Genetic crosses and chi-squared in CCEA A2 2

The layout that collects every mark

Write all seven lines even when the answer is obvious. Examiners award the diagram, not the conclusion.

• Define your symbols first: let H = allele for a huntingtin repeat expansion (dominant), h = normal allele.
• Parental phenotypes, written in words.
• Parental genotypes, written underneath.
• Gametes, circled, showing every possible type.
• Punnett square or cross lines.
• Offspring genotypes.
• Offspring phenotypes and the ratio.

Sex linkage

Sex-linked alleles must be written on the chromosome: X^H X^h, not Hh. Losing the X and Y notation loses marks even when the ratio is right.

Remember the consequence students are usually asked to explain: males have only one X chromosome, so a single recessive allele is expressed because there is no second allele to mask it. That is why conditions such as haemophilia and red-green colour blindness are more common in males.

When the ratio is not what you expected

A dihybrid cross that should give 9:3:3:1 but does not is usually one of three things: autosomal linkage (the genes are on the same chromosome, so parental combinations are over-represented), epistasis (one gene masks another), or a lethal allele reducing one class.

Say which you think it is and justify it from the numbers. 'The observed ratio has far more parental types than recombinants, suggesting the genes are linked and only separated by crossing over' is a full answer.

Chi-squared, step by step

The conclusion carries a mark of its own. If the calculated value is less than the critical value, the difference is not significant, it is due to chance, and the null hypothesis is accepted. If it is greater, the difference is significant, unlikely to be due to chance alone, and the null hypothesis is rejected.

• State the null hypothesis: there is no significant difference between observed and expected results.
• Calculate expected numbers from the predicted ratio and the total offspring.
• For each class, compute (O − E)² / E, then sum.
• Degrees of freedom = number of classes − 1.
• Compare with the critical value at p = 0.05 for those degrees of freedom.

An answer that states a chi-squared value but never compares it with the critical value at p = 0.05 cannot score the conclusion mark, however accurate the arithmetic.

Natural selection questions attached to genetics

A2 2 often follows a cross with an extended response on selection in an unfamiliar species. Name the selection pressure, state that variation already exists in the population because of mutation, explain that individuals with the advantageous allele survive and reproduce more, and conclude that the allele frequency increases over generations.

Avoid 'survival of the fittest' and avoid any wording that implies organisms change on purpose — those phrases regularly cost marks.