Genetics 275 Examination February 10, 2003. Do all questions in the spaces provided. The value for this examination is twenty marks (20% of the grade for the course). The value for individual questions is one except where indicated. 1. In corn, the R allele encodes red aleurones (seed kernel component) and its recessive allele r determines colorless aleurones. A cross was made between a diploid r/r plant and a trisomic R/r/r plant, using the diploid as the female parent. Among the progeny there were 1282 red kernels and 2451 colorless kernels. Explain why this approximation of a 1:2 ratio is expected, if n+1 pollen grains are nonfunctional. You may wish to use a diagram or Punnett square. MALE GAMETES WOULD BE PRODUCED IN THE PROPORTION OF:2Rr : 1 rr : 2 r : 1 R. IF N + 1 GAMETES ARE NONFUNCTIONAL IN THE MALE, THEN THE FUNCTIONING GAMETES ARE R AND r, IN A RATIO OF 1 : 2. THEREFORE, IN A CROSS TO rr WE EXPECT THE RATIO OF 1 RED TO 2 COLORLESS. 2. On the basis of the information in question #1, what ratio of red to colorless kernels would be expected from a cross of r/r females to R/R/r males? WITH THE MALE PARENT RRr, GAMETES ARE EXPECTED AS 2 R: 1RR: 2Rr: 1 r. IF THE Rr AND RR GAMETES FAIL TO FUNCTION, IN A TESTCROSS WE WOULD EXPECT A RATIO OF 2 RED TO 1 COLORLESS. 3. Recall that a test cross is when you cross a heterozygous individual to an individual that is homozygous recessive for the same genes. You have examined the test cross ratio obtained from a particular heterozygous individual and find it to be 1 wild type: 3 mutant. If the original heterozygous parent had been selfed or crossed to another individual of identical genotype what F 2 ratio would have been obtained among the offspring of this cross? a) 1:1:1:1 b) 3:1 c) 7:9 d) 9:7 e) 9:3:3:1 f) 2:1 CORRECT ANSWER IS D. 4. You have examined another testcross ratio and found it to be 1 wild type: 7 mutant. What phenotypic F 2 ratio would you obtain from selfing the heterozygous individual that produced this test cross ratio? a) 1:1:1:1 b) 1:1:1:1:1:1:1:1 c) 27:9:9:9:3:3:3:1 d) 9:3:3:1 e) 27:37 f) 7:1 CORRECT ANSWER IS E.
5. (2 marks) Hemophilia (a blood-clotting deficiency disease) is a sex-linked (Xlinked) trait in humans. If a hemophiliac Turner female is born to normal parents, in which parent and at what meiotic division did the non-disjunction event occur that produced the Turner female? THE TURNER RECEIVED THE HEMOPHILIA ALLELE FROM A HETEROZYGOUS MOTHER, THEREFORE THE NON-DISJUNCTION MUST HAVE OCCURRED IN THE MALE PARENT, WHEREIN THE MALE GAMETE CONTAINED NEITHER AN X NOR A Y. ONE CAN T TELL FOR CERTAIN WHICH MEIOTIC DIVISION THE NON- DISJUNCTION EVENT OCCURRED IN. IT COULD HAVE OCCURED IN EITHER MEIOTIC DIVISION. IF IT OCCURRED DURING MEIOSIS I, THE NULLO-GAMETE WAS UTILIZED AND THE XY CONTAINING GAMETE WAS NOT UTILIZED. IF IT OCCURRED DURING MEIOSIS II, THEN THE XX GAMETE WAS NOT USED AND THE NULLO-X GAMETE WAS OR THE YY GAMETE WAS NOT USED AND THE NULLO-Y GAMETE WAS. 6. Why is that, if viable, a homozygous inversion has full fertility but the same inversion as a heterozygote exhibits semi-sterility? A HOMOZYGOUS INVERSION HAS NO CONTORTIONS DURING MEIOTIC PAIRING, THEREFORE WHEN RECOMBINATION OCCURS THERE ARE NO DELETION NOR DUPLICATION CHROMATIDS PRODUCED. HOWEVER, IN ORDER TO MAXIMIZE PAIRING WITH A NORMAL CHROMOSOME IN AN INVERSION HETEROZYGOTE THE PAIRING LOOP FORMS AND THIS CREATES THE SEMI-STERILITY WHEN THE RECOMBINATION EVENT OCCURS WITHIN THE LIMITS OF THE INVERSION. 