Predicting Phenotypic and Genotypic Ratios
Worked Example
Fruit flies (Drosophila melanogaster) were crossed in a laboratory study looking for inheritance patterns for two characteristics, wing length and body colour. We can assume that these characteristics are unlinked.
The alleles for these characteristics are as follows:
V = long wings
v = short (vestigial) wings
B = brown body colour
b = black body colour
A black-bodied, heterozygous long-winged fly was crossed with short-winged, homozygous brown-bodied flies. Predict the phenotype ratio of their offspring.
Step 1: Write out the parental genotypes
Vvbb × vvBB
Step 2: Identify the gamete genotypes that each parent could produce
These are the allele combinations that each parent can produce in meiosis
Vb vb × vB
Step 3: Complete a Punnett square to show the genotypes of the offspring
Step 4: Identify the phenotypes of the offspring
Conclusion: The offspring would be 100% brown-bodied and 1:1 long to short-winged
Test crosses
- A test cross can be used to deduce the genotype
- The individual in question is crossed with an individual that is expressing the recessive phenotype
- This is because an individual with a recessive phenotype has a known genotype
- The resulting phenotypes of the offspring provide sufficient information to suggest the genotype of the unknown individual
- For a monohybrid test cross:
- If no offspring exhibit the recessive phenotype then the unknown genotype is homozygous dominant
- If at least one of the offspring exhibit the recessive phenotype then the unknown genotype is heterozygous
- For a dihybrid test cross:
- If no offspring exhibit the recessive phenotype for either gene then the unknown genotype is homozygous dominant for both genes
- If at least one of the offspring exhibit the recessive phenotype for one gene but not the other, then the unknown genotype is heterozygous for one gene and homozygous dominant for the other
- If at least one of the offspring exhibit the recessive phenotype for both genes then the unknown genotype is heterozygous for both genes
Worked Example
Worked example: Test crosses
- Rabbits have a single gene for ear length that has two alleles:
- D, a dominant allele that produces long ears
- d, a recessive allele that produces shorter ears
- A breeder has a rabbit called Floppy that has long ears and they want to know the genotype of the rabbit
- There are two possibilities: DD or Dd
- The breeder crosses the long-eared rabbit with a short-eared rabbit
- A rabbit displaying the recessive short ear phenotype has to have the genotype dd
Test Cross Possibility Table
- The predicted ratio of phenotypes of offspring – 1 long ears
- The predicted ratio of genotypes of offspring – 1 Dd
Test Cross Possibility Two Table
- Predicted ratio of phenotypes of offspring – 1 long ears : 1 short ears
- Predicted ratio of genotypes of offspring – 1 Dd : 1 dd
- The breeder identifies the different phenotypes present in the offspring
- There is at least one offspring with the short ear phenotype
- This tells the breeder that their rabbit Floppy has the genotype Dd
- If Floppy was genotype DD none of the offspring would have short ears
Worked Example
Worked example: Hard Question
A farmer wishes to maximise his yield of soybean oil from his crop. Oil is extracted from the seeds which typically measure 6-12 mm in diameter. In one species of soybean, Glycine max, two characteristics are governed by the following pairs of unlinked alleles.H = high oil content in the seeds; h = low oil content in the seeds
E = four seeds in a pod; e = two seeds in a pod
The farmer crossed two soybean plants, both with high oil content and four seeds per pod. This cross resulted in 381 offspring plants being produced. The 381 F1 offspring had a phenotypic spread as shown in the table below.
Use the data in the table to deduce the genotypes of the parent plants in this cross.
Both parents were high-oil content with 4 seeds per pod, which are the dominant traits for each pair of alleles. Neither parent can be homozygous recessive for either oil content or the number of seeds.
Possible allele combinations for 4 seeds per pod: EE, Ee
Therefore, the possible parental genotypes were: HHEE, HHEe, HhEE, HhEe
The smallest number of F1 offspring was 23 for low oil content, 2 seeds. These must have the homozygous recessive genotype, hhee.
The ratios of other phenotypes, relative to the homozygous recessive phenotype, were:
low oil, 4 seeds : hhee = 70:23 = 3.04 : 1 ≅ 3:1
high oil, 2 seeds : hhee = 73:23 = 3.17 : 1 ≅ 3:1
high oil, 4 seeds : hhee = 215:23 = 9.3 : 1 ≅ 9:1
The phenotypes of the offspring display an approximate 9:3:3:1 ratio. This characterises a cross between two double-heterozygous parents where the genes are unlinked, as in this case
Conclusion: Both parents' genotype was HhEe / double heterozygous
Exam Tip
Make sure before you start a test cross you think about the following: how many genes are there, how many alleles of each gene are there, which is the dominant allele, what type of dominance is it and is there linkage or codominance between genes?
Even though we learn about fruit flies (Drosophila melanogaster) being studied in detail which led to the discovery of gene linkage, many of their characteristics are unlinked and can be studied on a large scale in this manner.
