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Unformatted text preview: Molecular Genetics BSCI410
Homework 4; Lectures 1318  answers
Homework due at the beginning of class on Thursday, Nov. 11, 2010.
The exam will be Nov. 18.
Homework and review questions will be discussed again at the review on Nov. 16.
Homework questions. Please provide your answers on a separate sheet.
Examine the following pedigree.
A1,2 B1,2 A1,3 B1,3 A1,2 B1,2 A1,2 B1,3 1. (1 point) The A1 alleles in the two brothers are identical by state (this just means that they are
both A1). Can you infer that they are also identical by descent?
B) You can conclude that A1 is identical by descent.
Notice that A2 had to come from the father (and therefore, A1 is maternal in both cases).
2. (1 point) The B1 alleles in the two brothers are identical by state (this just means that they are
both B1). Can you infer that they are also identical by descent?
C) You can conclude B1 is not identical by descent.
Notice, for the brother on left, that B2 had to come from the father (and therefore, B1 is
maternal). For the brother on the right, B3 is maternal so B1 is paternal. Thus, B1 is
maternal in one case and paternal in the other case.
(Questions 35) Consider two populations, 1 and 2, that differ at two unlinked loci, A and B. In
each population a specific allele is fixed at each locus (i.e. all individuals in population 1 are
homozygous for A1 and B1  they have the genotype A1,1 B1,1  while all individuals in
population 2 have the genotype A2,2 B2,2).
You allow a large and equivalent number of individuals from the two populations  for example,
500 males and 500 females from population 1 and 500 males and 500 females from population 2 to mate at random (and they do mate at random).
paternal A1, B1
paternal A2, B2 maternal A1, B1
A1/A1 B1/B1
A1/A2 B1/B2 maternal A2, B2
A1/A2 B1/B2
A2/A2 B2/B2 Thus ¼ will be A1/A1 B1/B1, ½ will be A1/A2 B1/B2 and ¼ will be A2/A2 B2/B2. 1 Molecular Genetics BSCI410
Homework 4; Lectures 1318  answers
Homework due at the beginning of class on Thursday, Nov. 11, 2010.
The exam will be Nov. 18.
Homework and review questions will be discussed again at the review on Nov. 16.
3. (1 point) Is locus A at HardyWeinberg equilibrium in the "G1" generation?
Yes, from the forgoing, you can see that for A, the allele frequencies are p = q = 0.5 and the
genotype frequencies 0.25, 0.5 and 0.25 correspond to p2, 2pq and q2.
4. (1 point) What is the expected frequency of each of the four possible gametes transmitted from
the G1 to the G2. The possible genotypes are A1 B1, A1 B2 , A2 B1 and A2 B2
3/8 A1 B1, 1/8 A1 B2, 1/8 A2 B 1, 3/8 A 2 B2
1/4 of the G1 will be A1/A1 B1/B1; they will generate only A1 B1 gametes (2/8)
1/4 of the G1 will be A2/A2 B2/B2; they will generate only A2 B2 gametes (2/8)
1/2 of the G1 will be A1/A2 B1/B2; they will generate equal numbers of the four possibilities (1/8 for each)
Summing those yields the answer above. 5. (1 point) Do the two alleles A1 and B1 show genetic association (linkage disequilibrium) in this
G1 generation?
Yes! There is clearly and excess of A1 B1 and A2 B2 alleles.
6. Su p po se you carry ou t a map ping cross between two parent s trains P1 and P2 as described in
lecture 15, and you have 16 F2 individuals in your map ping po pulation. The parent s trains P1 and
P2 have numerous pol ymor phic differences between them. You assa y 3 RFLP genetic markers (A,
B and C) to determine the genot y pes of each of the F2 individuals. The data you obtain after PCR,
res triction digestion and gel electro phoresis is shown below.
a. Write the genot y pe s for each of the 16 individuals (116) for each marker (A, B, and C). For
example: 1 – P1/P2 or P1 (homoz ygous) or P2 (homoz ygous) (0.5 p t)
1
A 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 P1 P1/P2 P1 P1/P2 P1 P1/P2 P1/P2 P2 P1 P1/P2 P1/P2 P1/P2 P2 P1/P2 P1 B P2 P1/P2 P1/P2 P1/P2 P1/P2 P1/P2 P1 P2 P1 P1/P2 P1 P2 P1/P2 P2 P1/P2 P1 C P1 P1/P2 P1 P1/P2 P1/P2 P1/P2 P2 P1/P2 P1 P1 P1/P2 P1/P2 P2 P2 P1/P2 P1/P2 b. What is the observed recombination frequency between A and B based on the data?
11/32 = 0.344
c. What is the observed recombination frequency between A and C?
5/32 = 0.156
d. What is the observed recombination frequency between B and C?