7. How can one account for the existence of individuals in the human population with Lepore hemoglobin? SORRY ABOUT THIS ONE. YOU MUST REMEMBER A DETAIL HERE. LEPORE HEMOGLOBIN IS WHEN AN ANEMIA RESULTS FROM UNEQUAL CROSSING OVER BETWEEN ANCIENT BUT NOW EVOLVED BETA AND DELTA GENES. THUS, THE RESULT IS A FUSION BETWEEN DELTA AND BETA AND NO COMPLETE GENE OF EITHER TYPE. NORMAL INDIVIDUALS HAVE A COMPLETE DELTA AND A COMPLETE BETA GENE ON EACH MEMBER OF THE HOMOLOGOUS PAIR. 8. For humans, why do you suppose that aneuploid karyotypes involving the sex chromosomes are much more compatible with viability than somatic chromosome aneuploids? THE LIKELY ANSWER IS THAT HUMANS HAVE EVOLVED TO COPE WITH AN INEQUALITY IN SEX CHROMOSOME NUMBER, SINCE FEMALES HAVE TWO AND MALES HAVE ONE.
9. If 5% of the meiocytes in an organism exhibit a chiasma between two genes, how many map units separate them? This is a question calling on your background information about meiosis. Give your reasoning. SINCE A CHIASMATA REPRESENTS THE PHYSICAL MANIFESTATION OF A C/O EVENT, THE RESULT IS THAT FOR A MEIOCYTE WITH A CHIASMATA THERE ARE TWO RECOMBINANT CHROMATIDS AND TWO NON- RECOMBINANT CHROMATIDS. THUS, THE PERCENT RECOMBINATION FOR THIS MEIOCYTE IS 50%, AND SINCE ONE % RECOMBINATION = 1 MAP UNIT, THIS WOULD MEAN THAT THE TWO GENES ARE 2.5 MAP UNITS APART. 10. Present one possible reason why adjacent-2 segregation patterns are either much less frequent or even absent compared to adjacent-1 segregation patterns during meiosis with a translocation heterozygote. AGAIN, THIS IS AN EVOLUTIONARY CONSIDERATION. HOMOLOGOUS CENTROMERES HAVE EVOLVED TO SEGREGATE SO IT IS MUCH MORE UNLIKELY THAT ADJACENT-2 SEGREGATION WILL OCCUR, SINCE THIS INVOLVES NON-SEGREGATION OF HOMOLOGOUS CENTROMERES. ADJACENT-1 SEGREGATION HAS HOMOLOGOUS CENTROMERES SEGREGATING NORMALLY. 11. Assume that there are five sequential steps in the formation of wild-type fur in mice (agouti), and that each step in this pathway is mediated by a different gene product. Mutations in any one of the genes responsible for mediating any of the five steps will interrupt this pathway if they are homozygous and then the mice are albino. You cross two true-breeding albino lines of mice (that is they are homozygous for whatever pigment-producing genes they carry) and the F 1 mice are all agouti. What ratio of agouti mice to albino mice would you expect to see in the F 2? a) 3:1 b) 9:7 c) 15:1 d) 7:1 e) 13:3 f) 12:3:4 g) 9:3:3:1 CORRECT ANSWER IS B. 12. (2 marks) Explain the karyotypic (chromosomal) difference between familial and non-familial Down syndrome. Be sure to give chromosome number and how this number was caused. NON-FAMILIAL DOWN SYNDROME INDIVIDUALS HAVE 47 CHROMOSOMES COMPARED TO THE NORMAL NUMBER OF 46, AND THE DIFFERENCE IS THAT THEY HAVE 3 COPIES OF CHROMOSOME 21 AND REPRESENT AN EXAMPLE OF A VIABLE HUMAN TRISOMIC. THIS CONDITION IS MOST OFTEN CAUSED BY NON-DISJUNCTION IN ONE OF THE PARENTS TO PRODUCE A GAMETE WITH TWO COPIES OF 21. MOST FAMILIAL DOWN INDIVIDUALS HAVE 46 CHROMOSOMES CONSISTING OF
