In Drosophila, a Heterozygous Female for the X-Linked Recessive Traits A, B, and C Was

Name ______

Bio 3070 Genetics F06

Examination 1 Dr. E. M. Golenberg October 9, 2006

Write your name on every page of this exam. You have 14 problems on this test. Please read the questions carefully and understand what is asked of you. Organize your thoughts and your work, so that the graders can clearly see what you are doing. Answer all question on these pages. If you need more room, write on the back of the same page. Put your name on all of the pages. If you come to a question and you don’t know how to attack it immediately, don’t panic. Skip the question and go on. When you are done with the rest of the test come back to it. You can get partial credit. Any evidence of cheating will result in your failing this exam. If you need it for the exam the following are 2 2 2 2 theoretical chi-squared values. ( 0.05, df=1= 3.84;  0.05, df=2= 5.99;  0.05, df=3= 7.82;  0.05, df=4= 9.49)

The first 9 problems are short answer problems and are worth 5 pts. or 10 pts. each. Each can be answered by a single sentence and or calculation.

1. Define pleiotropy. (5 pts.)

Pleiotropy is the condition where a single gene directly influences the phenotype of several traits or characters in an individual.

2. The following pedigree is for myopia in humans. Determine the most probable genotype for each individual based on your prediction of the pattern of inheritance (dominant/recessive, autosomal/X-linked).(10 pts.) Name ______

Autosomal recessive. aa A-(probably AA) Aa Aa Aa Aa Aa Aa aa A- A- A- aa A-

3. Using the binomial distribution equation, determine the probability of a family of five offspring having two children afflicted with sickle-cell anemia (an autosomal recessive disease) when both parents are heterozygous for the sickle-cell allele.. (5 pts.)

5!/2!3! (.25)^2 (.75)^3=0.264

4. You cross a true-breeding colorless individual with a true-breeding red individual. All F1 offspring are red. When you cross two F1 individuals, you get 70 reds and 58 colorless. The reciprocal cross gives similar results. Test whether this distribution fits a 3:1 ratio. (10 pts.)

X2= (96-70)2/96 + (32-58)2/32=7.04+21.13=28.2 df=1. Significantly differs from expected. Reject 3:1 ratio. Name ______

5. You set up the following cross: AA Bb Cc Dd Ee Ff X Aa Bb cc Dd Ee ff. All genes are autosomal and unlinked. BB is lethal. D and d are codominant. What proportion of the living offspring are expected to have a dominant A, dominant B, dominant C, codominant D (heterozygous) , recessive E, and recessive F phenotype? (5 pts.) 1 X 2/3 X 1/2 X 1/2 X 1/4 X1/2=1/48

6. Assume that A and C are linked and are separated by 10 cM. Answer problem 5 again. (10 pts.)

This would mean that from the first parent, we would get gametes AC or Ac 90 % of the time. However, the recombinants would also be Ac or AC. The same goes for the other parent in terms of the effect of recombination. The ratio remains the same.

7. At what stage are chiasmata first seen? (5 pts.)

Diplonema Name ______

8. Define bivalent. (5 pts.)

A bivalent is a structure that is a pair of homologous chromosomes found in meiosis I.

9. In a cross between a dihybrid individual and a tester, the offspring were of the following phenotypic classes (A means dominant A trait, a means recessive; B means dominant B trait, and b means recessive). 117 A B. 109 a b, 120 A b, 108 a B. Are the loci linked and if so what is there map distance? (5 pts.) The loci are unlinked. Name ______

10. In Drosophila, a heterozygous female for the X-linked recessive traits a, b, and c was crossed to a male that phenotypically expressed a, b, and c. The offspring occurred in the following phenotypic ratios:

+ b c 460 a + + 450 a b c 32 + + + 38 a + c 11 + b + 9 No other phenotypes were observed. a. What is the genotypic arrangement of the alleles of these genes on the X chromosome of the female? (5 pts.) b. Determine the correct sequence and construct a map of these genes on the X chromosome. (15 pts.) c. What progeny phenotypes are missing? Why? (10 pts.) a. + b c and a + + b. The gene order is a b c. The distance between a and b is 7 mu. The distance between b and c is 2 mu. c. The double crossing over class ++c and ab + are missing. The expected number is 1.4. There may be interference or the sample size is too small. Name ______

