1) The citric acid cycle is catalyzed by the enzymes indicated by lower case let
ID: 262205 • Letter: 1
Question
1) The citric acid cycle is catalyzed by the enzymes indicated by lower case letters.
You
have mutants (1-10) that cannot produce oxaloacetate (OAA) from pyruvate. Each mutant has a single point mutation. You supplement the strains with one of the intermediate products (CT, DCT, AKG, SCA, F, M, or OAA) in the pathway to see which strains can produce OAA (+ indicates can produce, - indicates cannot produce OAA), but mixed up the labels to the supplements, so the supplements were randomly labeled A-G.
media supplement
mutant
A
B
C
D
E
F
G
1
–
–
–
+
–
–
–
2
–
+
–
+
+
+
+
3
+
+
–
+
+
+
+
4
–
–
–
+
–
–
–
5
–
+
–
+
+
–
+
6
–
+
–
+
–
–
–
7
–
+
–
+
+
–
–
8
–
–
–
+
–
–
–
9
+
+
–
+
+
+
+
10
–
+
–
+
+
–
–
A. Which mutant strain(s), if any, has/have a mutation in enzyme b?
B. You isolate two additional mutants. Strain 11 has the same media supplementation pattern as strain 1. Strain 12 has the same supplementation pattern as strain 5. Let an “+” indicate complementation and a “-“ indicate no complementation.
How should Q1, 2, and 3 be filled? Q1 __________ Q2 ___________ Q3 _________
1
2
3
4
5
6
7
8
9
10
11
12
1
2
Q1
3
4
5
6
7
8
Q2
9
10
Q3
11
12
C. You isolate additional recessive mutants that have a defective enzyme b. You label these mutants A, B, C, D, and E. You amplify the region corresponding to enzyme b and digest the PCR product with EcoR1. You find the following pattern during a gel electrophoresis.
How many EcoRI sites are there in the amplified product? What are the mutations (point mutation, deletion, insertion, etc.) in the different mutants? Indicate whether the mutants are homozygous or heterozygous for their alleles.
media supplement
mutant
A
B
C
D
E
F
G
1
–
–
–
+
–
–
–
2
–
+
–
+
+
+
+
3
+
+
–
+
+
+
+
4
–
–
–
+
–
–
–
5
–
+
–
+
+
–
+
6
–
+
–
+
–
–
–
7
–
+
–
+
+
–
–
8
–
–
–
+
–
–
–
9
+
+
–
+
+
+
+
10
–
+
–
+
+
–
–
pyruvate Citric acid (Krebs) cycle oxaloacetate (OAA) citrate (CT)· D-isocitrate ? ?-ketoglutarate ? succinyLCoA- fumarate ? malate (DCT) (AKG) (SCA)Explanation / Answer
Table interpretation for ANSWER A:
No supplement is given: C (This means the compound has to be one that is not useful in the pathway or will not degrade to the final product. Of the given options of supplements, this is OAA, the final product).
Only one intermediate is needed: The intermediate in this case has to be Malate : Cases 1 and 4 (D is Malate)
Two intermediates positive: Case 6 (D is malate, so B is Fumarate)
Three intermediates positive: Cases 7 and 10 (D and B +E; which is Succinyl Coenzyme A)
Four intermediates positive: Cases 5 and 8 (in case 8, with fumarate it gives negative, however, addition of malate makes it positive; in case 5, fumarate is also positive. So on comparing 5 with 8, addition of g makes the mutant positive for OAA. So G has to be the next intermediate, which is AKG)
Five positive intermediates: Case 2: F is DCT
Six positive intermediates: Cases 3 and 9: A is CT
The intermediates are
CT
F
OAA
M
SCA
DCT
AKG
The mutation in enzyme b should lead to inefficiency in production of alpha-keto glutarate (AKG). Thus, the mutants that will need AKG supplementation should be screened for. These are mutants 2,3,5,8 and 9. Of these, we ned to find that strain in which DCT has not been supplemented. This strain is mutant 5. Now mutant 5 has no supplementation of DCT which means it makes DCT, but probably has a mutant/inefficient enzyme b which does not carry out further formation into AKG, thus demanding AKG supply in mutant 5. Therefore, mutant 5 has mutated enzyme b
Table and Media interpretation for ANSWER B:
Q1: Ingredients given: All except C (OAA). So, complementation will occur, where media composition of 9 complements media of strain 2)
Q2: Ingredients given: D, E, F, G (AKG through M, except F). But our understanding says that strain 8 has a functional enzyme d (and non functional enzyme e) and thus, even when devoid of Fumarate, could produce OAA. However in both cases enzyme a is probably mutated. However, DCT is supplied to make up for mutated enzyme a, so complementation will occur.
Q3: Table 1 in the question shows that strains 10 and 12 only require the final four intermediates, and probably can produce the earlier ones by catalysis through active, functional enzymes a and b. So even in this case complementation will happen.
Answer C:
Total two EcoR1 bands are present on the Wild type strain; the cutting of first site that will linearize the DNA and the second that will give two bands.
Mutant A has insertion and mutant D has addition, both seen by additional bands in between and at peripheries, respectively. There is no mutation in strain B
In strain C, deletion has occurred at cutting site.
In strain E, a point mutation has caused one of the cutting sites to mutate and therefore change in conformation and band weight.
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