (DD) is mated to an individual homozygous recessive for the dwarf phenotype (dd), what proportion of the offspring will be tall? a. 0% b. 25% c. 50% d. 75% e. 100% 2. In the cross mentioned above, how many of the offspring will be heterozygous? a. 0% b. 25% c. 50% d. 75% e. 100% 3. Flower color, is controlled by alleles at the W locus (W red, w white). The results of a WW x ww cross are pink F 1 flowers. When the F1 plants with pink flowers are self-crossed, the proportions of the resulting F2 generation are 25% red, 50% pink, 25% white. What is the relationship between the W and w alleles? a. W is dominant to w b. w is dominant to W c. W and w are partially dominant d. W and w are codominant

2 4. If heterozygous tall plants with pink flowers (DdWw) are self-crossed, what proportion of the offspring will be tall with pink flowers? a. 25% b. 37.5% c. 50% d. 75% e. 100% 5. In Labrador retrievers, two of the loci controlling coat color (black, chocolate and yellow) are the E locus and the B locus. At the B locus, black (B) is dominant to chocolate (b). At the E locus, homozygous recessive alleles at the E locus (ee) results in yellow pups regardless of genotype at the B locus. a. E is epistatic to B b. B is epistatic to E 6. If a chocolate lab is mated to a yellow lab, is it possible to have black pups? a. Yes (if you went with the information from class; there is a third allele) b. No c. Not enough information (if you went strictly off the information from question Research at an institution in the US has isolated a mutation in mice that results in the absence of a tail in homozygous recessive individuals. At the same time, researchers in the Netherlands also identify a recessive tailless mutation. The researchers exchange breeding stock and cross US tailless mice with Netherlands tailless mice. The resulting offspring all have tails. Do the mutations complement each other? a. Yes b. No c. Not enough information can be generated from one mating. 8. Which of the following is an example of a temperature-sensitive effect? a. phenylketonuria b. Tay-Sach s disease c. Point (snout, ear and paw) color in Himalayan rabbits d. None of the above 9. Which of the following is an example of a nutritionally sensitive effect? a. phenylketonuria b. Tay-Sach s disease c. Point (snout, ear and paw) color in Himalayan rabbits d. None of the above 10. When a gene that is normally expressed is turned off when it is moved to a portion of the chromosome that is largely heterochromatin, this is an example of a a. tumor suppressor b. permanent mutation c. heterochromatin translocation

3 d. positional effect 11. The (2n + 1) rule allows an individual to calculate the number of a. alleles controlling a trait b. the number of genes controlling a trait c. the predicted outcome of a genetic cross d. none of the above 12. Traits exhibiting continuous variation are often a. polygenic b. quantifiable c. influenced by additive alleles d. a and b e. a and c f. all of the above 13. Erythroblastosis fetalis is a condition caused by incompatibility of a. secretor status b. parental status c. ABO status d. Rh factor status 14. Lethal alleles can be which of the following? a. dominant b. recessive c. conditional d. all of above 15. From a dihybrid cross, it was concluded that the 2 genes were linked with a recombination frequency of 20 percent. How many of the resulting offspring were nonparental (recombinant)? a. 40% b. 80% c. 20% d. 10% e. 60% 16. Which of the following is NOT a characteristic of Mendelian genetics? a. independent segregation b. simple dominance/recessiveness c. linkage d. genes occur in pairs True/False (1 point each)

4 F 17. Phenotype always reflects genotype. T 18. It is possible to have more than two alleles for one locus. T 19. Mitochondria and chloroplasts replicate their own DNA T 20. Traits that segregate independently are not linked. T 21. Mendel s unit factors are now known to be genes. T 22. The alleles for the genes controlling presence of the A/B/0 antigens are codominant T 23. Substance H is involved in blood type T 24. An individual may have no more than two alleles for a locus. T 25. The greater the distance between 2 linked genes, the greater the recombination frequency. F 26. If two genes are following Mendelian patterns of inheritance, the genes must be linked. Problems (points as assigned) 27. (4 pts) In Angus cattle, two coat colors are possible, red and black, controlled by a single locus with two alleles (B black, b red). Over several breeding seasons, a black bull, A286, was mated to 200 red cows, who gave birth to 104 F 1 black calves, 96 F 1 red calves. For each color, half were heifer calves. When the 100 heifer calves grew to maturity, they were all mated to a red bull. The 52 black F 1 cows gave birth to 25 black F 2 calves and 27 red F 2 calves. The 48 F 1 red cows gave birth to 48 F 2 red calves. a. Based on the results of the above matings, what is the relationship between B and b? B is dominant to b. b. What is the genotype of the original black bull, A286? Bb 28. (4 pts)the presence of scurs, which are small, moveable pseudohorns is another trait controlled by two alleles at a single locus in cattle (S n normal, S c scurred). A286 possessed scurs,

