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Genetics / D.J. Cove.

  • Cove, D. J.
Date:
1971
Catalogue details

Licence: Attribution-NonCommercial-NoDerivatives 4.0 International (CC BY-NC-ND 4.0)

Credit: Genetics / D.J. Cove. Source: Wellcome Collection.

  • Front Cover
  • Title Page
  • Table of Contents
  • Back Cover
    206/228 (page 198)
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    198 Answers It can be concluded that stocks A and В must be homozygous for a recessive mutant allele of the same gene. Similarly С and D, but that the two genes must be different, i.e. (ii) Now, considering differences leading to the production of red or mauve pigment, only В x D gives red, and so it can be proposed that these stocks are homozygous for a recessive mutant allele (c) of a third gene, whose wild-type allele is necessary for the conversion of red to mauve pigment and which is hypostatic to the a and b genes. The scheme now becomes : ü + stock A and В a + + b stock С and D + b stock E could be — a b a + + a + с A В Cl —h Cl С + ¿4*'' b с С D + b+ -\-b с The possible biochemical basis might be : colourless precursor > colourless precursor red precursor > mauve. blocked in — a blocked in blocked in h ^ b < individuals individuals individuals 13. (i) Neither parent requires adenine, but ^ of the progeny (58 out of 220) do. Two genes appear to be involved, and these must be unlinked. There are two possible explanations :
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