Download Problem Set 1A

Problem Set 2B
9-26-06
Name and Lab Section:_____________________
1. Define each of the following rearrangements (mutations) (use one phrase or sentence for
each). Then describe what kind of chromosomal structure you might see in cells in meiotic
prophase I if those cells are heterozygous for each of these rearrangements (one phrase or one
sentence). If possible, give enough detail to distinguish each rearrangement from the others in
your description of what you might see.
A. deletion: The loss of a seqment of DNA. In prophase I a person might see a
loop-out from one of the paired chromosomes (two of the four chromatids).
B. duplication: A portion of a chromosome is duplicated, so its present twice. A
person might see a loop-out that would look the same as in A above. (Note: it
might not be possible to distinguish whether you are looking at a deletion or a
duplication, just by looking at the paired chromosomes, unless there is a distinctive
banding pattern.)
C. inversion: The DNA sequences (or genes) in a portion of a chromosome have
rearranged such that they are in the opposite order (run in the opposite direction)
from what they originally were. A person might see a loop-out of the
chromosomes, but in this case (as opposed to the situation with deletions or
duplications) both chromosomes would be involved in the loop. Thus all four
chromatids would be involved in the loop.
D. translocation: Movement of genetic material between nonhomologous
chromosomes. A person might see a cross structure formed that involves four
chromosomes.
2. Draw a pair of homologous chromosomes as they would be matched during prophase I if
they had a paracentric inversion. (be sure to give yourself plenty of space so drawing is clear)
Let the original gene order on these chromosomes be: L, M, N, centromere, O, P, Q, R, S, T.
Let one of the chromosomes contain an inversion that includes P, Q, and R. Remember that
the chromosomes are duplicated at this time, and thus contain sister chromatids. (For the exam
you should be able to do this for deletions and duplications too.)
3. Now assume that a crossover occurs between genes Q and R. Draw 4 possible
chromosomes (with letters included) that might arise from this by the end of meiosis II (there
might be one of each of these chromosomes in 4 different gametes).
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Note that the break that occurred at anaphase I in the dicentric chromosome, could have
occurred anywhere between the two centromeres.
4. Compare legitimate recombination to illegitimate recombination. Which is more common?
Legitimate recombination is recombination between two DNA sequences that share
regions of high similarity, as opposed to illegitimate recombination, which is
recombination between two DNA sequences which share very little sequence similarity.
Legitimate recombination is the most common.
5. Most cases of autosomal aneuploidy in humans involve the small chromosomes. Why is
this? Since larger chromosomes contain more genes, aneuploidy in larger chromosomes
will result in gene dosage problems with a larger number of genes, which is more likely
to result in an embryo not even completing development.
6. Red-green color blindness is an X-linked recessive disorder. Bob has a 47,XXY karyotype
(Klinefelter syndrome) and is color blind. His 46,XY brother, Tom, is also color blind. Both of
their parents have normal color vision. Where did the nondisjunction occur that caused Bob to
be color blind and have Klinefelter syndrome?
Probably in meiosis II in their mother. The reasoning: Since Bob and Tom’s father
is not color blind, the X-linked color blind allele must be carried on one of their mother’s X
chromosomes. Tom got that color blind allele from one of his mother’s X chromosomes. Bob
got two X’s from his mom, both of which were carrying the color blind allele, so those two X’s
had to have come from nondisjunction in his mother, probably during meiosis II. Note that
nondisjunction in meiosis I would give rise to a cell with a regular X and a “color-blind” X,
which would have lead to Bob being a carrier with normal sight. A much less-likely
possibility would be mitotic nondisjunction in the mother.
7. Why does recombination in the inversion regions appear to be suppressed in individuals
heterozygous for paracentric and pericentric inversions?
A crossover within a paracentric inversion produces a dicentric and an acentric
recombinant chromatid. The acentric fragment is lost, and the dicentric fragment breaks,
resulting in chromatids with large deletions that lead to nonviable gametes or embryonic
lethality. A crossover within a pericentric inversion produces recombinant chromatids
that have duplications or deletions. Gametes with these recombinant chromatids also do
not lead to viable progeny. So even though there is the same amount of recombination,
those recombinant products never make it into viable offspring.
8. Give two ways that translocations can produce phenotypic effects.
If a translocation break point is within a gene, the translocation will disrupt that
gene, inactivating it. The translocation of a gene, causing it to become adjacent to a new
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region of a chromosome, changes the linkage pattern. That is, it changes the types of
sequences that are near the gene, which can lead to position effects, which are changes in
expression of the gene.
9. Draw a temperature profile graph for 2 cycles of PCR, giving an approximate temperature
at each stage, and tell in one short phrase what happens at each of those stages. You could
write each of the phrases alongside each step in the profile.
10. Use an entire sheet of paper for this drawing to give you plenty of room. It doesn’t need to
be neat, only clear.
Draw three base pairs of a DNA molecule that has the sequence 5’-GCA-3.’ Draw one
strand down the page on the left, and the other strand base-paired with it just to the right. Use
the same amount of detail as we used in class. Show the negative charge on each phosphate
group, label the 5’ and 3’ carbon positions with numbers, and show the hydrogen bonding
between paired bases with dashed lines (even though they don’t need to be placed accurately
with respect to specific atoms). Label the 5’ and 3’ ends of each strand. The only nitrogen
atoms you need to show are the ones in the bases that form the covalent bonds with the
deoxyriboses.
Place a small star next to each carbon atom that would have a hydroxyl (-OH) if the left
strand were an RNA molecule.
11. In the 1920’s Griffith did an experiment with bacteria and mice that showed there exists a
“transforming” material that can change the genetic characteristics of an organism. What was the
characteristic that was passed from one bacterium to another?
The ability to kill mice, a phenotype that could also be seen as the ability to produce
smooth colonies.
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What did he do to ensure that the bacteria which originally had the characteristic weren’t
merely passed through the critical experiment?
He killed them by heating them.
12. What is euchromatin? (one phrase or sentence) What is heterochromatin? (one phrase or
sentence) In which type of material does most transcription occur?
Euchromatin is chromatin that undergoes the normal process of condensation and
decondensation in the cell cycle. Heterochromatin is DNA that remains in a condensed
state throughout the cell cycle. Most transcription occurs in euchromatin.
13. What is a nucleosome and what are two of its functions? Name the 8 proteins classically
considered to be an essential part of the nucleosome core.
A nucleosome is a core particle of chromatin structure consisting of 8 histones
around which about 145-150bp of DNA are wrapped. Two of its functions are to aid in
the physical organization of DNA into a shorter structure, and to help control expression
of genes. The eight proteins are: two copies each of histones H2A, H2B, H3 and H4
14. In this problem you will do four very simple drawings (a-d).
a. Draw a DNA molecule as it would look before it is denatured, preparatory for PCR
amplification, using only lines with arrows at the ends to show the 5’ and 3’ ends. Here is that
molecule to get you started. Ignore the letters A and B for the moment.
A
B
b. Now draw that DNA as it would look during the heat denaturation step in the first
cycle of a PCR.
c. Now draw that DNA as it might look during the annealing (or hybridization) step
just after two PCR primers have annealed to it. Let the primers anneal at positions A and B
and draw them as short arrows oriented with their 3’ ends pointing in the direction of the
arrow.
d. Now draw those DNA molecules as they would look after the first extension step in
which Taq polymerase had extended from the primers. Let any newly synthesized DNA be
shown as a wavy line with an arrow at one end to signify the 3’ end. Cool! You’ve got more
DNA than you started out with!
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