5 but none of the 200 P 1 cows did. Of the F 1 calves, all of the bull calves were scurred, but none of the 100 heifer calves were scurred. The red bull mated to the F 1 heifers was also scurred. When the F 2 calves were born, there were 50 each bulls and heifers. All of the bull calves were scurred, but only 25 of the heifer calves were scurred. a. Is the scurs trait sex-limited? No b. Is the scurs trait sex-influenced? Yes c. Is scurs dominant in males? Yes d. Is scurs dominant in females? No 29. (6 pts) The developmental mutation, bicoid (bcd-) is recessive in mice. Normal bicoid protein must be present for embryos to form normally; if bicoid is not produced, the embryos will die. However, development can occur in homozygous recessives. Consider a cross between a female heterozygote (bcd+/bcd-) and a male homozygote bcd-/bcd-). a. How can males homozygous for the mutation survive? There is a maternal effect. Male embryos contain normal bicoid mrna from the egg, which is then translated into normal bicoid protein. b. Predict the outcome (normal vs. failed [lethal]) embryogenesis in the F 1 offspring of the cross above. All of the embryos will develop normally because of the maternal contribution of bicoid mrna then translated into bicoid protein. c. If the F 1 offspring are crossed, predict the outcome (normal vs. failed [lethal]) embryogenesis in the F 2. (Describe the genetic conditions that would allow survival, those that would not. No more than two short sentences are sufficient to answer this) Any embryo born to a bcd-/bcd- mother would fail (no normal bicoid mrna to contribute). Any embryo born to a mother with at least one copy of bcd+ would develop normally. 30. (4 pts) Describe an example of infectious heredity in flies. There were two examples given in class. The first was a sensitivity to CO 2 anesthesia in flies infected with the sigma virus. Infected flies do not recover from anesthesia while uninfected flies do. The sensitivity and the virus are passed on to subsequent generations. The second example is a temperature-sensitive alteration in sex ratios of fruit flies born to females infected with a protozoan. When reared at room temperature, there is a predominance of females born. The condition is passed only to females. 31. (4 pts) Below is a table showing inheritance of leaf pigmentation in a particular variety of four o clock plants.

6 Ovule Source Pollen Source White Green Variegated White White Green W, G, var Green White Green W, G, var Variegated White Green W, G, var a. What source (ovule or pollen determines the color of leaves in this variety of four o'clocks? Ovule b. What is the source of DNA controlling leaf pigmentation? Chloroplast (I will accept organelle) 32. (8 pts) The diagram on the next page is highlighting the key steps of meiosis, focusing on a single chromosome and 2 genes (A and B) that are linked on that chromosome. Using what you ve learned about linkage and the process of meiosis, complete the following: (2 pts) In the blanks next to each primary germ cell, write the genotype. These answers are written in on the separate Powerpoint document (1 pt) Identify (draw an arrow to) the tetrad structure. These answers are written in on the separate Powerpoint document (.5 pt each) Fill in the gametic genotypes, and identify what percentage of each would result (recombination frequency = 20%). These answers are written in on the separate Powerpoint document (1 pt) The mating illustrated in this scenario, where one individual is mated with an all-recessive individual (recessive phenotype for all traits of interest) is known as what type of cross? Test cross 33. (10 pts) The genes for body type and coat color are 15 map units apart. Dwarf (dw) body type is recessive to long (wild-type) body; likewise, albino (a) coat is recessive to brown (wildtype) coat. A normal individual (long body, brown coat) is mated to an albino, dwarf individual, and all of the F1 offspring are wild-type. Note: Use the horizontal line(s) format to show genotypes, for example: _dw a_ (you can use DW and A instead of + if you prefer) + + (2 pts) What is the genotype of the normal individual in this mating? (2 pts) What is the genotype of the F1 offspring? + + dw a (2 pts) Explain why recombination does not have an effect on the outcome of this mating. Both parents are homozygous at both loci.

7 Next, an F1 female from the mating above is crossed with a recessive (dwarf, albino) male to generate the F2 offspring. (2 pts) What percentage of the F2 offspring will be dwarfed with brown coat? 7.5% (2 pts) What are the recombinant genotypes? + a and dw + dw a dw a 34. (12 pts) 3-point mapping problem. Three recessive traits in fruit flies are being studied, short wings (sw), black body (b), and yellow eyes (y). The wildtype (dominant) traits for each gene are long wings, grey body, and red eyes. In order to set up an informative test cross, an initial mating was conducted between a short-winged female (wildtype for other 2 traits, and homozygous at each locus) and a black-bodied, yellow-eyed male (wildtype for wing type, homozygous). All of the resulting F1 offspring were wild type, as expected. Using horizontal line format to indicate genotypes (as in previous problem), diagram this cross and the resulting F1 (4 pts). sw + + x + b y F 1 : sw + + sw b y + b y NOTE: The order of genes does NOT matter at this point.

8 Subsequently, an F1 female was mated to a male recessive for all 3 traits. Out of 1000 offspring counted, the following numbers were observed: F2 phenotype Black body, short wings 28 Black body 146 Short wings 310 Yellow eyes 32 Short wings, black body, yellow eyes 3 Black body, yellow eyes 322 Short wings, yellow eyes 154 Wildtype 5 (4 pts) What is the correct gene sequence? b sw y OR y sw b (sw is in the middle) (4 pts) What is the map unit distance between the body type (b) gene and the wing type (sw) gene? 6.8 map units