Investigating Genetic Ratios (AQA A Level Biology): Revision Note
Exam code: 7402
Investigating genetic ratios
Genetic diagrams (e.g. Punnett squares) predict offspring genotypes and phenotypes using known parental genotypes and meiosis
In sexual reproduction, each allele from a homologous pair has an equal chance of entering a gamete and being inherited
Predictions from genetic diagrams are based on probability—actual offspring ratios may differ due to random fertilisation of gametes
Model organisms like Fast Plant® and Drosophila are used to study inheritance because:
They have short life cycles meaning rapid breeding
Traits are controlled by single genes that determine easily identifiable traits
For these experiments, the assumption has been made that gamete fertilisation occurs randomly
Monohybrid inheritance in Fast Plant®
A particular type of Fast Plant® is bred from the species Brassica rapa for the purposes of research
Usually, these plants produce a purple pigment called anthocyanin in their stems
This trait is controlled by a single gene with two alleles:
A = dominant, anthocyanin present (purple stem)
a = recessive, no anthocyanin (green stem)
A monohybrid cross can be carried out to study the inheritance of this single pair of alleles
Example 1: a homozygous cross using Fast Plant®
A homozygous dominant Fast Plant® is crossed with a homozygous recessive Fast Plant®
Parental phenotypes | Purple stem × Green stem |
|---|---|
Parental genotypes | AA × aa |
Gamete genotypes | A (from AA) a (from aa) |
Offspring genotypes | All Aa |
Offspring phenotypes | All purple stem |
Examiner Tips and Tricks
When both parents are homozygous, no Punnett square is needed — offspring genotypes are predictable directly.
Example 2: a heterozygous cross using Fast Plant®
A monohybrid cross can also be used to predict the phenotypes of offspring that would be produced from two heterozygous Fast Plants® interbreeding
Parental Phenotypes | Purple stem × Purple stem |
Parental Genotypes | Aa × Aa |
Gamete Genotypes | A or a from each parent |
Offspring Genotypes | 1 AA : 2 Aa : 1 aa |
Offspring Phenotypes | 3 purple stem : 1 green stem (3:1 ratio) |
Examiner Tips and Tricks
Use a Punnett square when both parents are heterozygous and always include genotype and phenotype labels in diagrams
Monohybrid inheritance in Drosophila
Small fruit flies from the genus Drosophila have been used in genetic experiments for many years
Their wing length is controlled by a single gene with two alleles:
L = dominant, produces long or wild type wing
l = recessive, produces the stunted or vestigial wing
A monohybrid cross can be carried out to study the inheritance of this single pair of alleles
Example: a heterozygous cross in Drosophila
A monohybrid cross between two heterozygous Drosophila can be completed as follows:
Parental phenotypes | Wild type wing × Wild type wing |
|---|---|
Parental genotypes | Ll × Ll |
Gamete genotypes | L or l from each parent |
Offspring genotypes | 1 LL : 2 Ll : 1 ll |
Offspring phenotypes | 3 wild type wing : 1 vestigial wing (≈75% : 25% ratio) |
Examiner Tips and Tricks
Remember when dealing with genetic diagrams and questions in the exam, you always know the genotype of the individual displaying the recessive phenotype; they have to be homozygous recessive!
Probability of inheritance
Genetic diagrams (like Punnett squares) help predict the chance of offspring inheriting certain phenotypes
These predictions are shown as ratios, which represent the probability of each phenotype appearing
Example: a dihybrid cross
In a dihybrid cross, a 9:3:3:1 ratio is typical when two individuals heterozygous for both genes are crossed
This ratio means there are 4 possible phenotypes in the offspring:
Phenotype | Expected number (out of 16) | Percentage probability |
|---|---|---|
W | 9 | 56.25% |
X | 3 | 18.75% |
Y | 3 | 18.75% |
Z | 1 | 6.25% |
Percentage probabilities are often used when dealing with the probability of inheriting genetic diseases
Worked Example
A man (Mike) and a woman (Sarah) want to have their own biological baby. They are concerned as Sarah's father has haemophilia, a blood clotting condition.
The allele for a blood clotting factor (F or f) is found on the X chromosome. Females have two copies of the allele whereas men only have one. It is a recessive sex-linked disease.
Neither Mike nor Sarah suffer from the condition.
Calculate the chance that they will have a baby who suffers from haemophilia. Give your answer as a percentage.
Step 1: Work out Sarah's genotype
As her father was a sufferer he must have the genotype XfY
Her father must have given her his Xf chromosome for her to be the female sex
As she does not suffer from the recessive disease it can be said that she must also have the dominant allele
Her genotype is XFXf
Step 2: Work out Mike's genotype
As Mike does not suffer from the disease he must have the dominant allele on his single X chromosome
His genotype is XFY
Step 3: Cross the two genotypes using a genetic diagram
Parental phenotypes: carrier female x normal male
Parental genotypes: XFXf XFY
Parental gametes: XF or Xf XF or Y
The predicted ratio of genotypes in offspring -
1 XFXF : 1 XFXf : 1 XFY : 1 XfY
The predicted ratio of phenotypes in offspring -
1 female with normal blood clotting : 1 carrier female : 1 male with normal blood clotting : 1 male with haemophilia
Ratio 1 : 3
Step 4: Use the predicted ratio to obtain the percentage probability
The ratio is 1:3
1 + 3 = 4 so need to find 1 as a percentage of 4
(1 ÷ 4) x 100 = 25
There is a 25% chance or probability that their baby will have haemophilia
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