How to use these genetics practice questions
These questions cover core genetics rather than one particular examination board. Answer each question before opening the explanation in the question panel. After checking the answer, record the exact fact or distinction you missed. A correct guess is not the same as secure knowledge.
A useful routine is:
- choose the answer and explain it aloud in one sentence;
- identify the strongest distractor and state why it is wrong;
- turn any missed distinction into a flashcard;
- revisit the missed cards after a delay rather than rereading the whole topic.
Here is how a question session on a genetics revision board appears in MySummaries:
A woman who is heterozygous for an autosomal recessive condition has a child with a man who is homozygous unaffected. What is the probability that their child will be a carrier?
The woman is Aa and the man is AA. Their possible children are AA or Aa in equal proportions, so the probability of a carrier child is 50%. The strongest distractor is 25%, which would apply to the probability of an affected child when both parents are heterozygous, not to this cross.
The first question tests whether you can distinguish an affected child from a carrier child. This is a common source of errors because the same parental genotypes can produce different probabilities depending on what the question asks.
1. Autosomal recessive inheritance
Cystic fibrosis is inherited as an autosomal recessive condition. Two unaffected carrier parents have a child. What is the probability that the child will be affected?
The parental cross is Aa × Aa. The four equally likely genotype outcomes are AA, Aa, Aa and aa. One of four is aa, so the probability of an affected child is 25%. The strongest distractor is 50%, which is the probability of being a carrier, not the probability of being affected.
The key step is to write the parental genotypes before calculating. “Unaffected carrier” means heterozygous in an autosomal recessive condition; it does not mean homozygous unaffected.
2. X-linked recessive inheritance
A woman is a carrier of an X-linked recessive condition and the father is unaffected. What is the probability that a son will be affected, assuming the child is male?
A son receives the Y chromosome from his father and his single X chromosome from his mother. A carrier mother has a 50% chance of passing on the affected X chromosome, so 50% of sons are expected to be affected. The strongest distractor is 25%, which combines the 50% chance of a male child with the 50% chance of inheriting the affected X when asking about all children, rather than sons specifically.
Read “assuming the child is male” carefully. It removes the sex probability from the calculation. For the probability among all children, the corresponding risk would be 25%, provided the father is unaffected and the mother is a carrier.
3. Meiosis and independent assortment
Which event during meiosis is the direct cause of independent assortment of homologous chromosomes?
At metaphase I, each homologous pair aligns independently of the other pairs. This creates different combinations of maternal and paternal chromosomes in the gametes. The strongest distractor is the separation of sister chromatids at anaphase II: this separates copied chromatids but does not establish the independent orientation of homologous pairs.
Do not confuse independent assortment with crossing over. Independent assortment results from chromosome-pair orientation at metaphase I; crossing over exchanges DNA between homologous chromosomes during prophase I.
4. Crossing over
Crossing over between non-sister chromatids of homologous chromosomes occurs during which stage of meiosis?
Crossing over occurs during prophase I, when homologous chromosomes pair and exchange corresponding DNA segments at chiasmata. The strongest distractor is metaphase I, when homologous pairs align at the equator; exchange has already occurred by that stage.
A strong answer distinguishes the timing and the structures involved: homologous chromosomes pair in meiosis I, and non-sister chromatids exchange DNA. Sister chromatids are copies of the same chromosome and are not the usual partners in homologous crossing over.
5. DNA replication
Which statement best describes semi-conservative DNA replication?
The two parental strands separate, and each acts as a template for a new complementary strand. Each daughter molecule therefore contains one old strand and one new strand. The strongest distractor is the first option, which describes conservative rather than semi-conservative replication.
When revising replication, link the model to the mechanism: complementary base pairing allows each original strand to act as a template. DNA polymerase adds nucleotides to the growing strand and cannot simply create a strand without a template.
6. Transcription and translation
What is the main product of transcription in a eukaryotic cell?
Transcription uses a DNA template to produce an RNA molecule, including a pre-mRNA transcript in a eukaryotic cell before processing. Translation then uses the mature mRNA sequence to direct polypeptide synthesis. The strongest distractor is a polypeptide chain, which is the product of translation rather than transcription.
