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IB Biology – Theme D: Continuity and Change

DNA, cell division, gene expression (HL), inheritance, homeostasis, natural selection.

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12 questions  ·  40 marks  ·  answers included  ·  IB Biology SL & HL (2023+ syllabus)
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This worksheet covers:

  • DNA
  • Cell division
  • Gene expression (HL)
  • Inheritance
  • Homeostasis
  • Natural selection

Questions marked (HL) are for Higher Level students only; all others are SL and HL.

Questions

Section A – DNA and cell division

1

Outline the roles of DNA helicase and DNA polymerase in the semi-conservative replication of DNA.

Show answer
DNA helicase unwinds the DNA double helix and breaks the hydrogen bonds between complementary bases, separating the two strands to form a replication fork. DNA polymerase then binds free nucleotides to the exposed bases on each template strand, in the correct complementary base-pairing order, joining them to form a new strand; each new DNA molecule therefore consists of one original (template) strand and one newly synthesised strand.
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2

A human somatic cell has $2n=46$ chromosomes. State the number of chromosomes and the number of chromatids present at each of the following stages:(a) metaphase of mitosis(b) the end of meiosis I(c) the end of meiosis II (in a gamete)

Show answer
(a) 46 chromosomes, 92 chromatids (each chromosome has already replicated into two sister chromatids).(b) 23 chromosomes, 46 chromatids (homologous pairs have separated, but sister chromatids are still joined).(c) 23 chromosomes, 23 chromatids (sister chromatids have separated into individual chromosomes).
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3

Explain how non-disjunction during meiosis can result in a gamete with an extra copy of chromosome 21, leading to Down syndrome after fertilisation.

Show answer
Non-disjunction occurs when homologous chromosomes (in meiosis I) or sister chromatids (in meiosis II) fail to separate properly during anaphase. This produces one gamete with two copies of chromosome 21 instead of one, and another gamete with none. If the gamete with two copies of chromosome 21 fuses with a normal gamete at fertilisation, the resulting zygote has three copies of chromosome 21 (trisomy 21), causing Down syndrome.
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4

Outline two ways in which meiosis increases genetic variation in offspring compared with mitosis.

Show answer
Crossing over between non-sister chromatids of homologous chromosomes during prophase I exchanges sections of DNA, creating new combinations of alleles on each chromosome. Independent assortment of homologous pairs during metaphase I means each pair orients randomly, so gametes receive a random mixture of maternal and paternal chromosomes, producing many different possible chromosome combinations. (Mitosis produces genetically identical daughter cells, with no crossing over or independent assortment of homologous pairs.)
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Section B – Gene expression and inheritance

5

(HL) Outline the process of transcription, including the role of RNA polymerase and the removal of introns to produce mature mRNA.

Show answer
RNA polymerase binds to a promoter region on the DNA and unwinds a section of the double helix. It moves along the template strand, joining free RNA nucleotides in a sequence complementary to the template (with uracil replacing thymine), producing a pre-mRNA transcript. This pre-mRNA contains both exons (coding sequences) and introns (non-coding sequences); the introns are removed and the exons joined together (splicing) to produce the mature mRNA that leaves the nucleus for translation.
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6

A gene mutation causes a single base substitution in the DNA coding for an enzyme, changing one amino acid in the primary structure of the protein.Explain how this mutation could result in a non-functional enzyme.

Show answer
The change in the primary structure (amino acid sequence) can alter the pattern of bonds (e.g. hydrogen bonds, disulfide bridges, hydrophobic interactions) that determine the enzyme’s tertiary structure, changing its overall three-dimensional shape. If this changes the shape of the active site, the substrate may no longer bind (no longer complementary), so the enzyme can no longer catalyse the reaction and is non-functional.
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7

In pea plants, the allele for round seed shape ($R$) is dominant to the allele for wrinkled seed shape ($r$). Two heterozygous ($Rr$) plants are crossed.Determine the expected phenotypic ratio of the offspring, showing your reasoning.

Show answer
Cross: $Rr \times Rr$. Gametes from each parent: $R$ or $r$. Combining gives genotypes $RR:Rr:Rr:rr$, i.e. $1\,RR : 2\,Rr : 1\,rr$. Since $R$ is dominant, $RR$ and $Rr$ are round and $rr$ is wrinkled, giving a phenotypic ratio of $3$ round $:1$ wrinkled (75 percent round, 25 percent wrinkled).
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8

(HL) Explain how epigenetic modification, such as DNA methylation, can silence gene expression without changing the base sequence of the DNA.

Show answer
Methyl groups are added to cytosine bases (often in promoter regions) of the DNA by enzymes; this does not change the sequence of bases. The added methyl groups can prevent transcription factors and RNA polymerase from binding to the promoter, or promote a more tightly packed chromatin structure, so the gene cannot be transcribed and is effectively switched off (silenced), even though the underlying DNA sequence is unchanged.
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Section C – Homeostasis and natural selection

9

Explain how negative feedback involving insulin and glucagon maintains blood glucose concentration within narrow limits after a meal.

Show answer
After a meal, blood glucose concentration rises above the set point; this is detected by beta cells in the pancreas, which secrete insulin. Insulin causes liver and muscle cells to take up glucose and convert it to glycogen (glycogenesis), and increases uptake of glucose by body cells generally, lowering blood glucose back towards the set point. If blood glucose falls too low (e.g. between meals), alpha cells secrete glucagon, which stimulates the liver to break down glycogen into glucose (glycogenolysis) and release it into the blood, raising blood glucose back towards the set point. In each case, the change is detected and a response is produced that counteracts (opposes) the original change, which is negative feedback.
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10

Explain how vasodilation and sweating help to reduce body temperature when core temperature rises above normal.

Show answer
Vasodilation: arterioles supplying skin capillaries widen, increasing blood flow near the skin surface, so more heat is lost from the blood to the environment by radiation and convection. Sweating: sweat glands secrete more sweat onto the skin surface; as this sweat evaporates it takes latent heat from the skin, cooling the body. Both responses increase heat loss from the body, helping to bring core temperature back down towards the normal set point.
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11

Using the example of the evolution of antibiotic resistance in a population of bacteria, explain the process of natural selection.

Show answer
Within a bacterial population there is genetic variation (arising from mutation), so a few individuals may already carry an allele giving resistance to an antibiotic. When the antibiotic is applied, it acts as a selection pressure: susceptible (non-resistant) bacteria are killed, while resistant bacteria survive. The surviving resistant bacteria reproduce, passing the resistance allele to their offspring, so the proportion (frequency) of the resistance allele in the population increases over successive generations, and the population as a whole becomes more resistant to the antibiotic.
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12

The bones in the forelimb of a human, whale and bat have a similar underlying arrangement, despite being used for very different functions (grasping, swimming, flying).Explain what this observation of homologous structures suggests about the evolutionary relationship between these species.

Show answer
The similarity in the underlying skeletal structure, despite the very different functions, suggests that humans, whales and bats share a common ancestor that had this same basic limb arrangement. Over time, as each lineage evolved, the same basic structure became adapted (modified by natural selection) for different functions in each species; structures that are similar due to shared ancestry, even when their function differs, are called homologous structures, and this pattern is evidence for evolution from a common ancestor.
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Related worksheets

More worksheets in IB Biology (SL & HL):

IB Biology – Theme A: Unity and DiversityIB Biology – Theme B: Form and FunctionIB Biology – Theme C: Interaction and InterdependenceIB Biology – Data-Based Questions (Paper 1B)

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