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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 F2 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 F2 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 NONDISJUNCTION 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 NONRECOMBINANT 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 F1 mice are all agouti. What ratio of agouti mice to albino mice
would you expect to see in the F2?
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.