LS4 Problem Set 3, 2010
Positional Cloning, RFLPs, microsatellites
Answers
Problem 1. Microsatellite allele A is 300bp (largest, closest to top of gel), allele B is
200bp, allele C is 150bp, and allele D is 50bp (smallest, closest to bottom of gel).
Genotype the family (which alleles does each individual have)
Mom: BC (carrier, allele C has mutation)
Dad: AD (carrier, allele A has mutation)
First Child: AC (since first child affected with recessive phenotype, these must be the
mutant alleles)
Fetus: AB (carrier, allele A from father has mutation)
What is the probability that the unborn child will have trouble with bladder
control?
0% - The fetus is a carrier of the mutation.
If this child grows up, gets married, and has a child, what is the probability that
their child will have trouble with bladder control?
0.5 * .12 * 0.5 = .03
[0.5 chance that carrier (fetus is known to be a carrier) will pass on mutated allele] *
[0.12 chance that random individual in population is a carrier] * [0.5 chance that carrier
will pass on mutated allele]
If they have one child that has trouble with bladder control, what is the probability
that they will have another child with bladder control problems?
0.5*0.5 = 0.25
If they had a child homozygous recessive for BC mutation, then they must both be
carriers.
Problem 2. A rare recessive X-linked mutation has been found which is caused by a
SNP in exon 1. This SNP causes a RFLP in which EcoRI will not cut the mutant
fragment. EcoRI normally cuts to give two bands: 200 bp and 600 bp.
a) Fill out the genotypes for each known individual and all possible genotypes for the
unkown. Assume unrelated individuals do not carry the mutation and do not have the
disease (must be homozygous dominant).
1
3
2
4
8
1. Rr
2. RY
3. RY
4. Rr
5. RY
6. RR or Rr
7. rY
8. rY
9. RR or Rr
5
6
7
9
b) Draw the appropriate bands on the gel for individuals whose genotypes are known
and explain the possible bands for individuals of unknown genotypes according to
possible genotypes.
6 and 8 can be RR (600 and 200) or Rr (all three bands)
Ladder
(1)
(2)
800
bp
600
bp
200
bp
1
2
3
4
5
6
7
8
9
Problem 3. In mouse Hind III restriction digests, a certain probe picks a simple RFLP
consisting of two alternative alleles of 1.7 kb and 3.8 kb. A mouse heterozygous for a
dominant allele for bent tail and also heterozygous for the above RFLP is mated to a
wildtype mouse that shows only the 3.8 kb-fragment. Thirty percent of the bent-tail
progeny are homozygous for the 3.8kb allele, and 70 percent are heterozygous for the
3.8 and the 1.7-kb forms.
a. Is the bent-tail locus linked to the RFLP locus? Explain your answer.
B=bent tail, b= normal tail, r1=3.8 kb, r2=2.7 and 1.1 kb
The cross is Bbr1r2 x bbr1r1
Progeny: 30% Bbr1r1, 70% Bbr1r2
Eliminating the contribution of the wildtype parent, the progeny
Are 30% Br1 , 70% Br2
If independent assortment exists, a 1:1 ratio of r1:r2 will be seen; r1 and r2 are clearly
not assorting independently. Therefore, the two genes are linked with 30 m.u. between
them.
b. What RFLP bands do you predict among the wild-type offspring, and in what
proportions?
The cross Br2/br1 x br1/br1
70% br1/br1 or 70% wildtype have only a 3.8 Kb band
30% br2/br1
30% wildtype have 3.8, 2.7, and 1.1 Kb bands.
Problem 4.
mwm
mom
dad
child 1
4 Kb
_____
_____
_____
3 Kb
_____
_____
2 Kb
_____
_____
1 Kb
_____
_____
genotype
+/-
+/-
child 2
child 3
child 4
child 5
_____
_____
6 Kb
5 Kb
_____
+/+
_____
_____
_____
_____
_____
_____
-/-
-/-
_____
_____
_____
_____
+/+
+/-
Problem 5.
a)
b) Microsatellite “B” is linked to the mutation, since one allele (B) is homozygous in the
mutants more often than would be expected by chance.
c) There are two recombinant alleles out of 24 total alleles, so there was about 2/24 or
8.33% recombination. The mutant is therefore about 8.33 cMorgans away from the
mutation.
d) Microsatellites C and D are linked to each other, since they are not segregating
independently from one another—the C allele tends to segregate with the D allele, and
the c allele segregates with the d allele. There are two recombinants with a genotype
different from the original parental genotype (flies #11 and #12), with respect to these
two microsatellites, so the two microsatellites are about 8.33 cM apart from one another.
