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Continuity and changeIB Biology SL: Topic test

20 questions, 54 marks

IB Biology SL

Continuity and change topic test

Total 54 marks

Name

Class

Date

  1. 1
    A biotechnology company compares two DNA polymerase enzymes used in PCR. Enzyme Taq polymerase, commonly used because it is heat-stable, introduces on average 1 incorrect base per 10,000 bases copied. Enzyme Pfu polymerase, which has a proofreading function, introduces on average 1 incorrect base per 1,000,000 bases copied.
    (a)
    Both DNA polymerases copy a DNA template into a new complementary strand by matching each template base to its complementary partner. Which pairing correctly matches a DNA template base to the base added by the polymerase to the new strand?
    [1 mark]
    • ATemplate guanine (G) paired with new cytosine (C)
    • BTemplate guanine (G) paired with new adenine (A)
    • CTemplate adenine (A) paired with new guanine (G)
    • DTemplate thymine (T) paired with new thymine (T)
    (b)
    Which statement best explains the much lower error rate of Pfu polymerase compared with Taq polymerase?
    [1 mark]
    • APfu polymerase works at a lower temperature, which always reduces errors
    • BPfu polymerase has a proofreading function that can detect and remove incorrectly paired bases before continuing replication
    • CPfu polymerase only copies half of the DNA template, so fewer errors accumulate
    • DPfu polymerase unwinds the DNA double helix more slowly, which is unrelated to base-pairing accuracy
    (c)
    State the role of complementary base pairing in allowing accurate copying of a DNA template, and explain why a strand of DNA copied with too many errors could be unreliable for uses such as DNA profiling.
    [2 marks]

    Total for question 1: 4 marks

  2. 2
    A short section of the coding (sense) strand of a gene has the base sequence TAC GGA CCT AAA. A point mutation changes the seventh base (C) in this coding strand to a T, giving TAC GGA TCT AAA. (Codon table extract: UAC = tyrosine; GGA = glycine; CCU = proline; UCU = serine; AAA = lysine.)
    (a)
    What is the sequence of amino acids coded by the original (unmutated) coding strand sequence?
    [1 mark]
    • ATyrosine–glycine–serine–lysine
    • BMethionine–glycine–proline–lysine
    • CTyrosine–glycine–proline–lysine
    • DTyrosine–proline–glycine–lysine
    (b)
    What effect does this particular point mutation have on the resulting polypeptide?
    [1 mark]
    • ANo effect, because the mutation is silent (degeneracy of the code)
    • BA frameshift, so every amino acid after the mutation is changed
    • CTranslation stops immediately at the mutated codon
    • DA single amino acid substitution, changing proline to serine at that position
    (c)
    Explain, using the term degeneracy, why some point mutations do not change the amino acid sequence of a protein even though they change the base sequence of the gene.
    [2 marks]

    Total for question 2: 4 marks

  3. 3
    A laboratory compares the number of new mutations detected in skin cell DNA samples from two groups of volunteers. Group X (outdoor workers with high, long-term UV exposure) show an average of 42 new somatic mutations per million bases sequenced. Group Y (indoor workers with low UV exposure), matched for age, show an average of 11 new somatic mutations per million bases sequenced.
    (a)
    Calculate how many times higher the mutation rate is in group X than in group Y, and identify the type of mutagen most likely responsible for the difference.
    [3 marks]
    (b)
    Distinguish between the consequences for the volunteers of these mutations occurring in their skin (somatic) cells, compared with the same kind of mutations occurring in cells that give rise to gametes (germ cells), and explain why mutation is described as random.
    [4 marks]

    Total for question 3: 7 marks

  4. 4
    In a plant-breeding programme, a horticulturalist crosses two pure-breeding varieties of pea plant: one with purple flowers (dominant allele) and one with white flowers (recessive allele). All F1 offspring have purple flowers. The horticulturalist then self-pollinates several F1 plants to produce an F2 generation of 320 plants, of which approximately one quarter are expected to have white flowers.
    (a)
    Using a genetic diagram or clearly described genotypes, explain the ratio of purple to white flowers expected in the F2 generation, and state how many of the 320 F2 plants would be expected to have white flowers.
    [6 marks]
    (b)
    Explain the roles of meiosis and self-fertilization in producing the variety of genotypes seen among the F2 plants, and discuss one advantage sexual reproduction (as used in this cross) has over asexual reproduction for a plant breeder seeking new flower colour combinations.
    [6 marks]

