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Genetics Exam #1
February 27, 2006
(5) Define and distinguish between heterochromatin and euchromatin.
heterochromatin is the condensed, gene poor DNA found mainly near centromeres and telomeres
euchromatin is the less condensed, gene rich DNA where most genes are transcribed
(5) Define and distinguish between centromere and telomere.
centromeres are regions of the chromosomes with DNA sequences recognized by kinetochores
telomeres are the terminal regions of single stranded DNA that regulate the ability to divide
(3) Define sry.
master sex determination gene found on the Y chromosome
(3) Define heterokaryon.
cell that contains nuclei or chromosomes from more than 1 species (typically mouse and human)
(3) Define inversion.
condition where some of the genes along a chromosome are found in reversed order
(3) Define autosome.
any of the chromosomes in the body other than the sex chromosomes X and Y
(3) Define Barr body.
condensed, inactivated X chromosome found (almost) exclusively in females
(3) Define cyclin.
protein familywhose abundance varies according to the cell cycle and regulate the cell cycle
(3) Define histone.
protein component of chromatin, family of basic proteins that stabilize DNA
(3) Define synteny.
chromosomal map of genetic and physical linkages comparing chromosomes across species
(3) Define karyotype.
ordered presentation of metaphase chromosomes used for analyzing chromosome structure
(3) Define chimeric gene.
gene generated during a translocation that fuses 2 (or more) different genes into 1
(10) List and describe the 5 phases of meiosis.
leptotene- chromosomes begin to condense
zygotene- homologous chromosomes pair up
pachytene- continued pairing/condensation; crossovers form
diplotene- chromosomes start separating; chiasma become visible
diakinesis- chromosomal pairs are separated; chiasma resolved
(10) For peas, round pods are dominant over wrinkled pods. If a heterozygous round pod plant is
self crossed, what is the probability of obtaining an odd number of wrinkled pod plants out of 6
offspring?
for simple dominance and recessives, P(dominant)= 0.75 and P(recessive)=0.25
to have an odd number of wrinkled pea plants out of 6, requires 1, 3, or 5 plants be wrinkled
P(odd) = P(1) + P(3) + P(5)
= [6!/(1!5!)](.25)1(.75)5 + [6!/(3!3!)](.25)3(.75)3 + [6!/(5!1!)](.25)5(.75)1
= 0.3560 + 0.1318 + 0.0044
P(odd) = 0.4922 or 49.22%
(10) When a black spotted snail is crossed to a solid white snail, all of the offspring are black
spotted. When these individuals are crossed to themselves, 57% of offspring are black spotted,
19% are brown spotted, 18% are solid black, and 6% are solid white. Explain what is going on
genetically.
From the F1 result, black is dominant and spotted is dominant. In the F2 generation, 2 new
phenotypes appear: brown spotted and solid black. Based on the 9:3:3:1 ratio, there are likely 2
separate genes involved, with black dominant over brown and spotted dominant over solid. In
the double recessive homozygotes, the two genes show epistasis, where the 'new' phenotype of
white appears.
(10) Genes A and B are 32 centimorgans apart. If an AaBb trans-heterozygote is testcrossed,
what are the frequencies of each phenotype?
for it to be a trans heterozygote, the original cross must be Ab/aB x aabb
if a and b are 32 centimorgans apart, the total recombination frequency = 32%, or 16% each
recombinants would be AB and ab, while parentals would make up 100-32 = 68, or 34 % each
therefore: Ab 34% aB 34%
AB 16% ab 16%
(10) If you performed the test cross above and obtained 23 AB phenotypes, 21 ab phenotypes, 27
Ab phenotypes, and 29 aB phenotypes, are these results consistent with your predictions? Why
or why not?
there are 23+21+27+29 = 100 total individuals so would expect 34Ab, 34 aB, 16 AB and 16 ab
we observed 27 Ab, 29 aB, 23 AB and 21 ab
2 =  (observed-expected)2/ expected
= (27-34)2/34 + (29-34)2/34 + (23-16)2/16 + (21-16)2/16
= 1.441 + 0.735 + 3.063 + 1.563
= 6.802 with (4-1) 3 degrees of freedom, giving a P value of 0.09 or so
therefore these numbers ARE consistent for a recombination distance of 32 centimorgans
(10) Using the information in the table below, prove whether gene A or gene B is closer to the
centromere in Neurospora, a yeast that produces ordered tetrads. Explain your answer.
# asci
231
222
69
66
63
60
1-2
AB
ab
AB
ab
Ab
aB
3-4
AB
ab
ab
AB
AB
ab
5-6
ab
AB
AB
AB
aB
AB
7-8
ab
AB
ab
ab
ab
Ab
1st division
1st division
2nd division for both
2nd division for both
2nd division for B only
2nd division for B only
total = 231+222+69+66+63+60 = 711 asci
0.5(69+66)/711 = distance to A
0.5(69+66+63+60)/711 = distance to B
therefore A must be closer to the centromere than B since A has less 2nd division segregations