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1. The figure indicates the results of a Southern blot: all three lanes would contain
genomic DNA that was cut into thousands of individual fragments, only one of which
detects the CF gene on each homolog chromosome. An example from the lectures:
homozygote
heterozygotes
Ans: essentially the same number regardless of whether a wild type or mutant CF allele
was present in the genome (e).
Wild type
CF probe
2.
mutant
R
R
R
R
R
R
R
P
F
-
M
origin
The probe identifies two different size fragments; the shorter
fragment (in red) is associated with cystic fibrosis in this
instance:
+
In this family, the father and mother are both
heterozygoes; the child has inherited a mutant allele from one parent, and a wild type
allele from the other parent, so the fetus is also a heterozygote:
The fetus, therefore is phenotypically normal, but can pass the disease allele on to his
progeny. Within the general population, however, the frequency of the disease allele is
rare (< 5% of the population would be expected to be heterozygotes). Therefore, there
would be little chance the child would give rise to a child with
G
A
T
C
the disease. Ans: none of the above (e).
C
3. All of the DNA sequencing reactions require all 4
deoxynucleotide triphosphates to synthesize DNA.
Ans: all of the above (e).
-
4. The sequence of the strand being synthesized is read from
the bottom of the gel up:
5’-GATACTGTTACC-3’; the sequence of the TEMPLATE
strand would therefore be:
5’-GGTAACAGTATC-3’ Ans: (a)
+
5. W→V→S→U→T→Q→R
The immediate precursor to compound Q is
compound T. by the way, since none of the
mutants grow on W, the link that it would be
a precursor in the pathway is if W increased
in mutant 5,
Ans: (a)
Mutant number
1
2
3
4
5
6
Q
+

+
+
+
+
Chemical Compounds
R S T U V W
+  + +  
+     
+ + + +  
+     
+ + + + + 
+  +   
6. W→V→S→U→T→Q→R
5 3 1 6 4 2
A mutation in gene 5 would lead to accumulation of the immediate precursor to V. Ans:
none of the above (e).
7. The rhodopsin gene family is discussed on pages 242-244.
The green-receiving protein is most closely related to the red-receiving protein; not all
genes are on the X chromosome;
Ans: red/green colorblindness can be due to unequal crossing over between genes (c).
8. 1 and 3 do not complement:
3 and 4 do not complement;
Therefore 1,3,4 are in same complementation group.
2 and 6 do not complement;
1 complements 5 and 5 complements 6, so 5 exists in its
own complementation group.
Ans: * would not show complementation, and ** would
show complementation. (d)
mutations
1
2
3
4
5
6
1
–
2
+
–
3
–
4
*
–
–
–
5
+
6
–
**
–
+
–
9. Ans: 3 complementation groups (1,3,4); (2,6); 5 (c)
10. If the genome is considered to be random with regard to nucleotide composition, then
a six cutter would be expected to cut (1/4)6 = 1 out of every 4096 nucleotides
(3.2 x 106)/4096 = 781.25
Ans: (b)
STS marker
BAC Clone
A
B
C
D
E
F
1
+
+
-
2
+
+
3
+
-
4
+
+
-
5
+
+
6
+
+
-
7
+
-
11. In this instance, a + sign indicates that the clone CONTAINS the nucleotide sequence
present in the STS marker because they hybridize- therefore the sites have to be on the
same piece of DNA.
A: 3 and 4 are together; B: 2 and 7 are together; C: 1 and 5 are together
D: 1 and 6 are together, therefore 615; E: 4 and 6 are together, therefore 34615
F: 2 and 5 are together, therefore 3461527 Ans: (e)