Bios 344
Molecular Biology and Genetics
Study Questions on Recombination Mapping and Restriction Mapping
1.
A, B, and C are dominant alleles; a, b, and c are recessive. The cross was made ABC/abc 
abc/abc. Among 1000 progeny the different phenotypes appeared in the following numbers:
ABC
2.
347
Abc
1
abC
8
AbC
145
aBC
2
ABc
9
abc
336
aBc
152
(a)
Deduce the map order and calculate the map distances between markers.
(b)
Do the double crossovers show interference?
A test cross between Drosophila females heterozygous for 3 pairs of markers (A/a, B/b, and D/d)
and males homozygous for the recessive alleles gave the following results:
ABD
23
abd
26
ABd
51
abD
48
AbD
426
aBd
425
Abd
0
aBD
1
How were members of the allelic pairs distributed in the members of the appropriate chromosome
pair of the heterozygous females? What is the sequence of these linked genes in their
chromosomes? Calculate the map distances between the genes.
3.
In Drosophila, assume three pairs of alleles, +/n, +/o, and +/p. Genes n, o and p are all recessives
and sex-linked. They occur in the order n-o-p in the X-chromosome, with n being 12 map units
from o and o being 10 units from p. The coefficient of coincidence for this region of the X++p
chromosome is 0.5. From a cross between females of the genotype no+ and the wild type males,
predict the kinds and frequencies of phenotypes that would be expected to occur in a progeny of
2000 individuals.
4.
Curved wing and brown eyes are recessive characteristics in Drosophila. The genes for these
characteristics are located about 30 units apart in chromosome 2. Suppose that you wish to obtain a
double-mutant stock that is curved, brown. You have available for breeding purposes a singlemutant stock having brown eyes and normal wings, and a double-mutant stock which has normal
eyes but which is homozygous for curved wings and also homozygous for gene d, a sex-linked
recessive mutation. (The starting strains are homozygous at all loci.) Outline the steps you would
take in obtaining as efficiently as possible a true-breeding stock for curved wings and brown eyes,
which is also homozygous for the wild-type D allele. For each step show the genotypes of the
parents to be used in the cross, the genotypes of progeny that will be useful in succeeding steps
and the probability of obtaining such progeny. Remember that there is no recombination in male
Drosophila.
5.
You have two phage mutants, one temperature sensitive and the other suppressor sensitive. You
want to measure the frequency of recombination between the two mutations. Under what
conditions do you do the recombination? Under what conditions do you measure the frequency of
recombinants produced?
6.
Suppose that you do a cross between one phage which is a-b-c+ and another which is a+b+c-. You
select for recombinants which are a+c+. Either b- or b+ phage can grow under the selective
conditions.
7.
(a)
Among the a+c+ recombinants selected, you find that 0.1% are b+, 99.9% b-. What
conclusions can you draw?
(b)
What additional experiment would permit you to distinguish between the possible conclusions?
(c)
What conclusions can you draw if most of the a+c+ recombinants are b+?
(d)
What if the number of b- and b+ were about equal?
You have three strains of phage T4, each carrying a single mutation:
Strain
Mutation
1
a
2
b
3
d
In mixed infection with strains 1 and 2, approximately 0.0001 of the progeny are wild-type
recombinants. In mixed infection with either 2 and 3 or 1 and 3, approximately 0.01 of the progeny
are wild type recombinants.
(a)
What can you tell about the map positions of the three mutations from the data already
given?
(b)
What further experiments would be the most effective way of determining completely the
map positions of these three mutations relative to each other? (Your experiments may
include recombinations between the three strains given, to produce new strains. They may
not include modification of any of the strains by mutation or introduction of any additional
strains; i.e., the only strains that can be used are the original 3 and the strains produced by
recombinations between the original 3.
(c)
Give an example of the sort of data that might be obtained from the experiments in (b) and
state the conclusions about map order that you would draw from those data.
8.
