Download BL414 Genetics Spring 2006 page Test 2

BL414 Genetics Spring 2006
Test 2
page 1
150 pts. total, 15% of final grade
March 20, 2006
Your name:__________Answer Key___________________________________
1) (2.5pts) T or F: ___T_______ Bacterial genomes do not have many repetitive
sequences, most of their genome is unique.
2) (2.5pts) T or F: ____F______ Genetic linkage in corn can be analyzed using
asci tetrad analysis.
3) (2.5pts) T or F: _____F_____ The Holliday model is the currently accepted
model for all recombination events.
4) (2.5pts) T or F: ____F______ During recombination, genetic material is
exchanged between a male and a female.
5) (5pts each) What are the three steps of DNA replication?
a) ________Initiation___________
b) ________Elongation___________
c) ________Proofreading___________
6) (5pts) In eukaryotic chromosomes, DNA replication begins at what
sites?____Origin of replication___________
7) (5pts each) Name four proteins that are involved in DNA synthesis and
describe their function.
Possible answers:
Helicase: unwinds DNA at the replication fork
Single-stranded binding proteins (SSPs) stabilize unwound single-stranded DNA
Gyrase: relieves rotational stress on DNA at the replication fork
Primase, primosome or polymerase : creates RNA primers
DNA polymerase : elongation of the DNA strand
3’-to-5’-exonuclease: corrects mismatches bases
PolI or RPA: removes RNA primers
8) (10pts) The molecules below are used in an in vitro DNA sequencing reaction,
one is dideoxyadenosine and one is deoxyadenosine: label them correctly
(2.5pts).
a) (2.5pts) Circle the functional group that is different between them.
b) (5pts) Explain briefly how each one functions in the sequencing reaction.
BL414 Genetics Spring 2006
Test 2
page 2
150 pts. total, 15% of final grade
Dideoxyadenosine will terminate the DNA synthesis reaction where there is an
adenosine in the sequence, because additional NTP’s cannot be added to it’s 3’
position because there is no hydroxyl group there. Deoxyadenosine will be
added to the chain where there is an adenosine but will allow the continuation of
elongation of synthesis of the DNA strand, by addition of an NTP to its 3’OH
group.
9) (5pts) Describe one difference in DNA synthesis between bacteria and
eukaryotes.
Possible answers:
1. The rate of DNA replication is much faster in bacteria than in eukaryotes.
2. Okazaki fragments are 1000-2000 base pairs long in bacteria, but much
shorter in eukaryotes, about 100-200 base pairs long.
3. Primase creates primers in bacteria that are solely RNA, and eukaryotes
have a multi-enzyme primosome complex that creates a primer consisting
of RNA and DNA.
4. Pol I 5’-3’ exonuclease removes RNA primers in bacteria one nucleotide at
a time, and in eukaryotes a complex called the primosome including
replication protein A (RPA) unfolds the RNA fragment as an intact unit,
BL414 Genetics Spring 2006
Test 2
page 3
150 pts. total, 15% of final grade
stabilizes it, and recruits an endonuclease activity to cleave it away from
the DNA.
10) (5pts) Name the structure shown below. It occurs during recombination.
__________ Holliday junction_______
11) (10pts) Describe the differences and similarities between yeast centromeres
and those of higher eukaryotes, including the following terms in your
description: regional centromeres, alpha satellite, point centromeres and
localized centromeres.
Yeast and higher eukaryotes both have localized centromeres which have a
single specific site on the chromosome. Yeast centromeres are called point
centromeres, and consist of a short, specific DNA sequence within a centromeric
core particle. Higher eukaryotic centromeres are within a much longer stretch of
DNA, up to hundreds of kilobases, and are called regional centromeres. The
repeating DNA sequence in the regional centromere is called alpha satellite DNA
and consists of a 170 base pair tandem repeat, which is repeated up to 500015,000 times.
BL414 Genetics Spring 2006
Test 2
page 4
150 pts. total, 15% of final grade
12) (10pts) The enzyme shown below is involved in maintaining the ends of
chromosomes.
a) (5pts) Name the enzyme_________telomerase____________
b) (5pts) Name the component of the enzyme indicated by the
arrow______guide RNA__________
BL414 Genetics Spring 2006
Test 2
page 5
150 pts. total, 15% of final grade
13) (10pts) A large extended family in an isolated region of eastern Kentucky has
a history of a rare blood disorder blumonia. An analysis of the family
pedigrees for three generations along with DNA sequence testing for a DNA
marker “S” on Chromosome 18 results in an lod score of 2.87.
a) Is the blumonia gene linked to the marker “S”?
