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A-Level Biology – Genetics, Variation and Evolution

DNA and protein synthesis, meiosis, inheritance, Hardy-Weinberg, speciation.

12 questions  ·  44 marks  ·  answers included  ·  A-Level Biology (Edexcel/AQA/OCR style)
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This worksheet covers:

  • DNA structure and protein synthesis
  • Meiosis and variation
  • Patterns of inheritance (including chi-squared testing)
  • The Hardy–Weinberg principle
  • Speciation

A calculator is allowed. Allow about 50 minutes.

Questions

Section A – DNA, protein synthesis and meiosis (skills)

1

(a) State the base-pairing rule in DNA.(b) A section of the template (antisense) strand of DNA has the base sequence $\text{TACGGCATT}$. Deduce the sequence of bases on (i) the mRNA transcribed from this strand, and (ii) the complementary DNA strand.

Show answer
(a) Adenine always pairs with thymine (A–T), and guanine always pairs with cytosine (G–C), by hydrogen bonding (complementary base pairing).(b) (i) mRNA (using U in place of T, complementary to the template strand): $\text{AUGCCGUAA}$.(ii) Complementary DNA strand: $\text{ATGCCGTAA}$.
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2

Explain the roles of mRNA and tRNA in translation at a ribosome.

Show answer
mRNA carries the genetic code (as a sequence of codons, copied from DNA during transcription) to the ribosome, specifying the order in which amino acids should be joined. tRNA molecules each carry a specific amino acid and have an anticodon that base-pairs (by complementary base pairing) with a complementary codon on the mRNA at the ribosome, ensuring the correct amino acid is added to the growing polypeptide chain in the sequence specified by the mRNA.
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3

A gene mutation changes one base in a DNA triplet from $\text{CTC}$ to $\text{CTT}$. Both triplets code for the amino acid leucine.(a) State the type of mutation shown.(b) Explain why this mutation is described as ‘silent’ and has no effect on the protein produced.

Show answer
(a) A substitution mutation (specifically a point/base substitution).(b) The genetic code is degenerate, meaning more than one triplet (codon) can code for the same amino acid; since both $\text{CTC}$ and $\text{CTT}$ code for leucine, the amino acid sequence of the resulting protein is unchanged, so the mutation has no effect on the protein’s structure or function — it is ‘silent’.
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4

Explain two ways in which meiosis produces genetic variation among gametes.

Show answer
Independent assortment: during metaphase I, homologous chromosome pairs line up randomly at the equator, so maternal and paternal chromosomes are distributed into gametes in random combinations, producing many different possible combinations of chromosomes in the gametes.Crossing over (recombination): during prophase I, non-sister chromatids of homologous chromosomes pair up and can exchange sections of DNA at chiasmata, creating new combinations of alleles on a single chromosome that were not present in either parent chromosome.
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Section B – Inheritance and chi-squared testing (exam-style)

5

Cystic fibrosis is caused by a recessive allele, $f$. Two parents, both unaffected but each a carrier (genotype $Ff$), have children.(a) Complete a genetic cross (Punnett square, described in words) to show the possible genotypes of their offspring.(b) State the probability that a child of these parents is affected by cystic fibrosis, and the probability that a child is an unaffected carrier.

Show answer
(a) Parents $Ff \times Ff$. Gametes from each parent: $F$ or $f$. Possible offspring genotypes: $FF$, $Ff$, $Ff$, $ff$ (in a 1:2:1 ratio).(b) Probability affected ($ff$) $=\dfrac{1}{4}$. Probability unaffected carrier ($Ff$) $=\dfrac{2}{4}=\dfrac{1}{2}$.
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6

In pea plants, tall ($T$) is dominant to dwarf ($t$), and round seed ($R$) is dominant to wrinkled seed ($r$); the two genes assort independently. A cross between two dihybrid plants ($TtRr \times TtRr$) is expected to produce offspring in a 9:3:3:1 ratio (tall round : tall wrinkled : dwarf round : dwarf wrinkled).Out of 160 offspring, calculate the expected number of each phenotype.