The simplest memory aid is DNA to RNA by transcription, then RNA to polypeptide by translation. In a written answer, avoid saying that DNA is “translated” or that mRNA is “replicated”.
7. Mutations and the genetic code
A single nucleotide substitution changes a codon, but the encoded amino acid remains the same. What type of mutation is this?
A silent mutation changes the nucleotide sequence without changing the amino acid because the genetic code is degenerate: several codons can specify the same amino acid. The strongest distractor is a missense mutation, in which the substitution changes one amino acid to another.
The mutation name describes the effect on the coding sequence or protein, not simply the fact that a base changed. A substitution can be silent, missense or nonsense. An insertion or deletion changes the reading frame when its length is not a multiple of three.
8. Gene regulation
In the lac operon of Escherichia coli, what is the immediate effect of lactose-derived allolactose binding to the repressor?
Allolactose binds the lac repressor and changes its shape, reducing its affinity for the operator. RNA polymerase can then transcribe the structural genes when the other regulatory conditions permit. The strongest distractor is that the repressor binds more tightly: the inducer has the opposite effect.
This question tests mechanism rather than terminology. State what binds to what, then give the consequence: inducer binds repressor, repressor leaves or cannot bind the operator, and transcription becomes possible.
9. Hardy–Weinberg calculations
In a population at Hardy–Weinberg equilibrium, the frequency of a recessive phenotype is 0.09. What is the expected frequency of heterozygotes?
The recessive phenotype frequency is q² = 0.09, so q = 0.3. Therefore p = 0.7, and the heterozygote frequency is 2pq = 2 × 0.7 × 0.3 = 0.42. The strongest distractor is 0.21, which is pq before applying the factor of two for the two heterozygote arrangements: Aa and aA.
For these calculations, write the sequence explicitly: q², then q, then p, then 2pq. The model assumes a large population, random mating, no mutation, no migration, no selection and no genetic drift. If those assumptions are not met, the expected frequencies may not apply.
A short-answer check
Multiple-choice practice can conceal whether you can construct an explanation. Try answering this prompt in three or four sentences, using the terms “template”, “complementary base pairing” and “semi-conservative”.
A marked written response can show which part of the explanation is missing:
Explain how semi-conservative DNA replication produces two DNA molecules with the same base sequence as the original molecule.
7/9The two DNA strands separate and each acts as a template. Free nucleotides join by complementary base pairing, with adenine pairing with thymine and cytosine with guanine. DNA polymerase joins the new nucleotides to make two DNA molecules, each with one original strand and one new strand.
The mechanism is accurate and the definition of semi-conservative replication is present.
Missed
−State that the original strands separate by breaking hydrogen bonds between complementary bases.
−Explain that the sequence is preserved because each template determines the complementary sequence of the new strand.
Model answerThe hydrogen bonds between complementary bases are broken, separating the two parental DNA strands. Each strand acts as a template, and free nucleotides are added by complementary base pairing: A pairs with T and C pairs with G. DNA polymerase joins the nucleotides to form the new strands. The products are two DNA molecules, each containing one parental strand and one newly synthesised strand, so each has the same base sequence as the original molecule.
That is a MySummaries paper, filled with genetics material. Yours is written from your own notes. Start free
The important correction is not simply adding more words. It is making the causal chain complete: strand separation permits templating; complementary pairing determines the new sequence; semi-conservation describes the composition of each product.
Turn errors into the next revision session
After these questions, sort errors by concept rather than by question number. For example, questions 1–3 may reveal confusion about genotype probabilities, while questions 4–5 may show that the stages of meiosis are being mixed up.
The weak-area view makes that pattern visible:
A useful rule is to repair the narrowest weak point first. If the problem is 2pq rather than all of population genetics, revise that calculation, answer two new examples, and then schedule the card again.
Repeated errors should become a small remediation task rather than another full page of notes:
You lost this mark twice: a recessive phenotype has frequency q² = 0.09. What are q, p and the expected heterozygote frequency 2pq?
How MySummaries helps
MySummaries can turn your genetics notes into a revision board, generate question practice from that material, mark written explanations against the board, and place repeated errors in a remediation tray. It can also schedule the resulting facts as flashcards, so a missed distinction such as 2pq versus pq returns for later practice.