In: Biology
You set up a dihybrid cross between true-breeding tall pea plant with white flowers and a true-breeding short pea plant with purple flowers. Your F1 generation looks uniformly tall with purple flowers. The results of the F2 generation counts are shown in the table below:
Phenotype | Number of progeny |
Tall/purple flowers | 850 |
Short/purple flowers | 350 |
Short/white flowers | 87 |
Tall/white flowers | 313 |
Your calculated Chi square test statistic for this analysis is:
a. 8.31
b. 3.37
c. 4.19
d. 16.3
e. 13.4
Let the dominant gene for tallness and purple flower color be C and P respectively while the correspondant genes for shortness and white flower color are c and p respectively. Let true breeding tall pea plants with white flowers (CCpp) is crossed with true breeding short pea plant with purple flowers (ccPP). The F1 generation obtained are tall with purple flowers (CcPp). Now the F1 are selfed to obtain F2 generation, the results of the cross is shown below:
Parents: CcPp (tall purple) CcPp (tall purple)
gametes: CP cP Cp cp CP cP Cp cp
F2 generation: CCPP (tall purple), CCPp (tall purple), CcPP (tall purple), CcPp (tall purple), CCPp (tall purple), CCpp (tall white), CcPp (tall purple), Ccpp (tall white), CcPP (tall purple), CcPp (tall purple), ccPP (short purple), ccPp (short purple), CcPp (tall purple), Ccpp (tall white), ccPp (short purple), ccpp (short white) So the phenotypic ratio is 9 (tall purple) : 3 (short purple) : 1 (short white) : 3 (tall white)
The numbers of progeny of different phenotypes are: tall/purple flowers (850); short/purple flowers (350); short/white flowers (87) and tall/white flowers (313). Now we calculate chi-square test for this analysis which is as follows:
Phenotype | Observed number(o) | Expected number(e) | d=(o-e) | (o-e)2 | (o-e)2/e |
Tall/purple flowers | 850 | 900 | -50 | 2500 | 2.77 |
Short/purple flowers | 350 | 300 | 50 | 2500 | 8.33 |
Short/white flowers | 87 | 100 | -13 | 169 | 1.69 |
Tall/white flowers | 313 | 300 | 13 | 169 | 0.56 |
Total | 1600 |
Chi-square test value,
So the correct option is (e)