Problem 6.
a) Individuals 3, 4, 7, and 10 are homozygous for the disease allele, and therefore they
will all have two copies of the allele that can’t be digested, so the gel will have one bands
(1 kb). Individuals 1, 2, 6, 8 and 9 must be heterozygous, so they carry both alleles and
3 bands will be seen on a gel when they’re genotyped (1 kb, 800 bp, and 200 bp).
b) There is not enough information to deduce the genotype of indidvidual 5—she could
be heterozygous for the allele, in which case there would be 3 bands on a gel when she
was genotyped, or homozygous for the dominant allele, in which case there would be
two bands (8000 bp and 200 bp).
Problem 7.
a)
0.8 X 0.8 = 0.64
b)
0.2 X 0.2 = 0.04
c)
0.8 X 0.2 = 0.16
d)
out of 18 chromosomes, 2 are recombinants (C and D) = 2/18 = 11%
Problem 8.
a) This is a most likely a dominant phenotype.
b) There is evidence of linkage. Notice that the iB often occurs in individuals affected by
the disease.
c) If we call the wild-type allele of the nail-patella gene “n” and the dominant mutant
allele “N”, both of the grandfather’s chromosomes had the i and n alleles. The
grandmother had one chromosome with the i and n alleles, and one with the iB and N
alleles.
d) In generation 2, individuals 5 and 8 appear to be recombinants. In generaton 3,
individual 3 is a recombinant.
Problem 9. If an individual is homozygous for an allele at a microsatellite, all of their
gametes will have the same allele. The microsatellite therefore can not be used as a
marker to distinguish the parental and recombinant gametes, and so will not be useful
for mapping studies. If a locus has few alleles and low heterozygosity, many individuals
in a population will share the same genotypes.
Problem 10
A) Given the sequence fragment from the standard allele below, write the corresponding
sequence for the Super Sweaty allele
Standard allele:
5’-ATGGCACCCGATCGATTCGCTACTGA-3’
Super Sweaty allele:
5’-ATGGCACCCGCTCGATTCGCTACTGA-3’
B) Autosomal recessive
C) Fill in the bands on the gel below for as many family members as possible. MW
1
2
3
4
5
6
7
8
9
10
11
12
1000bp
700bp
500bp
NOTE: lane 6 should have two bands as well, just like person 4 and 14
D) Individuals 3, 5, 7, 8, 10, 13 can be either heterozygous or homozygous.
13
14
Problem 11
WT
parent
Microsatellite
Obese
parent
—
—
—
RFLP
—
—
F1
Ob Ob
1
2
Ob Ob
3
4
Ob WT WT WT WT WT
5
1
2
3
4
5
—
—
—
— ——————
— ——————
—
—
—
—
—
—
—
—
—
—
—
—
—
— — —
— — —
— — —
—
—
—
—
A) Draw the missing bands for both the microsatellite and RFLP for the F1 individual
The F1 is a heterozygote for both the RFLP and microsatellite so all of the bands should
be seen on the gel.
B) Based on the data, what is the map distance between the gene that controls body
weight and the microsatellite?
Obese Individual 4 is a recombinant. None of the others are. Any recombination that
would have occurred would have been in the F1 parent, not the obese parent (since the
obese parent was homozygous at all loci). Since there were 10 individuals in the F2
generation, each of which could have received one recombinant chromosome, the
recombination rate is 1/10, so the gene is 10 map units away from the microsatellite.
C) What is the map distance between the body weight gene and the RFLP?
Obese individuals 2 and 4 and wild type individual 4 are recombinants, so the
recombination rate was 3/10. The RFLP is 30 map units from the gene.
D) Are the microsatellite and RFLP linked to each other? If so what is the map distance
between them?
Yes. In obese individual 2 and wildtype individual 4 there was a recombination between
the RFLP and the microsatellite, so the map distance between these two markers is
2/10, or 20 map units. You could also figure this out from knowing that the RFLP is 30
map units from the gene and the microsatellite is 10 map units from it. So the RFLP and
microsatellite could have been 20 map units apart (same side of the gene) or 40 map
units apart (opposite sides of the gene), but the recombination rate between them and
the pattern of recombinants suggest that they are on the same side.