11. Drosophila melanogaster has one pair of sex chromosomes (XX or XY) and three pairs of autosomes, referred to as chromosomes 2, 3, and 4. A genetics student discovered a male fly with very short legs. Using this male, the student was able to establish a pure breeding stock of this mutant and found that it was recessive. She then incorporated the mutant into a stock containing the recessive gene black (body color located on chromosome 2) and the recessive gene pink (eye color located on chromosome 3). A female from the homozygous black pink short stock was then mated to a wild-type male. The F1 males of this cross were all wild type and were then backcrossed to the homozygous b p sh females. The F2 results appeared as shown in the table that follows. No other phenotypes were observed. Wild Pink Black Short Black Pink Short Females 63 58 55 69 Males 59 65 51 60 a. Based on these results, the student was able to assign short to a linkage group (a chromosome). Which one was it? Include a step-by-step reasoning. (15 pts.) b. The student repeated the experiment, making the reciprocal cross, F1 females backcrossed to homozygous b p sh males. She observed 85 percent of the offspring fell into the given classes (above), but 15% or the offspring were equally divided among b + p , b + +, + sh p and+ sh + phenotypic males and females. How can these results be explained and what information can be derived from the data? (15 pts.)

a. Black and short always appear together or wildtype with wildtype. In terms of pink, you have 123 pink –wildtype, 106 wildtype- short, 122 wildtype-wildtype, and 129 pink short. This is close to independent assortment. There is no sex-specific distribution. Males don’t recombine. Altogether, short must be on the same chromosome as black, chromosome 2. b. The distance between black and short must be 15 map units. (There is recombination in the females.) Name ______

12. In birds, the male contains two like sex chromosomes and is designated ZZ. The female contains one Z as well as a nearly blank chromosome designed W (females are ZA). In parakeets, two genes control feathe color. The presence of the dominant Y allele at the first gene results in the production of a yellow pigment. The dominant B allele at the second gene controls melanin production. When both genes are active, a green pigment results. If only the Y gene is active, the feathers are yellow. If only the B gene is active, a blue color is exhibited. If neither gene is active, the birds are albinos. Therefore, with our conventional designation, phenotypes are produced as follows: Y - B - green Y - b b yellow y y B - blue y y b b albino

A series of crosses established that one of the genes is autosomal and one is Z- linked. Based on the results of the following cross shown here, where both parents are true-breeding, determine which gene is Z-linked. P: green male X albino female F1: 1/2 green males, 1/2 green females F2: 6/16 green males 2/16 yellow males 3/16 green females 1/16 yellow females 3/16 blue females 1/16 albino females Support your answer by establishing the genotypes of the P parents and working the cross through the F2 generation. (30 pts.) You can tell right away from the F2 males that Y is on Z. All males get the Y from the mother, but not always the B. This implies that the mother is heterozygous for Bb which means it must be autosomal. The crosses and genotypes are as follows: P: green male YY BB X albino female yW bb F1: 1/2 green males Yy Bb, 1/2 green females YW Bb F2: 6/16 green males Y- B - 2/16 yellow males Y - bb 3/16 green females YW B- 1/16 yellow females YW bb 3/16 blue females yW B- 1/16 albino females yW bb Name ______

13. Pigeons may exhibit a checkered or plain pattern. In a series of controlled matings, the following data were obtained:

F1 Progeny P Cross Checkered Plain a) checkered X checkered 36 0 b) checkered X plain 38 0 c) plain X plain 0 35

Then, F1 offspring were selectively mated with the following results (the P cross giving rise to each F1 pigeon is indicated in parentheses): Progeny

F1 X F1 Crosses Checkered Plain checkered (a) X plain (c) 34 0 checkered (b) X plain (c) 17 14 checkered (b) X checkered (b) 28 9 checkered (a) X checkered (b) 39 0 How are the checkered and plain patterns inherited? Select and define symbols for the genes involved, and determine the genotypes of the parents and offspring in each cross. (25 pts.)

The trait is a simple one gene two allele system with Checkered (Ch) dominant over plain (ch ch ). a. Ch Ch X Ch Ch b. Ch Ch X ch ch c. ch ch X ch ch d. Ch Ch X ch ch e. Ch ch X ch ch f. Ch ch X Ch ch g. Ch Ch X Ch ch Name ______

14. During oogenesis in an animal species with a haploid number of 6, one dyad undergoes nondisjunction during meiosis II. Following the second meiotic division, the involved dyad ends up intact in the ovum. How many chromosomes are present in (a) the mature ovum, (b) the second polar body, and (c) the first polar body? (d) Following fertilization by a normal sperm, how many chromosomes are present? (e) What chromosomal condition is created? (25 pts.)

a. 7 b. 5 c. 6 d. 13 e. trisomy.