GENETICS 603
Exam 1
Oct 26, 2008
Name
I. What is “attenuation”? Would it work in eukaryotes? Justify your answer.
It is a mechanism for transcriptional regulation of some amino acid biosynthetic pathways in
bacteria where several codons in the leader of the mRNA call for that amino acid. If it (and its
activated tRNA) is plentiful transcription ceases, but if too little is present, a different folding
pattern allows transcription (and translation) to proceed.
It can only work in prokaryotes where transcription and translation are coupled.
II. A. List stepwise the reagents (other than buffers) and enzymes you would use to create a
cDNA library from a specific tissue like mouse liver or plant leaf cells.
(many alternatives are possible!)
oligo-dT beads for separation of mRNA from cell extracts
oligo-dt primers and reverse transcriptase for first strand cDNA synthesis
RNAaseH to nick the mRNA strand
DNA polymerase to extend and replace the RNA primer fragments
DNA ligase to connect second strand fragments
ligate Eco or other linkers or primer binding sites to create useful ends
(can just blunt end ligate)
digest with Eco to create Eco ends
mix with Eco digested puc plasmid, re-anneal and ligate with ligase
transform into competent E. coli.
b. Suggest 3 different ways to try to determine if your library has a copy of a cDNA for
anthranilate synthase, the next to the last step in tryptophan biosynthesis. (Tell what assumptions
you are making in order to use each method).
1. Add to an anthranilate minus E. coli strian and look for growth on medium without
tryptophan; assumes the native amino acid sequence produced in E. coli will function without
modification.
2. Hybridize to an anthranilate clone from another species: assumes such a clone is available
3. PCR using highly conserved sequences from anthranilate synthase from related species to
obtain amplicon of predicted size; assumes sufficient sequence information available
4. Sequence large numbers of clones and compare the sequence to those in GenBank; assumes
correct annotation of genes identified
etc.
IV. Assume that resistance to a specific disease in peas is inherited as a single gene defect and
that you have purebreeding resistant and susceptible parents to work with. However, due to
“escapes” in field inoculation tests, only 70% of the inoculated susceptible parent plants develop
disease.
A. Predict the F1 and F2 phenotypic ratios that would be observed in field tests from a cross of
the purebreeding resistant and susceptible parents under the assumption that:
(where res. means no disease detected, susc. is disease detected)
1) resistance is dominant F1 all resistant
F2 82.5 % Res : 17.5% sus
2) resistance is recessive F1 30 no disease:
70 disease
F2 52.5% sus : 47.5% res
B. If 6 F2 plants are observed in each case, what is the probability of seeing at least one of them
will show the susceptible phenotype? (Filled in formulas are ok as an answer)
1)
1- (0.825)6
2) 1- (0.475)6
C. Suppose you observe in the field that 60 F2 plants were disease free and 40 diseased. Would
these data support either a single gene dominant or recessive model of resistance? Show your
work.
For dominant resistance
R
S
observed
60
40
expected
82.5 17.5
(O-E)
22.5 -22.5
2
(O-E) /E
6.25 28.9
Chi Square, 1 df: 35
Prob of a deviation this large or larger
due to chance approx 0
resistance is 1 gene recessive
R
S
60
40
47.5 52.5
-12.5 12.5
3.3
2.98
6.2
Probability of a deviation this large or
larger by chance is between 1 and 2%
Reject H0 unless using “highly sig”
D. Assume that the R gene is dominant and that the mutation rate to the recessive susceptible
allele is 1 in 100,000 gametes. If 500,000 hybrid plants are grown from seeds from the RR by rr
cross what is the probability 1) none will be susceptible or 2) at least one will be susceptible
(show answer in terms of a formula only).
(using the poison where expect 5 to occur:)
1) e-5
2) 1-e-5
VI The following description is from OMIM entry 162200
NEUROFIBROMATOSIS, TYPE I; NF1: Neurofibromatosis is an autosomal
dominant disorder characterized particularly by cafe-au-lait spots and fibromatous
tumors of the skin. Among 18 cases of neurofibromatosis with hypertension, Kalff
et al. (1982) found pheochromocytoma in 10 (3&4, possibly 1). Age at diagnosis
ranged from 15 to 62 years (7). At least 2 of the pheochromocytoma patients had
renal artery stenosis. Three had small-bowel and/or stomach neurofibromata (3&4.
1?). Data from 247 pregnancies in 105 women with NF1 revealed 44 first
trimester spontaneous abortions. Mutation analysis in NF1 has been hampered by
the large size of the gene (350 kb with 60 exons), the high rate of new mutations,
lack of mutational clustering, and the presence of numerous homologous loci (6).
