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7.1.5Genetics, Populations, Evolution and Ecosystems (A Level only)

Investigating Genetic Ratios — AQA A Level Biology 7402 Revision Notes & Topical Questions

5 min read 9 parts Lesson 3 of 25 1 quiz

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7.1.5 Investigating Genetic Ratios — AQA A Level Biology 7402 notes: Predict and test genetic ratios in breeding experiments, including autosomal, sex-linked, codominant and epistatic inheritance, and explain why observed ratios differ from expected ones.Genetic crosses give expe…

  • expected ratios
  • Drosophila melanogaster
  • Exam-style question 1
  • Exam-style question 2
  • Exam-style question 3
  • Monohybrid cross

Section 1 of 9Why study ratios?

Genetic crosses give expected ratios of phenotypes in the offspring, based on Mendel's laws: alleles segregate when gametes are formed, and, for genes on different chromosomes, alleles assort independently.

Breeders and geneticists compare the ratios they actually observe with the predicted values to find out how a characteristic is inherited.
3:1
Typical ratios are a 3:1 ratio of dominant to recessive phenotypes from a monohybrid cross between two heterozygotes, a 1:1 ratio from a test cross,
  • and a 9:3:3:1 ratio in a dihybrid cross between two double heterozygotes.
Deviations from these ratios provide clues about linkage, codominance, epistasis or lethal alleles.

Section 2 of 9Organisms used and method

Organisms used for genetic crosses have a short life cycle, produce many offspring and are easy to keep. Drosophila melanogaster (fruit fly) has a generation time of about 10 to 14 days, and its phenotypes, such as eye colour and wing form, are easy to see.

Organisms used and method
Other examples are the garden pea, maize (grain colour), and mice or Brassica (rapid-cycling brassica) seeds.
A typical investigation is:
1
Choose parents of known phenotype, usually pure-breeding (homozygous).
2
Cross them to produce the F1 generation, then cross the F1 with each other (or test cross) to produce the F2.
3
Anaesthetise the flies (or collect seeds), count the offspring of each phenotype in the F2, taking care to record all of them and avoiding bias.
4
Calculate the observed ratio and compare with the predicted ratio using a chi-squared test (next lesson).

Ethical issues include humane treatment and anaesthesia of animals.

Section 3 of 9Predicting ratios: a worked sequence

To predict the ratio:

1
Define symbols for alleles, for example R (round) and r (wrinkled).
2
Write the parental genotypes and the gametes each can produce.
3
Use a Punnett square to combine gametes.
4
Read off genotype and phenotype ratios, considering dominance.

For example, Rr × Rr gives 1 RR : 2 Rr : 1 rr, a phenotype ratio of 3 round : 1 wrinkled. Rr × rr gives 1 Rr : 1 rr, so 1:1. For a sex-linked gene on the X chromosome, such as white eyes in Drosophila, a heterozygous red-eyed female (XRXr) crossed with a red-eyed male (XRY) produces all red-eyed females and half red- and half white-eyed males, a ratio different from that of an autosomal gene.

Section 4 of 9Modified ratios

SituationCrossResult
Codominance (e.g. human blood group alleles IA and IB)Heterozygote × heterozygote1 : 2 : 1 phenotype ratio
Sex linkageXRXr × XRYRatio differs between sexes
Autosomal linkageGenes on the same chromosomeParental types more frequent than expected 9:3:3:1
EpistasisDihybrid, one gene masks the otherRatios such as 9:3:4 or 9:7
Lethal alleleHomozygous genotype dies2:1 ratio

Epistasis occurs when one gene influences the expression of another, for instance when the product of one gene is needed to produce a pigment from the product of another. In a dihybrid cross a 9:3:4 ratio can show recessive epistasis.

Section 5 of 9Why observed results differ

Observed numbers rarely match the predicted ratio exactly.

1
Fertilisation is random,
  • so by chance a small sample will give a ratio different from the expected one;
    • the larger the sample, the closer the results come to the predicted ratio.
2
Differences may also arise from linkage, selective survival of certain genotypes (some phenotypes may be less viable), incorrect assumptions about dominance, errors in classification of phenotypes,
  • or sex linkage.
3
A statistical test is used to decide if the difference is due to chance.
4
If it is, the results are consistent with the predicted ratio; if not, the hypothesis about the mode of inheritance should be rejected and another suggested.

Section 6 of 9Interpreting breeding data

Given numbers of offspring, simplify the ratio by dividing by the smallest number.

Interpreting breeding data
For example, 612 round : 188 wrinkled is close to 3.3 : 1, which is close to 3:1, consistent with a monohybrid cross between heterozygotes.
For a test cross, a 1:1 ratio shows that the unknown parent is heterozygous, whereas all dominant offspring suggest the parent is homozygous dominant.
Working out the genotype of the parents from the offspring ratios is a favourite exam question: first identify the dominant allele from a cross of two parents with the same phenotype producing offspring with a different phenotype,
  • then deduce genotypes.

Section 7 of 9Exam-style questions

  1. In Drosophila, grey body (G) is dominant to black body (g). Two heterozygous flies are crossed and give 400 offspring. State the expected numbers of each phenotype and explain why the observed numbers may differ. [4]4 marks

  2. A cross between a red-eyed female and a white-eyed male gives offspring that are all red-eyed. The F1 are crossed. Explain the expected F2 if the gene is X-linked, with red dominant. [4]4 marks

  3. A cross between two heterozygous plants gives 9 : 3 : 4. Suggest an explanation. [2]2 marks

Section 8 of 9Key terms

TermMeaning
Monohybrid crossCross involving one gene with two alleles
F1 / F2First and second filial generations
Test crossCross of an unknown genotype with a homozygous recessive
Sex linkageInheritance of a gene located on a sex chromosome
CodominanceBoth alleles are expressed in the heterozygote
EpistasisInteraction in which one gene affects the expression of another
Autosomal linkageGenes on the same chromosome inherited together

Section 9 of 9Exam technique

1
Expecting the exact predicted numbers in a small sample; chance variation is normal.
2
Forgetting that sex-linked genes give different ratios in males and females.
3
Writing that a test cross uses a heterozygous parent; the tester is homozygous recessive.
4
Thinking that a 9:3:3:1 ratio will result when genes are linked.
5
Confusing genotype ratio with phenotype ratio.

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