Download 1. Which gene could be X-linked? If it is a male, then only one X

1. Which gene could be X-linked? If it is a male, then only one X chromosome would be
present and it should segregate into ½ of the sperm cells…Ans: Gene S (c)
2. Which gene could be Y-linked? Exactly the same logic! The Y chromosome would
segregrate the same as the X…Ans: Gene S (c)
3. Which gene is autosomal and homozygous? It would be expected to be seen in all
cells, somatic and germ line…Ans: Gene R (b).
4. If you examine alleles for genes Q and T, they do not segregate randomly…Q1/T2
usually segregate together and Q2/Q1 usually segregate together….except…(see below)
ASO
Somatic Cell
Sperm number
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Gene Q
Q1 Q2
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Gene R
R1 R2
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Gene S
S1 S2
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Gene T
T1 T2
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Ans: 1/20 = 0.05 = 5 map units (a)
5. The sequence of the STRAND BEING
SYNTHESIZED is read 5’ to 3’ from the bottom of the
gel, since the smallest fragments (first synthesized and
terminated) migrate fastest during electrophoresis:
5’-ACGTGCACGTGC-3’
3’-TGCACGTGCACG-5’ would therefore be the
TEMPLATE STRAND.
A
G
C
T
Ans: 5’-GCACGTGCACGT-3’ (b)
6. False. Deoxyadensoine, an normal precursor, would be in all four of the reactions.
Dideoxyadenosine, the terminator, is present in only one fo the reactions.
7. False. The light is released as a result of an enzymatic reaction which requires ATP.
The ATP is generated from the pyrophosphate released from the incorporation of a
deoxynucleotide into a DNA chain.
8. The flowgram is generated by sequentially running the four nucleotide precursors over
the fiber optic block containing the amplified DNA molecules, and seeing which
nucleotide produces light output. If two bases are present in the amplified DNA, then the
light output should be twice that of one base, if three bases you should see three times the
light output, etc. The base calls would therefore be:
5’TCAGCGTAAG…3’
Ans: (c).
9. The last compound in the pathway supports the growth of the greatest number of
mutants; continuing this logic the following pathway could be constructed:
F → C → D → A→ B → E→ Z
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6
5 2
4 3
1
In this pathway, the immediate precursor to compound Z would be E. Ans: (a).
10. A mutant in the gene for enzyme six would be unable to convert F to C, but would be
wild type for all downstream enzymes. Continuing this logic, the above mutants could be
ordered in the putative pathway as indicated. If this were the case, then mutant 2 would
block the pathway between D and A, and precursor D could accumulate.
Ans: e) none of the above.