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A particular variety of corn has a gene for kernel color and a gene for height w

ID: 18566 • Letter: A

Question

A particular variety of corn has a gene for kernel color and a gene for height with the
following phenotypes:

CC, Cc; purple kernels
cc; white kernels

TT, Tt; Tall stems
tt; dwarf stems

Two true breeding plants are crossed; one is true breeding for purple kernels and dwarf stems, and the other is true breeding for white kernels and tall stems. The F1 offspring all have purple kernels and tall stems. Two of these F1 plants are then crossed.

(a) Diagram a Punnett square showing the cross between the two F2 plants. In each square, indicate the phenotype of that genotype. Indicate the ratios of the phenotypes and the genotypes.

(b) You are given one of the F2 plants that has purple kernels and tall stems. You want to determine what the genotype of this plant is, so you perform a testcross. What will be the genotype and phenotype of the test plant? Why?

(c) If you perform the test cross in part c and the F2 plant is heterozygous for purple kernels and homozygous for tall stems, predict the genotypic and phenotypic ratio of the test-cross offspring.

Explanation / Answer

a)  F1 generation = CcTt

     F1 Generation cross = CcTt xCcTt

                 CT        Ct          cT         ct   

        CT    CCTT   CCTt       CcTt      CcTt           

        Ct     CCTt    CCtt       CcTt      Cctt           

        cT     CcTT    CcTt       ccTT     ccTt            

        ct      CcTt     Cctt      ccTt       cctt

         CcTt = 5/16    CCtt = 1/16     CCTT = 1/16
CCTt = 2/16         CcTT = 1/16           Cctt = 2/16

ccTT = 1/16          ccTt = 2/16          cctt = 1/16

 

b) cctt, will be the test plant crossed with the purple/tall plant. This way you can rule out by ratios of the progeny whether the parent (purple/tall) is a homozygous or heterozygous

 

c) CcTT x cctt

          CT         CT       cT        cT                

   ct   CcTt     CcTt    ccTt     ccTt                 CcTt = 50%         ccTt = 50%

   ct   CcTt     CcTt    ccTt     ccTt

   ct   CcTt     CcTt      ccTt   ccTt

   ct   CcTt      CcTt    ccTt    ccTt

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