Biotechnology: Practice Questions (Biology 6.8)

Answer: (B). All fragments move toward the positive electrode. Charge per unit length is about the same, so the pull per unit length is similar; the gel’s mesh slows long fragments most. The 4000 bp fragment stays closest to the wells.

• (A) The 300 bp fragment has less total charge but also much less length to drag through the gel. It travels farthest.
• (C) A negative fragment is attracted to the positive electrode, not repelled. And more charge does not mean it moves less.
• (D) Equal charge per length is true, which is why separation depends on size; the gel is what separates them.

Answer: (A). At 95 °C the strands separate. Cooling to about 55 °C lets the short primers form hydrogen bonds with matching sequences at each end of the target. This is annealing.

• (B) describes extension, which is usually done at about 72 °C.
• (C) If the template strands paired again, primers could not bind and nothing would be copied. The many primer molecules bind first.
• (D) The polymerase is chosen to be heat-stable and works through every cycle.

Answer: (D). Total DNA = 10 µL × 0.010 µg µL⁻¹ = 0.10 µg. Fraction spread = 100 ÷ 500 = 0.20, so DNA on the plate = 0.10 × 0.20 = 0.020 µg. Efficiency = 156 ÷ 0.020 = 7800 = 7.8 × 10³ transformants per µg.

• (A) multiplies colonies by DNA (156 × 0.020) instead of dividing.
• (B) is the value per nanogram (0.020 µg = 20 ng; 156 ÷ 20 = 7.8), labelled with the wrong unit.
• (C) uses all 0.10 µg of DNA, but only a fifth of the mixture was spread.

(a) N = 5 × 2²⁸ = 5 × 268 435 456 = 1 342 177 280 ≈ 1.34 × 10⁹ copies.

(b) Need 2ⁿ ≥ 1 × 10⁸ ÷ 5 = 2 × 10⁷. log₂(2 × 10⁷) ≈ 24.3, so 25 cycles (round up). Check: 5 × 2²⁴ ≈ 8.4 × 10⁷ (too few); 5 × 2²⁵ ≈ 1.7 × 10⁸ (enough).

(c) Every cycle heats the mixture to about 95 °C to separate the strands. An ordinary polymerase would denature and stop working after the first cycle, so it would have to be added again every cycle. A heat-stable enzyme survives repeated heating.

(d) The primer can no longer base-pair well with its binding site, because the sequences are no longer complementary. Little or no copying starts from that end, so little or no product forms (no band, or a faint band, on a gel). The other primer alone can only make single-stranded copies that increase slowly, not exponentially.

Accept “about 1.3 × 10⁹” in (a).

(a) The child’s 600 and 380 bp bands are present in the mother. The 450 and 250 bp bands are not, so they must have come from the father.

(b) Man 1 is excluded: he has the 450 bp band but not the 250 bp band, so he could not have given the child that band. Man 2 has both 450 and 250 bp bands, so he cannot be excluded.

(c) The claim is too strong. The results are consistent with Man 2 being the father, but other men could also carry 450 and 250 bp bands, especially relatives. Only a few bands were compared. Testing many more DNA regions would make a chance match far less likely. (Man 2’s 520 bp band is not in the child, which is expected: a child inherits only one of each pair of a parent’s alleles.)

(a) A: a lawn (dense, continuous growth) of white bacteria. B: no growth. C: a lawn, almost all white. D: a small number of separate colonies, blue.

(b) Plate A shows that the bacteria were alive and able to grow. Plate B shows that kanamycin kills bacteria without the plasmid, so any growth on D cannot be explained by naturally resistant cells.

(c) On plate C, both transformed and untransformed cells grow, and the few transformed cells are hidden among millions of others. On plate D, kanamycin kills every cell that did not take up the plasmid, so each colony must have grown from a transformed cell.

(d) Blue: each colony comes from a cell carrying the plasmid, which includes the pigment gene. The genetic code is (almost) universal, so the bacterial cell’s RNA polymerase and ribosomes can transcribe and translate the gene into the same amino acid sequence.

Accept “a few blue colonies may be hard to see on C” for plate C if the lawn is described.

(a) X differs at 2 positions: identity = 22 ÷ 24 = 91.7%. Y differs at 5: 19 ÷ 24 = 79.2%. Z differs at 9: 15 ÷ 24 = 62.5%.

(b) X, because it has the fewest differences from W. Mutations build up over time in separated lineages, so fewer differences suggest a more recent common ancestor.

(c) These two changes probably arose once, by mutation, in a common ancestor of X, Y and Z after its lineage split from W’s, and were inherited by all three. (They are less likely to have arisen independently three times.)

(d) Any one: only 24 bases of one gene were compared, so chance could affect the counts; different genes can change at different rates; the same base can change more than once, hiding some differences; sequencing errors are possible.