IB Biology SL Theme D: Continuity and change questions

Answer: C. Taq polymerase extends primers by adding nucleotides, making copies of the region between them. The 300 bp product is shorter than the 600 bp fragment, so it moves farther through the gel. A: assigns DNA synthesis to helicase and reverses the gel-size relationship; helicase unwinds strands, while the shorter fragment travels farther. B: correctly identifies polymerase synthesis but reverses the size-dependent migration. D: assigns DNA synthesis to helicase, which separates strands rather than making DNA copies. Remember: PCR copies DNA with polymerase; gel electrophoresis separates fragments by size.

Answer: B. A base substitution can give rise to an SNP, but degeneracy of the genetic code means it may or may not change the amino acid. A one-nucleotide insertion within a coding sequence shifts the reading frame of subsequent codons, making loss of polypeptide function likely. Thus I and III are correct; II is not.

A: accepts the misconception in II that every substitution changes an amino acid, and omits the consequence of the insertion.
C: accepts II and fails to identify the base substitution as an SNP.
D: includes II; degeneracy means the replacement need not change the amino acid.

Remember: substitutions may be silent, while a one-nucleotide insertion can shift the reading frame.

Answer: C. The egg contributes X, while a Y-bearing sperm results in an XY zygote, associated with certain male-typical physical characteristics. An XY child receives the father’s Y chromosome, not his X chromosome, so cannot inherit R from him. The X chromosome carries more genes than the Y chromosome.

A: assumes the father passes his X-linked allele R to an XY child.
B: identifies the Y-sperm outcome but omits the true comparison of gene numbers.
D: assumes R can pass from father to an XY child and omits the sperm’s role in determining the zygote’s sex chromosomes.

Remember: an XY child receives its X chromosome from the egg and its Y chromosome from the sperm.

Answer: A. Type 2 diabetes involves reduced responsiveness of body cells to insulin. Insulin is being released, but its effect in lowering blood glucose is reduced, matching the pattern described.

B: identifies type 2 but substitutes insufficient insulin release for reduced cellular responsiveness; the case specifies normal release.
C: identifies reduced responsiveness but assigns it to type 1, which is associated with insufficient insulin production.
D: recalls insufficient insulin production in type 1, but that does not fit the normal insulin release in this case.

Remember: Type 1 involves insufficient insulin production; type 2 involves reduced responsiveness to insulin.

Answer: B. Biomagnification is an increase in pollutant concentration in consumer tissues at higher trophic levels. Both plotted concentrations rise from level 2 to level 5; at level 5, DDT is 0.8 mg kg−1 and mercury is 0.4 mg kg−1, so DDT is higher. A: recognizes the rising trends but reverses the comparison at level 5. C: follows the DDT trend but misses that mercury also increases, from 0.1 to 0.4 mg kg−1. D: recognizes both rising trends but reads the two level-5 values as equal. Remember: biomagnification is indicated by pollutant concentration increasing across higher trophic levels.

Answer: D. Warmer surface water can impede nutrient upwelling, so fewer nutrients reach the surface. The reduced nutrient supply lowers primary production, decreasing energy flow through marine food chains.

A: identifies the decline in nutrients and production but reverses the resulting change in energy flow. B: reverses the effect of reduced nutrient availability on primary production. C: reverses the effect of warmer surface water on nutrient upwelling, while keeping the later links consistent.

Remember: Lower primary production means less energy enters marine food chains.