    Total for question 4: 12 marks

  5. 5
    In a species of plant with a diploid chromosome number of 2n = 8, a researcher examines cells from the actively dividing root tip (undergoing mitosis) and separately examines cells from the developing anthers (undergoing meiosis to form pollen).
    (a)
    How many chromosomes are present in each daughter cell produced by mitosis in the root tip?
    [1 mark]
    • A4
    • B8
    • C16
    • D2
    (b)
    How many chromosomes are present in each of the four haploid cells produced by meiosis in the anther?
    [1 mark]
    • A8
    • B2
    • C4
    • D16
    (c)
    State one similarity and one difference between mitosis and meiosis in terms of the number of divisions and the genetic identity of the daughter cells produced.
    [2 marks]

    Total for question 5: 4 marks

  6. 6
    A doctor measures a patient's blood glucose concentration before and after a meal on two occasions. On occasion 1, blood glucose rises from 4.5 mmol dm⁻³ before the meal to 7.8 mmol dm⁻³ thirty minutes after the meal, then returns to 4.6 mmol dm⁻³ after two hours. On occasion 2 (a different patient), blood glucose rises from 5.0 mmol dm⁻³ before the meal to 12.4 mmol dm⁻³ thirty minutes after the meal, and is still 10.1 mmol dm⁻³ after two hours.
    (a)
    Which statement best interprets the data from occasion 2, in relation to homeostasis of blood glucose?
    [1 mark]
    • ANegative feedback control of blood glucose appears impaired, since the concentration remains high two hours after the meal rather than returning to close to the pre-meal level
    • BBlood glucose is being regulated normally, since it still rises after the meal
    • CThe set point for blood glucose has increased permanently for this patient
    • DThe pancreas has produced too much insulin, causing blood glucose to remain elevated
    (b)
    Which hormone is mainly responsible for returning blood glucose concentration to its set point after the rise seen following a meal in occasion 1?
    [1 mark]
    • AGlucagon
    • BThyroxin
    • CAdrenaline (epinephrine)
    • DInsulin
    (c)
    Outline the role of negative feedback in the normal regulation of blood glucose shown in occasion 1.
    [2 marks]

    Total for question 6: 4 marks

  7. 7
    A hospital tracks the percentage of Staphylococcus aureus bacterial samples resistant to a commonly used antibiotic over a 20-year period during which the antibiotic was used extensively. In year 1, 2% of sampled bacteria were resistant. By year 20, 38% of sampled bacteria were resistant. Resistant bacteria carry a mutant allele of a gene that alters the antibiotic's target protein so it is no longer affected by the drug.
    (a)
    Explain, in terms of natural selection, why the percentage of resistant bacteria increased over the 20 years.
    [3 marks]
    (b)
    This is described as an example of natural rather than artificial selection, even though it results from human antibiotic use. Explain this distinction, and suggest one way in which careful use of antibiotics could slow the rate of increase in resistance shown by the data.
    [4 marks]

    Total for question 7: 7 marks

  8. 8
    Over several decades, rising sea surface temperature and increasing atmospheric carbon dioxide concentration have affected a coral reef ecosystem. Marine biologists have recorded (i) an increase in the frequency of coral bleaching events, during which the mutualistic algae living inside coral polyps are expelled, and (ii) a gradual decline in the rate of calcification (growth of the coral skeleton) as seawater pH has fallen.
    (a)
    Using the concept of ecosystem stability, explain why the loss of live coral from bleaching events threatens the stability of the whole reef ecosystem, not just the coral itself.
    [6 marks]
    (b)
    Explain how rising atmospheric carbon dioxide concentration causes both coral bleaching (via rising sea temperature) and the decline in calcification (via falling seawater pH) described, distinguishing between these two separate mechanisms.
    [6 marks]

    Total for question 8: 12 marks

End of questions