A) Circle or underline and indicate with the associated number the part of the description that fits
any of the following terms:.
1. Lack of Penetrance 2. Co-dominance 3. Variable expressivity 4. Pleiotropy 5. Teratogen
6. Genetic heterogeneity 7. Age of onset 8. Phenocopy
B) Define any of the 8 terms that did not apply in the description.
codominance: a situation where alleles in a heterozygote are expressed in different cells (roan cattle)
or simultaneously and each contributes to the pehnotype (AB blood)
teratogen: an agent that causes birth defects by interfering with prenatal development
phenocopy: an environmental mimic of a genetic condition (thalidomide causes a mimic of
phocomelia)
VI. What is the C-value paradox? How is it explained?
The C-value paradox is that the DNA content of organisms does not necessarily conform to the
number of genes or evolutionary complexity.
It is generally explained by the occurrence of high levels of highly repeated sequences in those
organisms with the most DNA/genome.
VII In 1950 Hauschka et al reported that one of their inbred mouse strains being maintained by
sib matings each generation consistently produced more females than males in a ratio of 4
females: 3 males. When individual females were selected and crossed to a male from the same
generation, the following data were a sample of those produced:
cross
# female progeny
# male progeny
1
10
5
2
9
7
3
9
4
4
3
5
5
14
7
6
7
8
If the progeny of these crosses were used to create another generation, the same ratios were
found in those progeny. Develop a hypothesis to account for their observations.
The actual case is that the line has females segregating for an X-linked recessive lethal. Half the
females are XLXL so when mated to an XLY male give a 1 :1 sex ratio while those females that
are XLXl when mated to an XLY male give 1 XLXL :1XLXl :1 XLY in the surviving progeny.
VIII In the beetles golden body (go) is a recessive X-linked mutation, and brown eyes
(bw) is a recessive autosomal mutation. A female homozygous for golden body and red
eyes is mated to a male with brown eyes and a black body. A.) Predict the phenotypes
of their F1 offspring. B.) If the F1 progeny are intercrossed, what kinds of progeny will
appear in the F2, and in what proportions?
1) XgoXgo, Bw/Bw X XGoY, bw/bw -> all black body, red eye females and gold body, red eyed
males
2) ½ F2 females: 3/8 gold body, red eyes: 1/8 gold, brown : 3/8 black, red : 1/8 black, brown
½ F2 males: 3/8 gold body, red eyes: 1/8 gold, brown : 3/8 black, red : 1/8 black, brown
(or, overall, change all the 8s to 16s and include the sex as a phenotype)
IX. Given in maize: Bz_ bronze aleurone, C_ colored coat
bz/bz white aleurone cc white coat
Sh normal kernel
sh/sh shrunken
A plant heterozygous for all three genes was testcrossed and 1000 progeny scored for each traint
with the following results (only alleles from the heterozygous parent are shown),
Bz
bz
bz
Bz
bz
Bz
Bz
bz
C
c
C
c
C
c
C
c
Sh 479
sh 473
sh
15
Sh
13
Sh
9
sh
9
sh
1
Sh
1
Bz
bz
bz
Bz
bz
Bz
Bz
bz
Sh
sh
sh
Sh
Sh
sh
sh
Sh
C
c
c
c
C
c
C
c
A) Show the actual genotypes of the parents
Bz Sh C/bz sh c X bz sh c/bz sh c
B) Map the genes
Bz 2.0 Sh 3.0 c
C) Calculate the value of Interference.
2 DXOs observed ; 0.03 X 0.02 X 1000 = 0.6 expected
Coef. of Coincidence = 3.33 so I = -2.33
X. Southern blots of the human OBF1 gene were made using the NIGMShuman/rodent somatic
cell hybrid mapping panel No. 2 (Coriell Cell Repository, Camden, NJ). This mapping panel is
based on 24 human–rodent somatic cell hybrids of which all but two are monochromosomal with
respect to human chromosomes. Results are shown below (Hu, human genomic DNA, Mo,
mouse genomic DNA, Ha hampster genomic DNA. The remainder show the human
chromosomes present in the hybrid lines.
A) What kind of mapping is being done? Explain how it works.
Synteny mapping based on human/rodent stable cell lines where only a few (1 or 2
in ths case) human chromosomes are present in each cell line. When the DNA from
each line is digested and subjected to Southern blotting using a human DNA
sequence as a probe under high stringency conditions, the probe will stick only to
the homologous human genomic DNA.
B) What was the conclusion? OBF1 is on chromosome 11