It might be – it is indeterminate from the given data.
b) Why or why not?
The lod score is between -2 and 3, therefore it cannot be determined that the gene
and marker are definitely linked or unlinked.
c) What further research should be done?
More pedigrees should be collected from families with the disease. Since the lod
score is very close to 3 it may be that with further data, it can be proven that the
gene and marker are linked.
14) (5pts) Male C. elegans worms with the mutation lin-4 and are mated to rol-6
hermaphrodites. The lin-4 gene causes the worms to be delayed in
development, a type of mutation called heterochronic. The rol-6 gene causes
the worms to be unable to travel forward and only move in circles, a
phenotype called “roller.” Examine carefully the resulting number of progeny
for each phenotype.
Wildtype: 179
Roller and delayed: 23
Roller: 59
Delayed: 64
What do these results indicate about the linkage of lin-4 and rol-6?
If the genes were linked, all of the offspring would be wildtype, and mutants
would not be seen in the first generation. The appearance of mutants in the
typical 9:3:3:1 phenotypic ratio indicates that the genes are segregating
independently and located on different chromosomes.
BL414 Genetics Spring 2006
Test 2
page 6
150 pts. total, 15% of final grade
15) (5pts each) Label the indicated items (a-d) in the figure of higher order DNA
structures depicted below:
16) (25pts) You perform a three-point cross using the flowering plant foxglove,
looking at the following three mutations: the recessive mutation called white,
which produces white flowers, the mutation called peloria, which causes large
flowers at the apex of the stem and the dwarf mutation which causes short
stem length. For the cross you mate a white-flowered plant that is otherwise
phenotypically wild-type, to a plant that is dwarf and peloria, and otherwise
wildtype. The F1 generation from this cross is wildtype in appearance. You
mate the F1 generation to a triple mutant that is double recessive for white,
peloria and dwarf. The progeny are shown in the chart below with numbers
given for each phenotype.
Dwarf, peloria, white
Dwarf, peloria
Peloria, white
Dwarf, white
Peloria
56
172
5
51
43
BL414 Genetics Spring 2006
Test 2
page 7
150 pts. total, 15% of final grade
Dwarf
White
Wildtype
6
162
48
a) (5pts) What were the genotypes/allele arrangements of the parents in the
original cross?
These are the same as the genotypes with the largest numbers of offspring:
White (162)
Peloria, dwarf (172)
So the allele arrangements are
pel + dw and + wh +
(different linear order of the genes okay at this point)
b) (5pts) Determine the order of the linked genes on the chromosome.
The results of double crossover events have the lowest number of offspring,
so these are offspring with the arrangements:
pel wh + (5)
and + + dw (6)
So, comparing to the parental arrangements in part (a), mutants dw and wh are
still in trans configuration, but pel has switched in relationship to both of them,
so dw must be in the middle of the other two because its new position must be
due to a double crossover event.
Therefore the linear order is:
dw
pel
wh
and we now know the parental arrangement of alleles in correct linear order is:
dw
pel
+
+
+
wh
BL414 Genetics Spring 2006
Test 2
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150 pts. total, 15% of final grade
c) (10pts) Calculate the recombination frequencies between adjacent genes.
A crossover between pel and dw,
dw
pel
+
+
+
wh
would give us:
dw + wh and + pel +
these arrangements are observed:
Dwarf, white (51)
Peloria (43)
And the numbers due to double crossovers should also be included.
So the recombination frequency for dw and pel is:
(51 + 43 + 5 + 6)/ 543 x 100% = 19 %
Likewise we find that a crossover between pel and wh,
dw
pel
+
+
+
wh
gives us:
dw pel wh and + + + which are observed in the following numbers:
Dwarf, peloria, white (56) and wildtype (48), along with the double
crossovers we can calculate the recombination frequency for pel and wh:
(56 + 48 + 5 + 6) / 543 x 100% = 21 %
d) (5pts) Draw a genetic map showing the distances between the three loci.
BL414 Genetics Spring 2006
Test 2
dw
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150 pts. total, 15% of final grade
wh
pel
19 m.u.
21 m.u.