Download Southern Blot

Biology 22
Problem Set 2
Spring 2011
Due May 10th, 9:30 am
Hand in the solutions to each of the following problems. Put a box around each of your
answers and show your work for partial credit. You may consult your book, lecture notes,
other students and/or your instructor for assistance.
1. Amino acids 51-56 were compared for an E. coli metabolic enzyme from a wild type
strain and a strain with a double frameshift mutation having one nucleotide deletion
followed by one nucleotide insertion. The results are shown below.
Amino Acid Residue
51
52
53
54
55
56
Wild Type
NH2ALA PHE ASP THR PRO VAL -COOH
Double
NH2GLY SER ILE
LEU HIS
VAL -COOH
Mutant
a. Determine the base sequence of the mRNA for both the wild type and double
mutant polypeptides.
b. Using your wild type mRNA sequence, show which base was deleted and the
site from which it was removed.
c. Using your mutant sequence, show which base was added and where it was
inserted. (You only need to give one of several possible solutions.)
UUU = phenylalanine
UUC = phenylalanine
UUA = leucine
UUG = leucine
CUU = leucine
CUC = leucine
CUA = leucine
CUG = leucine
AUU = isoleucine
AUC = isoleucine
AUA = isoleucine
AUG = methionine (start)
GUU = valine
GUC = valine
GUA = valine
GUG = valine
PHE
LEU
ILE
MET
VAL
UAU = tyrosine
TYR
UAC = tyrosine
UAA = stop
UAG = stop
CAU = histidine
HIS
CAC = histidine
CAA = glutamine
GLN
CAG = glutamine
AAU = asparagine ASN
AAC = asparagine
AAA = lysine
LYS
AAG = lysine
GAU = aspartic acid ASP
GAC = aspartic acid
GAA = glutamic acid GLU
GAG = glutamic acid
UCU = serine SER
UCC = serine
UCA = serine
UCG = serine
CCU = proline PRO
CCC = proline
CCA = proline
CCG = proline
ACU = threonine THR
ACC = threonine
ACA = threonine
ACG = threonine
GCU = alanine ALA
GCC = alanine
GCA = alanine
GCG = alanine
UGU = cysteine
UGC = cysteine
UGA = stop
UGG = tryptophan
CGU = arginine
CGC = arginine
CGA = arginine
CGG = arginine
AGU = serine
AGC = serine
AGA = arginine
AGG = arginine
GGU = glycine
GGC = glycine
GGA = glycine
GGG = glycine
CYS
TRP
ARG
SER
ARG
GLY
2. Four auxotrophic mutant strains of E. coli (1, 2, 3, 4) are each blocked at a different
step of the same metabolic pathway. This pathway involves compounds L, M, N, O,
and P. Each of these compounds is tested for its ability to support growth of mutant
strains, with the results given below.
(+= growth, __ = no growth)
Compound Used as a
Supplement
L
M
N
O
P
__
__
__
Mutant 1
+
+
__
Mutant 2
+
+
+
+
__
__
__
__
Mutant 3
+
__
__
Mutant 4
+
+
+
a. Give the order of the compounds in the metabolic pathway and indicate the
specific step that is blocked in each mutant strain.
b. Several sets of double mutations were produced. For example, double
mutant 1, 2 has mutant forms for enzymes 1 and 2. Pairs of these double
mutations were combined in partial diploids. Would a partial diploid consisting
of each of the following pairs of double mutations grow on minimal medium
(without supplementation with L, M, N, O, and P)?
1. 1,2 and 1,4
2. 2,4 and 1,3
3. 2,3 and 1,2
3. Ten mutants of the rII region in bacteriophage T4 DNA were analyzed for
complementation. These mutations are known to fall into TWO cistrons. The results
are given in the table below where += complementation and __ = no complementation.
1
2
3
4
5
6
7
8
9 10
__
__
__
__
__
__
__
__
__
__
1
__
__
__
__
__
2
+
+
+
+
__
__
__
__
3
+
+
+
+
__
__
__
__
__
4
+
+
__
__
__
__
__
__
5
__
__
6
+
+
+
__
__
__
7
+
__
__
8
+
__
9
+
__
10
a. Indicate which mutations fall into each of the two cistrons.
b. What type of mutation can explain the results for mutants 1 and 5?
4. In each of the following cases, determine whether B-galactosidase and Permease
would be produced by the lac operon in the presence and absence of lactose as an
inducer. Write + if the enzyme is produced and __ if the enzyme is not produced.
B-galactosidase
No
Lactose
lactose
a.
b.
c.
d.
e.
I- P+ Oc Z-Y+/ I+ P- O+ Z+ Y+
I+ P+ O+ Z- Y+/ I- P+ O+ Z+ YI- P+ O+ Z+Y-/ IS P+ O+ Z- Y+
I- P+ O+ Z- Y+/ I+ P+ Oc Z+ YI+ P+ Oc Z+ Y+/ I+ P+ O+ Z- Y-
Permease
No
Lactose
lactose
5. Consider the following two human genes, each with two possible alleles. The PIP
gene, coding for prolactin-inducible protein, is known to be on the short arm of
chromosome 7. The location of the KEL gene, which codes for a specific red blood cell
antigen, is unknown. Consider the pedigree shown below. Inheritance of alleles KEL1
and KEL2 was determined by assaying for the production of KEL1 and KEL2 antigens.
The PIP1 and PIP2 alleles were studied with a Southern blot analysis of endonuclease
Taq1 cut DNA where PIP1 shows a 4 kb fragment hybridizing to the probe while PIP2
shows a 6 kb fragment hybridizing to the probe. The results are shown below.
a. List the genotypes for all individuals shown in the pedigree. Refer to
individuals by the numbers given in the figure, eg. III-2.
b. Is the KEL gene on chromosome 7? Explain your decision by giving specific
examples from the genotypes you derived in part a.
Genotype
KEL 2, KEL 2
I
1
KEL 1, KEL 2
2
II
1
2
III
1
I-1
Size
6 kb
4 kb
I-2
II-1
2
Southern Blot
II-2
III-1
III-2
3
4
5
III-3
III-4
III-5