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Transcript
The first midterm will consist of 20 four-answer multiple choice questions, six or seven
short written questions (totaling 25 points), and defining five of six terms (10 points). You
will have the full class period to complete the exam. The exam covers lecture material up to
and including the 8 February class and text material from chapters 1-6. Here are a few
sample questions.
Multiple Choice
1) Which sib-pair linkage outcome indicates the highest heritability in the following graph?
2) Phineas Photinus conducts a breeding experiment on Drospohila and calculates (based
on his phenotype counts of 1000 F1 individuals) that his two traits are located 120
centimorgans apart, this means:
a) they are on the same chromosome
b) they are on different chromosomes
c) they are on the same chromosome, but very far apart
d) Phineas goofed, the results are impossible
3) Microsatellites are
a) inactivated (methylated) chromosomes
b) inactivated (RNA "painted") chromosomes
c) repeated short sequences of DNA
d) small fragments of DNA
4) Allelic association study is to Linkage study as
a) disequilibrium is to epistasis
b) genome wide QTL search is to pleiotropy
c) identifying QTL effect is to locating QTLs in the genome
d) pleiotropy is to complex trait
5) An uncle is considered to be a ____________ relative of his nephews and nieces.
a) first degree
b) second degree
c) third degree
d) fourth degree
6) A heritability measure in the lab is virtually guaranteed to ________ heritability in the
wild.
a) over estimate
b) under estimate
c) fail to detect
d) falsely detect
7) A SNP is an example of
a) a frame shift mutation
b) transpositional control
c) genetic regulation
d) a genetic marker
8) The gene defect for both Huntington's Disease and Fragile-X syndrome consists of
a) a series of repeated nucleotide sequences
b) a mispairing of base pairs
c) a major deletion of an important segment of a gene
d) a metabolic block
9) RNA is synthesized from the DNA template during
a) transcription
b) translation-1
c) translation-2
d) transportation
10) The human genome contains about how many base pairs?
a) 3,000,000
b) 30,000,000
c) 300,000,000
d) 3,000,000,000
11) The frequency of crossing over between any two linked genes is
a) different between males and females.
b) determined by their relative dominance.
c) proportional to the distance between them.
d) the same as if they were not linked.
12) If an affected male has affected daughters and sons in about the same number as
unaffected daughters and sons, the trait is likely to be an
a) X-linked dominant trait
b) autosomal dominant trait
c) autosomal recessive trait
d) X-linked recessive
Definitions (2 pt. each)
1. Trinucleotide repeat
2. QTL linkage analysis
3. Transgenic
Short Answer Questions
1. What are microsatellites and how do they differ from SNPs? How are they useful in
behavioural genetics? What advantage does identifying a few SNPs that mark a haplotype
block have for genetic mapping of the entire genome? (3 points)
2. Hemophilia is an X-linked recessive disorder. Suppose a male afflicted with this disorder
mates with a female who is a carrier of this disorder. Give the genotypes, phenotypes, and
their expected frequencies among their male and female offspring. (4 points)
3. Answer the following: A. Explain the components of the ACE model and what MZ and
DZ twins tell us with respect to this model. (3 pt) B. Why is the equal environment
assumption commonly applied in twin design studies? Should it be? (1 pt)
ANSWERS
Multiple Choice Answers
1. B
2. D
3. C
4. C
5. B
6. A
7. D
8. A
9. A
10. D
11. C
12. B
Definitions
1. A codon; three base pairs in sequence that code for an amino acid (or stop signal).
2. Linkage analysis searching for linkages of small effect size of multiple genes at many
loci.
3. Containing foreign DNA; e.g., inserting DNA from one organism into the DNA of
another.
Short Answer
1. Microsatellites are 2 to 4 DNA base pairs repeating up to 100 times; SNP is a single
non-repeating change of a base pair. Both are useful genetic markers. Haplotype blocks
are units of SNPs that are highly correlated with one another, meaning that you only need
to genotype about half a million haplotype block SNPs rather than millions of SNPs to
scan the whole genome.
2. Let “H” represent the dominant (normal) allele and “h” the recessive (hemophilia)
allele. Let “X” represent X chromosome and “Y” the Y chromosome.
So, the mother is XH Xh (i.e., phenotypically normal, but genotypically homozygous -- a
carrier) and the father is Xh Y (i.e., phenotypically hemophilic, and genotypically
recessive for the trait).
Then, in a Punnet square:
XH
Xh
Xh
XH Xh
Xh Xh
Y
XH Y
Xh Y
Genotypes of female children are: 50% XH Xh (heterozygous), 50% Xh Xh
(homozygous recessive). Genotypes of male children are: 50% XH Y (dominant), and
50% Xh Y (recessive).
Phenotypes of female children are: 50% normal, 50% hemophilic. Phenotypes of male
children are: 50% normal, 50% hemophilic.
3. Part A: A = additive, or allelic, genetics; C = common environment, E = unique
environment. MZ twins share 100% of A and 100% of C, so differences are due to E (or,
correlation between MZ twins gives estimate of A+C). DZ twins share 100% of C and
50% of A, so correlation is estimate of 0.5A+C
3. Part B: Assumes environmentally caused similarity is about the same for MZ and DZ
twins; without this, can’t use the basic ACE model. It is probably safe to use this for most
traits.