In the tomato, red fruit ( R ) is dominant over yellow fruit ( r ), and yellow f
ID: 95615 • Letter: I
Question
In the tomato, red fruit (R) is dominant over yellow fruit (r), and yellow flowers (F) are dominant over white flowers (f). A cross was made between true-breeding plants with red fruit and yellow flowers, and plants with yellow fruit and white flowers. All F1 offspring are heterozygotes for both traits. The F1 female plants were then test crossed to males with yellow fruit and white flowers (homozygous recessive for both traits). The following results were obtained:
333 red fruit, yellow flowers
64 red fruit, white flowers
58 yellow fruit, yellow flowers
345 yellow fruit, white flowers
Total of 800
1. Conduct a Chi square analysis to see if the two genes are linked ( Check a solved problem S1, page 148 in the first edition text book, and page 150 in the second edition problem #1 for an idea)…follow the same steps)
a. Propose a hypothesis to be tested
b. Find the expected (predicted) number of offspring ( Hint what is the ratio of test cross results?)
c. Calculate and Interpret the Chi –square result
d. If the two genes (fruit color and flower color) are linked, calculate the map distance between the two genes.
2. Set a sample test cross, Explain the rationale behind a testcross, why is it necessary for one of the parents to be homozygous recessive for the genes of interest?
Explanation / Answer
Ans: a.
The null hypothesis is ‘The genes are not linked’
The alternative hypothesis is ‘The genes are linked’.
Ans: b.&c
In a pure dihybrid cross Homozygous (RRFF), Homozygous(rrff), expected ratio : All must be same RrFf.
Group
Observed
Expected
O-E
(O-E)2
(O-E)2 /E
red fruit, yellow flowers
333
800
-467
13689
272.61
red fruit, white flowers
64
0
64
7396
5.12
yellow fruit, yellow flowers
58
0
58
8464
4.2
yellow fruit, white flowers
345
0
345
87025
148.78
Then chi square value is 430.71
Degrees of Freedom = Number of Groups – 1 = 4 groups – 1 = 3
Now, at significance level 0.05 chi square value is very high (more than critical value), thus null hypothesis rejected. Thus the genes are linekd.
Ans: d
Map distance between them is-=((64+58)/800)*100= 15.25 cM.
2.
Sample test cross example: Where red fruit (R) is dominant over yellow fruit (r), and yellow flowers (F) are dominant over white flowers (f)
RRFf X rrff
In a test cross we try to identify the genotype of unknown samples which crossed with recessive homozygous one. Homozygous recessive is used because using only this will reveal all the allele present in unknown one will be expressing the genes in the next generation. As the known onw is recessive in both allele thus any gene present in heterozygous or homozygous condition can be traced out.
Group
Observed
Expected
O-E
(O-E)2
(O-E)2 /E
red fruit, yellow flowers
333
800
-467
13689
272.61
red fruit, white flowers
64
0
64
7396
5.12
yellow fruit, yellow flowers
58
0
58
8464
4.2
yellow fruit, white flowers
345
0
345
87025
148.78