Biology 101 Practice Exam — Free

Write the genotypes from the phenotype descriptions, because that is where most errors start. The black, short-tailed parent heterozygous for both genes is BbSs. The brown parent must be bb (brown is recessive, so it shows only when homozygous), and heterozygous for the tail gene makes it Ss: the second parent is bbSs.

Because the genes are on different chromosomes they assort independently, so treat each gene as a separate monohybrid cross and multiply the probabilities (the product rule).

Coat colour: Bb × bb gives one half Bb (black) and one half bb (brown). Probability of brown is $\tfrac{1}{2}$.

Tail length: Ss × Ss gives one quarter SS, one half Ss, one quarter ss. Probability of long tail (only ss) is $\tfrac{1}{4}$.

Combine:

A Punnett square gives the same answer: the bbSs parent makes gametes bS and bs in equal numbers, the BbSs parent makes BS, Bs, bS and bs in equal numbers, and exactly 1 of the 8 equally likely combinations is bbss.

Expected fraction with brown coat and long tail: 1/8 (option A).

Option B (1/16) is the familiar dihybrid $F_2$ figure for BbSs × BbSs, but the brown parent here cannot contribute a B allele, so the recessive class is twice as common. Option C (1/4) is the chance of a long tail alone, ignoring coat colour, and option D (3/8) is the fraction expected to be brown and short-tailed.

Transcription pairs each template base with its complementary RNA base: A with U, T with A, G with C, C with G. RNA polymerase reads the template in the $3' \rightarrow 5'$ direction and builds the mRNA $5' \rightarrow 3'$, so the mRNA is antiparallel to the template and its 5′ end lines up with the template's 3′ end.

Pair the bases one at a time:

• Template: 3′-T A C G G A C T T A T C-5′
• mRNA: 5′-A U G C C U G A A U A G-3′

Now read the mRNA in the $5' \rightarrow 3'$ direction in non-overlapping triplets, starting at the first AUG: AUG | CCU | GAA | UAG.

AUG is the start codon and encodes methionine; CCU encodes proline; GAA encodes glutamate; UAG is a stop codon, which signals release of the finished chain and does not add an amino acid. The polypeptide therefore has three residues.

A common error is to copy the template directly, swapping only T for U, which gives UAC GGA CUU AUC. That sequence has no start codon and would wrongly read as Tyr–Gly–Leu–Ile. Another is to count the stop codon as a fourth amino acid.

mRNA: 5′-AUGCCUGAAUAG-3′; the polypeptide has three amino acids, Met–Pro–Glu.

(a) Light beetles show the recessive phenotype, so they are the only genotype that can be read directly from the count; that is why the calculation starts with $q^2$:

Allele frequencies sum to 1, so $p = 1 - 0.40 = 0.60$.

(b) Under Hardy–Weinberg equilibrium the heterozygote frequency is $2pq$:

Check that the genotype classes add up: $p^2 \times 2000 = 0.36 \times 2000 = 720$ homozygous dominant, and $720+960+320 = 2000$.

(c) Selection acts on phenotypes, and only homozygous recessive beetles are light. Count where the d alleles are. The 320 light beetles carry $2\times320 = 640$ copies; the 960 heterozygotes carry 960 copies; that is $1600$ copies in total, which matches $q \times 2N = 0.40 \times 4000$. So $960/1600 = 60\%$ of the d alleles sit in dark-coloured heterozygotes, which look identical to DD beetles and are not eaten. Each generation the predator removes only the d alleles that happen to be in homozygotes, and as $q$ falls, an even larger share of the remaining d alleles is sheltered in heterozygotes, because $q^2$ shrinks faster than $2pq$. Selection against a recessive allele is therefore effective at first but becomes very slow once the allele is rare, and the allele is not eliminated within any reasonable number of generations.

(a) q = 0.40, p = 0.60; (b) 960 heterozygotes; (c) 60 % of the d alleles are protected in heterozygotes, so q declines slowly.