1 answer

Use the Chi^2 test to determine if l & r assort independently. Show your work. Use...

Question:

Number A wild-type trihybrid soybean plant Phenotype is crossed to a pure-breeding Pale soybean plant with the recessive Pale

Use the Chi^2 test to determine if l & r assort independently. Show your work.

Use the Chi^2 test to determine if l & t assort independently. Show your work.

Use the Chi^2 test to determine if r & t assort independently. Show your work

Number A wild-type trihybrid soybean plant Phenotype is crossed to a pure-breeding Pale soybean plant with the recessive Pale, oval phenotypes pale leaf (!), oval seed Pale, short (r), and short height (t). The results Pale, oval, short of the three-point test cross are Oval shown in the table. Traits not listed Oval, short are wild-type. Wild type 102 618 Short 1630

Answers

Cross:

Wild type (trihybrid) x Pale, Oval, Short
Ll Rr Tt x ll rr tt

Progeny from cross Ll Rr Tt x ll rr tt
Progeny Phenotype Genotype
(Gamete from Tryhybrid in bold)
No. of Progeny
Pale l R T/l r t 648
Pale, Oval l r T/l r t

64

Pale, Short l R t/l r t 10
Pale, Oval, Short l r t/l r t 102
Oval L r T/l r t 6
Oval, Short L r t/l r t 618
Short L R t/l r t 84
Wild Type L R T/l r t 98
1630

i) Use the Chi^2 test to determine if l & r assort independently.

If L and R assort independently, the Trihybrid parent would produce the LR, Lr, lR and lr gametes in equal ratios. Therefore, the Progeny would also have these phenotypic classes in equal ratios. Therefore, the expected number of progeny for each class is 407.5.

Progeny Class Observed (O) Expected (E) O - E (O - E)2 (O - E)2/E
LR 84 + 98 = 182 407.5 -225.5 50850.25 124.786
Lr 6 + 618 = 624 407.5 216.5 46872.25 115.024
lR 648 + 10 = 658 407.5 250.5 62750.25 153.988
lr 64 + 102 = 166 407.5 -241.5 58322.25 143.122
\chi ^{2} = 536.92

L and R do not assort independently, as the probability of observing these deviations due to chance alone is less than 0.005. For d.f = 3; Critical Value of Chi-square at the 95% confidence level is 7.815.

As 536.92 > 7.815, we can reject the null hypothesis that these loci are assorting independently.

ii) Use the Chi^2 test to determine if l & t assort independently.

Similarly to i), since the Chi-square value is greater than the critical value at the 95% confidence, we reject the hypothesis that the L and T alleles assort independently.

iii) Use the Chi^2 test to determine if r & t assort independently.

These two loci are also not assorting independently. This is because the Chi-square value for the devation from expected, is much greater than the significaant value at 95% confidence.

.

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