8/32 = 0.25
2 P1 Molecular Genetics BSCI410
Homework 4; Lectures 1318  answers
Homework due at the beginning of class on Thursday, Nov. 11, 2010.
The exam will be Nov. 18.
Homework and review questions will be discussed again at the review on Nov. 16.
e. Draw a genetic map for this data showing the order of A, B and C and the map units between
them. (You do not need to use a map ping function to correct for unseen double crossovers.) (2 p t s
to tal for b, c, d and e)
A  ~15 cM. C ~25 cM. B
f. Which F2 individual(s), if any, carry a double crossover? (0.5 p t)
#15 3 Molecular Genetics BSCI410
Homework 4; Lectures 1318  answers
Homework due at the beginning of class on Thursday, Nov. 11, 2010.
The exam will be Nov. 18.
Homework and review questions will be discussed again at the review on Nov. 16.
This pedigree shows a family affected by an autosomal dominant genetic disease.
Genotypes for three linked markers, A, B and C, are shown I 1 2 II 1 2 III
III 1 2 3 4 5 6 The genotypes are:
I1
I2
II1
II2 A1,2 B1,2 C1,2
A3,3 B3,3 C3,3
A1,3 B1,3 C1,3
A2,2 B2,2 C2,2 III1
III2
III3
III4
III5
III6 A1,2 B1,2 C1,2
A3,2 B3,2 C3,2
A1,2 B3,2 C3,2
A3,2 B3,2 C1,2
A3,2 B3,2 C3,2
A1,2 B1,2 C1,2 7. (1 point) Indicate the phase of alleles in individual II1 by showing his haplotypes.
There are four possibilities. They are
a) A1 B1 C1 / A3 B3 C3
8. (1 point) What is the order of these three markers? There are three possibilities:
ABC
9. (1 point) Ignoring all of the other loci, calculate a lod score for linkage to of the disease to A with
θ=0
6 * log10(2) = 1.8
4 Molecular Genetics BSCI410
Homework 4; Lectures 1318  answers
Homework due at the beginning of class on Thursday, Nov. 11, 2010.
The exam will be Nov. 18.
Homework and review questions will be discussed again at the review on Nov. 16.
10. (1 point) Ignoring all other loci, what value of θ would give the highest lod score for linkage of
the disease to B?
1/6 = 0.166
11. (1 point) What is the value of that maximal lod score (for linkage of the disease to B at the
value of θ that gives the highest possible lod score)?
log10((5/6)5(1/6)/(1/2)6) = 0.632
12. A geneticist succeeds in maintaining three colonies (A, B, and C) of humanmouse hybrid cells.
The only human chromosomes retained by the h ybrid cells are indicated by plus signs (+) in the
table:
H ybrid colony
A
B
C 1 2 3 +
+
+ +
+
– +
–
+ Human Chromosome
4
5
6
+
–
– –
+
+ –
+
– 7
–
–
+ The geneticist tes t s each of the colonies for the presence of 5 enz ymes (q, r, s, t, v) with the
following results: q is active only in colony C; r is active in all three colonies; s is active only in B
and C; t is active only in B; v is not active in any of the colonies. What can you conclude about the
locations of the genes res ponsible for these enz yme activities? (1 p t)
q is on chr. 7, r is on chr. 1, s is on chr. 5, t is on chr. 6
v is not on chromosomes 17
13. As suggested in the assigned reading, what are the advantages of the FISH ph y sical map ping
p rotocol compared with the method of linkage map ping for the initial determination of the
chromosomal location of a gene? (1 p t)
(from page 361 in the assigned Hartwell reading):
i) all clones can be mapped by FISH, but only clones that detect
polymorphisms can be mapped by linkage analysis.
ii) Linkage mapping requires the analysis of one locus in relation to another,
but FISH does not.
iii) FISH requires only a single sample on a microscope slide, but linkage
analysis requires genotype information from large set of individuals. 5 Molecular Genetics BSCI410
Homework 4; Lectures 1318  answers
Homework due at the beginning of class on Thursday, Nov. 11, 2010.
The exam will be Nov. 18.
Homework and review questions will be discussed again at the review on Nov. 16.
14. A pa per in last week's Nature repor t s the latest results on characterization of human genetic
diversit y b y DN A sequencing.
"A map of human genome variation from po pulationscale sequencing"
The 1000 Genomes Project Consor tium
2010. Nature 467: 10611073
ht t p://www.ncbi.nlm.nih.gov/pubmed/20981092
Look at the pa per and find something relevant to what we have been discussing in class that you
consider sur prising or interes ting. What is it?
Quo te the one or two sentences that bes t cap tures the point that interest s you and write between
200 and 400 words on wh y you find this result interes ting or sur prising. (3 points)
This will be graded by checking that you have indeed found something that is
relevant to the material in class and not already well known.
