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6-2-7-predicting-inheritance-identifying-epistasis

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Predicting inheritance: identifying epistasis

Epistasis

  • Epistasis occurs when two genes on different chromosomes affect the same feature

    • One gene affects the expression of another gene

  • In the same way that a deviation from expected phenotypic ratios suggests that genes may be linked, phenotypic ratios can also be used to identify epistasis

  • In a standard dihybrid cross between two double heterozygous parents we would expect to see a ratio of 9:3:3:1 in the offspring, but when epistasis is taking place the offspring ratios will differ from this; we might expect ratios such as:

    • 9:3:4

    • 12:3:1

    • 9:7

    • 13:3

Example: epistasis in pigeons

  • There are two genes involved in determining feather colour in pigeons; gene R/r and gene F/f

    • For gene R/r:

      • allele R codes for a pigment that produces grey feathers

      • allele r doesn’t produce a pigment, resulting in white feathers

    • For gene F/f:

      • allele F codes for the production of an enzyme that stops production of grey pigment even if the allele R is present

      • allele f doesn’t produce an enzyme

  • This means that, for example:

    • Genotypes RRff and Rrff = grey feathers

    • Genotype rrff = white feathers

    • Any genotype that contains the allele F will result in white feathers, regardless of R/r, e.g.

      • RRFF

      • rrFF

      • RrFf

Worked Example

A cross occurs between two pigeons that are heterozygous at both gene loci.

Predict the offspring ratios that result from this cross.

Answer:

Step 1: determine the parental genotypes

  • Both parents are heterozygous for R/r and F/f

  • Parental genotypes = RrFf x RrFf

Step 2 determine the parental gametes

  • Both parents will produce the gametes RF and Rf and rF and rf

Step 3: produce a Punnett square with the offspring genotypes

Dihybrid cross without epistasis, downloadable AS & A Level Biology revision notes

Step 4: determine the offspring phenotypes

  • Grey feathers = 2x Rrff and RRff

  • White feathers = RRFF, 2x RRFf, 2x RrFF, 4x RrFf, rrFF, 2x rrFf, rrff

Step 5: note the phenotype ratio

  • 13 white : 3 grey

Examiner Tips and Tricks

You are unlikely to be asked to complete large Punnett squares in examples that relate to epistasis, but you should be able to recognise common ratios that occur with epistasis, and use information provided to predict phenotype in examples of epistasis