Show answer
Total ratio parts $=9+3+3+1=16$. Each part $=160/16=10$. Expected numbers: tall round $=9\times10=90$; tall wrinkled $=3\times10=30$; dwarf round $=3\times10=30$; dwarf wrinkled $=1\times10=10$.
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7

The observed offspring numbers from the cross in the previous question were: tall round $=92$, tall wrinkled $=29$, dwarf round $=32$, dwarf wrinkled $=7$.(a) Calculate the chi-squared ($\chi^2$) value using $\chi^2=\sum\dfrac{(O-E)^2}{E}$.(b) The critical value of $\chi^2$ at $p=0.05$ for 3 degrees of freedom is 7.815. State the conclusion that should be drawn, with a reason.

Show answer
(a) $\chi^2=\dfrac{(92-90)^2}{90}+\dfrac{(29-30)^2}{30}+\dfrac{(32-30)^2}{30}+\dfrac{(7-10)^2}{10}=0.044+0.033+0.133+0.900=1.11$ (to 3 s.f.).(b) Since the calculated $\chi^2$ value (1.11) is less than the critical value (7.815) at $p=0.05$, the null hypothesis (that there is no significant difference between the observed and expected results) is accepted; the difference between observed and expected offspring numbers is not statistically significant and is likely due to chance, so the data support a 9:3:3:1 ratio consistent with independent assortment of the two genes.
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8

State the null hypothesis that would be tested by the chi-squared test in the previous question, and explain what is meant by ‘degrees of freedom’ in this context.

Show answer
Null hypothesis: there is no significant difference between the observed and expected numbers of offspring of each phenotype (any difference is due to chance). Degrees of freedom is the number of categories (phenotypic classes) that are free to vary once the total is fixed; it is calculated as the number of categories minus 1, so here $4-1=3$.
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Section C – Hardy–Weinberg and speciation (challenge)

9

In a population of 1000 individuals, a recessive allele $b$ causes a particular disorder when homozygous ($bb$). The frequency of individuals showing the recessive phenotype is 9%.Using the Hardy–Weinberg equations ($p^2+2pq+q^2=1$ and $p+q=1$):(a) Calculate the allele frequency $q$ of allele $b$.(b) Calculate the allele frequency $p$ of the dominant allele $B$.(c) Calculate the percentage of the population that are heterozygous carriers.

Show answer
(a) $q^2=0.09$, so $q=\sqrt{0.09}=0.3$.(b) $p=1-q=1-0.3=0.7$.(c) Heterozygote frequency $=2pq=2(0.7)(0.3)=0.42$, so $42\%$ of the population are carriers.
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10

State three assumptions that must hold for a population to be in Hardy–Weinberg equilibrium.

Show answer
Any three of: no mutations occur; mating is random (no selective mating); no gene flow (no migration into or out of the population); no natural selection occurs (all genotypes are equally likely to survive and reproduce); the population is very large (so allele frequencies do not change much by chance/genetic drift).
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11

Distinguish between allopatric and sympatric speciation, and outline how each can lead to the formation of a new species.

Show answer
Allopatric speciation occurs when a population is separated by a geographical barrier (e.g. a river, mountain range or ocean), preventing gene flow between the two resulting populations; each population is exposed to different selection pressures/mutations and, over time, accumulates enough genetic differences (through natural selection and genetic drift) that individuals from the two populations can no longer interbreed to produce fertile offspring, forming reproductively isolated new species.Sympatric speciation occurs without geographical separation, within the same physical area; reproductive isolation instead arises through other mechanisms (for example, behavioural changes, differences in breeding time/season, or polyploidy in plants), which reduce or prevent interbreeding between subgroups of the population until they diverge into separate species.
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12

Explain, using the theory of natural selection, how a population of insects could evolve increased resistance to a pesticide after repeated pesticide use, referring to variation, selection pressure and allele frequency.

Show answer
Within the insect population there is already genetic variation (arising from mutation), so a small number of individuals may already carry an allele conferring some resistance to the pesticide. When the pesticide (the selection pressure) is applied, susceptible individuals are killed, while resistant individuals are more likely to survive and reproduce, passing the resistance allele on to their offspring. Over successive generations, the frequency of the resistance allele in the population increases (and the frequency of the susceptibility allele decreases), so the proportion of resistant insects in the population rises — the population as a whole evolves increased pesticide resistance.
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