Question

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There is a flowering plant where plants can make either purple flowers or white flowers. A P1 cross was set up with true-breeding plants crossing plants with purple flowers was plants with white flowers. The F1 progeny all had purple flowers. The following results were obtained in the F2 generation: 155 purple 11 white How many genes are involved in plant pigment, and what is the genetic explanation for the modified ratio of offspring observed? (Mark 2 choices) Duplicate gene function 1 gene Gene interaction Complementation Recessive lethal allele Codominance Dominant epistasis 2 genes Incomplete dominance

Answer 1

Answers- 2 genes are involved in pigment production.

Genetic explanation- Duplicate gene function, Gene interaction

Here the F2 phenotypic ratio is = 155/11= 14.09 which is almost equivalent to 15:1 ratio of dihybrid cross.

Here two genes are interacting to produce the flower colour. If the alleles of both gene loci produce the same phenotype without cumulative effect then the 9:3:3:1 ratio is modified into 15:1 ratio. Duplicate gene interaction allows dominant alleles of either duplicate gene to produce the wild type phenotype. Only organisms with homozygous recessive of both genes produce the mutant phenotype.

Here, the mechanism by which colour is determined is an example of duplicate gene function. Here flower colour is dependent upon a biochemical reaction that converts a colourless precursor into a coloured product and this reaction can be performed by either the product of gene A or gene B. Thus having either of the A or B allele produce colour of the flower. But lack of either allele will produce a white flower.

So, if two plants with genotype AaBb are crossed with each other then the progeny genotypes- AABB,AABb,AaBB,AaBb,AAbb,Aabb,aaBB,aaBb will produce purple flower and the genotype aabb will produce white flower. Whenever a dominant allele is present at either locus, the biochemical conversion occurs and the resulting phenotypic ratio is 15:1 ( colour to non colour).

Enzyme A Cproduct of gene A) Precursor (colourless) - Product (coloured - Purple) Enzyme B (product of gene B)

P1 - X AABB (purple) i aa bb (white) F1 -> Aa Bb A B% х Аав АВ Ab АВ А. е в ол AABB AABN AABB Aa Bb | ААВ АА. АавАал | AaBB Aabb Jaa BB cable AaBb Aabb В ab Only white 15:1 = Purple : White