A
|
B
|
1
C
|
D
|
E
|
2
F
|
3
G
|
H
|
4
I
|
J
|
5
The five lines above represent five T4 chromosomes, each with a deletion defined by the two cross
lines. For instance, chromosome 1 has the region between A and B deleted. A point mutation, P, is
isolated which forms wild-type recombinants with phage 1, 2, 4 and 5, but not with 3. This implies
that the above indicated location is wrong for one of the deletions. You guess that the mistake may
be in phage 3, so you test its recombination with the other phages and find that it recombines with
1 and 5 but not with 2 or 4. (a) Was your guess right or wrong? (b) Suggest a change in the
location of one of the ends of one of the deletions that would correct the error and make the
position of each of the deletions consistent with all of the above data. NOTE: There is more than
one possible answer. (c) To test the validity of the change that you suggested in part (b), what
cross would you make that would produce no wild-type recombinants if the original order were
correct but will produce wild-type recombinants if your suggested change is correct? (d)
Assuming that the change that you suggested in (b) is correct, and that no other corrections are
necessary, where does the mutation P map?
9.
Limit digestion of phage  DNA with Eco RI yields 8 fragments A, B, C, D, E, F, G, H. Digestion
with Sal I yields 3 fragments, I, II and III. When each of these is digested with Eco RI, the fragments
produced are:
I.
A, C, F and one new fragment
II.
E, G, and one new fragment
III.
B, H
Partial digestion with Eco RI yields several large fragments. One of them, upon complete digestion
with Eco RI, yields fragments B and G. Another yields C, F, and H; and another yields A and F.
Construct the map for phage .
10. Suppose that you digest mouse DNA with each of three restriction enzymes: Bgl II, Xba I, and Eco RI.
The digests are subjected to electrophoresis and three Southern blots are made, each having all three
digests. P32-labeled probes are made from three different restriction fragments, each also obtained
from the mouse genome. Two of these are Eco RI fragments J and K, and the third is Xba I fragment P.
The three Southern blots are hybridized with the three labeled fragments and autoradiographs are
made, resulting in the data shown below. The number next to each band represents the size of the
DNA fragment in that band, in kilobases.
Blot 1. Probed with Eco RI fragment J:
Digest:
Bgl II
Eco RI
Xba I
12.2
10.0
9.6
7.1
6.2
4.8
Blot 2. Probed with Eco RI fragment K:
Digest:
Bgl II
Eco RI
Xba I
10.0
8.5
6.9
4.8
3.3
Blot 3. Probed with Xba I fragment P:
Digest:
Bgl II
Eco RI
Xba I
10.0
9.6
7.1
6.2
Assume that each fragment shown is the only one of its size generated by digestion of the mouse
genome by the enzyme that produced that fragment. Construct a restriction map of the region of
the mouse genome defined by these probes. There may be ambiguities in the map, but do it as
accurately as the data allow.
Bios 344
Molecular Biology and Genetics
Answers to Study Questions on Recombination Mapping and Restriction Mapping
1.
2.
(a)
B
(b)
yes
(a)
30%
10%
A
a
3.
A
2%
C
5%
D
d
b
B
1000 females, all wild type
1000 males, including:
786 ++p
94 +++
114 +0+
and
and
and
n o+
nop
n+p
6+op
and
n++
5.
(a)
(b)
Su+ host at permissive temperature
Su- host at restrictive temperature
6.
(a)
(b)
(c)
(d)
Either the sequence is acb, or the sequence is abc, with c very close to b.
Do the same cross, but select for b+ c+ and observe distribution of a alleles.
Either the sequence is bac, or it is abc, with a very close to b.
b equidistant between a and c.
7.
(a)
a and b close together, d farther away. Approximately same distance from a to d as from b to d.
8.
(a) wrong;
(b) move D to between E & G;
B
9.
(c) 2 x 4;
(d) between D & G
H
III
C
G
E
II
I
D
F
Eco RI fragments indicated on the outside.
Sal I fragments indicated on the inside.
A
10. |-------------8.5-------------|-----------4.8------------|-----------6.2----------|-------------12.2------------|---X B
E B
X E
X
B
X E
B
X
_______________________________________________________________________________________
------3.3--------------------------------10.0-------------------------7.1---------------------------------------------6.9----------------------------------9.6-------------------------------------------