9) A flower breeder notices that a cross between true-breeding red roses and true-breeding white roses...
9) A flower breeder notices that a cross between true-breeding red roses and true-breeding white roses makes all red-flowering F1 offspring. Interbreeding these F1 offspring, she gets 100 red-flowering roses, 40 pink-flowering roses, and 44 white-flowering roses in the F2 generation. What are the genotypes of the parents, F1 offspring, and F2 offspring? What is the probable mode of inheritance? Check your answer with the appropriate statistical test.
Homework Answers
Answer 9)
This is the case of recessive epistasis.Let there are two genes. R gene encodes red color. And I gene suppresses the R gene in homozygous recessive conditions.
Genotypes of parents : RRii ( red flower)
rr II ( white flower)
F1 offsprings have RrIi ( red flower)
Punnet square for F2:
| Gametes | RI | Ri | rI | ri |
| RI | RRII ( red) | RRIi ( red) | RrII ( red) | RrIi ( red) |
| Ri | RRIi ( red) | RRii ( pink) | RrIi( red) | Rrii ( pink) |
| rI | Rrii( red) | RrIi ( red) | rrII ( white) | rrIi ( white) |
| ri | RrIi ( red) | Rrii ( pink) | rrIi( white) | rrii ( white) |
The phenotypic ratio: Red (9): pink (3): white (4)
Chi square test;
| Phenotypes | observed (O) | expected (E) | (O-E)^2/E |
| Red | 100 | (9/16)×188=106 | 0.3396 |
| Pink | 40 | (3/16)×188=35 | 0.625 |
| White | 44 | (4/16)×188=47 | 0.1914 |
| Total | 188 | 188 | 1.156 |
Chi square experimental value is 1.156
The chi square value from table at p=0.05 and at degree of freedom 3-1=2 is 5.991
As our experimental value is less that value from table. That's why our hypothesis is accepted.
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