15. Sup po se you are examining the genome of a newl y discovered organism that has a single linear
chromosome. You create a genomic DNA library of 5 BAC clones (designated A through E) that
each carry 200 kb of the genomic DNA . You next determine 500 bp of DNA sequence at each end
of the 5 cloned inserts using Sanger sequencing and two universal primers that anneal on op po site
sides of the cloning site of the BAC vector. (That is, you ob tain the 500 bp of the insert sequence
that is located at each junction with the BAC vector. These are referred to as "BAC end
sequences.") Next, you create STSs out of each of the 10 sequences b y designing PCR primers that
can amplify these 10 sequences. The table below shows which STSs can be amplified from each of
the 5 BACs.
BAC clone STSs at ends other STSs A
B
C
D
E 1, 2
3, 4
5, 6
7, 8
9, 10 4, 5, 7, 10
2, 9
1, 8
1, 5
2, 4 a. Diagram
a
physical
map
of
this
region
that
is
consistent
with
the
data
in
the
table.
Be
sure
to
indicate
the
relative
order
of
the
BAC
clones
and
show
the
location
of
the
STSs.
(1
pt)
CDAEB and 6 8 1 5 7 10 4 2 9 3, or: BEADC and 3, 9, 2, 4, 10, 7, 5, 1, 8, 6
b. What
is
the
range
of
possible
lengths
of
your
contig
(provide
a
maximum
and
a
minimum
length
in
kb)?
(1
pt)
400600 kb
6 Molecular Genetics BSCI410
Homework 4; Lectures 1318  answers
Homework due at the beginning of class on Thursday, Nov. 11, 2010.
The exam will be Nov. 18.
Homework and review questions will be discussed again at the review on Nov. 16.
Review questions.
These questions are related to the homework questions.
However, they do not need to be turned in.
(Questions 15 – compare to homework questions 35):
Consider two populations, 1 and 2, that differ at two unlinked loci, A and B. In each population a
specific allele is fixed at each locus (i.e. all individuals in population 1 are homozygous for A1 and
B1  they have the genotype A1,1 B1,1  while all individuals in population 2 have the genotype
A2,2 B2,2).
First, you cross a single male from population 1 with a single female in population 2. Of course, all
of the F1 progeny are heterozygous at both loci, A1,2 B1,2.
1. Is locus A at HardyWeinberg equilibrium in the F1 generation (your answer would be the same
for locus B)?
No! All of the individuals are heterozygous!!
2. What is the expected frequency of each of the four possible gametes transmitted from the F1 to
the F2. The possible genotypes are A1 B1, A1 B2 , A2 B1 and A2 B2
The four should be in equal abundance.
3. Do the two alleles A1 and B1 show genetic association in this F1 generation?
(consider the haplotypes that are transmitted by this F1 generation to the F2 generation).
4. What is the expected frequency of each of the nine possible genotypes in the F2 progeny
(assuming random mating among the F1)?
The possible genotypes are:
A1,1 B1,1 ; A1,1 B1,2 ; A1,1 B2,2 ;
1/16 2/16 1/16
A1,2 B1,1 ; A1,2 B1,2 ; A1,2 B2,2, ;
2/16 4/16 2/16
A2,2 B1,1 ; A2,2 B1,2 and A2,2 B2,2
1/16 2/16 1/16
5. Is locus A at HardyWeinberg equilibrium in the F2 generation (your answer would be the
same for locus B)?
Yes. 7 Molecular Genetics BSCI410
Homework 4; Lectures 1318  answers
Homework due at the beginning of class on Thursday, Nov. 11, 2010.
The exam will be Nov. 18.
Homework and review questions will be discussed again at the review on Nov. 16.
6. Consider question 9 (the lod score for gene A). Now, assume that this disease is only 80%
penetrant (meaning that people with the causative genotype are get the disease 80% of the time).
What is the lod score for linkage to A with θ = 0 under this revised model?
Here you have to calculate the probability of each possible allele/disease combination.
allele
A1
A1
A3
A3 Disease
yes
no
yes
no Under M1, θ = 0
0.4
0.1
0
0.5 Under M0, θ = 0
0.2
0.3
0.2
0.3 Z = log10 [P(dataM1)/ P(dataM0)] = log10[(0.4)3(0.5) 3/ (0.2)3(0.3) 3] = 1.569
7. Consider the pedigree used for questions 711. Ignoring all other loci, what value of θ would
give the highest lod score for linkage of the disease to C?
0.333
8. What is the value of that maximal lod score (for linkage of the disease to C at the value of θ that
gives the highest po ssible lod score)?
log10((2/3)4(1/3) 2/(1/2)6) = 0.148 8 ...
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This note was uploaded on 11/07/2011 for the course BIO 325 taught by Professor Saxena during the Summer '08 term at University of Texas at Austin.
 Summer '08
 SAXENA
 Genetics, Molecular Genetics

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