A NORMAL 14, TWO NORMAL 21S AND ONE LARGE 14/21 TRANSLOCATION CHROMOSOME. THUS, EFFECTIVELY THE DOSE FOR 14 IS TWO AND FOR 21 IT IS 3. THIS CONDITION IS THE RESULT OF A RECIPROCAL TRANSLOCATION BETWEEN CHROMOSOMES 14 AND 21 AND THE SMALL 21/14 PIECE OF THE TRANSLOCATION USUALLY GOES MISSING IN FAMILIAL DOWN PEDIGREES. 13. (2 marks) Very briefly, if you were examining chromosomal aberrations microscopically, what diagnostic features involving homologous pairing would you look for to distinguish deletions, duplications, inversions, and translocations from each other before any genetic crossing-over events had taken place? ALL OF THESE ARE MOST EASILY DETECTED AS HETEROZYGOTES. DELETIONS: LOOK FOR A SMALL REGION WITH NO PAIRING PARTNER. THE NON-DELETED HOMOLOGUE WOULD HAVE A BULGE OR LOOP OPPOSITE THE DELETION. DUPLICATIONS: MUCH THE SAME AS DELETIONS EXCEPT THE BULGE IS NOW ON THE DUPLICATED CHROMOSOME. INVERSIONS: ONE LOOKS FOR A MORE COMPLICATED LOOP STRUCTURE BUT THIS TIME IT INVOLVES BOTH MEMBERS OF THE HOMOLOGOUS PAIR. TRANSLOCATIONS: ONE LOOKS FOR THE DIAGNOSITIC CROSS-LIKE CONFIGURATION OF PAIRING HOMOLOGUES. 14. (2 marks) Which type of chromosomal aberration is associated with the production of dicentrics and acentric chromosome fragments as the result of recombination and what type of cross-over event in heterozygotes carrying this aberration will produce a meiotic tetrad with no viable gametes at all? PARACENTRIC INVERSIONS PRODUCE THESE EVENTS WHEN A SINGLE C/O OCCURS WITHIN THE LIMITS OF THE INVERSION. A FOUR CHROMATID DOUBLE C/O WOULD PRODUCE TWO DICENTRICS AND TWO ACENTRICS, THUS NO VIALBLE PRODUCTS AT ALL. 15. (2 marks) Assume that you have a herd of goats that is heterozygous in trans for two autosomal genes on the same chromosome that influence coat color and texture. The a+ allele causes brown hair on the goat while the a allele causes white hair when homozygous. The b+ allele causes normal hair length but the b allele causes the long angora hair when homozygous. Thus, the herd all has brown fur and normal length hair(non-angora) and are a+/a and b+/b. You note a likely case of mitotic recombination in one of your goats. This goat has mostly brown hair of normal length but there is a twin-spot of mutant hair. One of the twin spots has angora brown hair and the other one has non-angora but white hair. Draw the chromosomal and gene
arrangement with one mitotic recombination event that would give a goat with mostly normal brown non-angora hair but with the twin spot described above. HAVE a + /a AND b + /b HETEROZYGOUS IN TRANS SO THAT THE CHROMOSOMES COULD BE a + b ON ONE HOMOLOGUE AND a b + ON THE OTHER HOMOLOGUE. THEN SHOW A C/O THAT IS CENTROMERE PROXIMAL TO BOTH LOCI AND YOU HAVE PRODUCED THE GENOTYPE AND EVENT THAT WILL GIVE THE TWIN SPOT. THERE ARE TWO VERSIONS OF RIGHT ANSWER FOR THIS ONE. AS LONG AS THE C/O IS CENTROMERE PROXIMAL TO BOTH GENES, IT DOESN T MATTER WHICH GENE YOU PUT FURTHER FROM THE CENTROMERE.