Inheritance and Genetics: Question 4
Syllabus 17.4
In a species of cotton plant, fibre colour is controlled by a single gene. Brown fibre is produced by the dominant allele, F; white fibre is produced by the recessive allele, f. A plant breeder has a brown-fibred cotton plant of unknown genotype and wants to find out whether it is homozygous or heterozygous for this gene.
(a) State the two possible genotypes of the brown-fibred plant. [1]
(b) Describe how the breeder could use a test cross to determine the genotype of this plant, including the result that would be expected for each possible genotype. [3]
(c) The breeder crosses the unknown brown-fibred plant with a white-fibred plant. Of the offspring produced, 48 have brown fibre and 51 have white fibre. Use a genetic diagram to explain what this result shows about the genotype of the original brown-fibred plant. [3]
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Worked solution
Part (a): Possible genotypes of the brown-fibred plant
Brown fibre is the dominant phenotype, so a brown-fibred plant carries at least one F allele. It could therefore be homozygous, FF, or heterozygous, Ff, both genotypes produce brown fibre, so appearance alone cannot distinguish them.
Part (b): Using a test cross
To find out which genotype the plant has, the breeder should carry out a test cross: crossing the unknown brown-fibred plant with a plant that is homozygous recessive, ff (white fibre). This is useful because the white-fibred parent can only contribute the recessive allele, f, so any dominant allele appearing in the offspring must have come from the unknown plant.
- If the unknown plant is FF, every offspring inherits an F allele from it (and an f allele from the white parent), giving genotype Ff in every offspring, so all of the offspring will have brown fibre.
- If the unknown plant is Ff, roughly half of its gametes carry F and half carry f. Combined with the f-only gametes from the white parent, about half the offspring will be Ff (brown) and about half will be ff (white), so the breeder expects roughly equal numbers of brown-fibred and white-fibred offspring.
Part (c): Interpreting the offspring numbers
Suppose the unknown plant is heterozygous, Ff, crossed with a white-fibred plant, ff:
| f | f | |
|---|---|---|
| F | Ff | Ff |
| f | ff | ff |
This gives offspring genotypes Ff and ff in a 1:1 ratio, i.e. an expected phenotypic ratio of 1 brown-fibred : 1 white-fibred.
The breeder’s actual results, 48 brown-fibred and 51 white-fibred offspring, are close to this expected 1:1 ratio (small differences between the two numbers are due to the chance nature of fertilisation, not an error). If the unknown plant had instead been homozygous, FF, none of the offspring could have had white fibre, which does not match the observed results. The data therefore show that the original brown-fibred plant was heterozygous, Ff.
Final answers
- (a) Possible genotypes: FF or Ff.
- (b) Test cross with a homozygous recessive (ff) white-fibred plant: all-brown offspring indicates FF; roughly equal brown and white offspring (1:1) indicates Ff.
- (c) The observed 48 : 51 brown-to-white result matches the expected 1:1 ratio from an Ff × ff cross, showing the original plant was heterozygous, Ff.