Question

An F1 tall, round seed pea plant is crossed with a dwarf, wrinkled seed pea plant. The following offspring are observed: Tall, round: 510 Tall, wrinkled: 495 Dwarf, round: 490 Dwarf, wrinkled: 505 a. What is the expected phenotype ratio? WW or ###.#, depending on what you think the answer is. b. What is the calculated chi-square value? Format your answer with 1 significant digit (...). Sure, you can try to guess it 10 times, but work it out so you make sure you know how to do this for the test. The chi-squared equation won't be provided on the test, so make sure you put it on the permitted 3x5 note card c. What is the critical chi-square value from the table provided in the book at the P=0.05 level of significance? Enter the number exactly as shown in this table d. Would you "reject" or "fail to reject the null hypothesis in favor of the alternative hypothesis? Just enter the phrase as shown above, but without the quotation marks.

Answer 1

Tall, round x   dwarf , wrinked cross resulted in
Tall, round 510
Tall, wrinkled 495
Dwarf round 490
Dwarf, wrinkled 505
Total offsprings = 2000

a. The expected the phenotype ratio is 1:1:1:1 Because it looks like mendelian dihybrid dominant cross. Each allele has equal chance to segregate.

b. When we say ratio 1:1:1:1,. we mean that out of total 2000, each type would occur at 500 numbers. So expected number is 500. observed is what is given in the question. So lets us do chi square test.
Chi square = (observed - expected)^2 / expected
= (510-500)^2/500 + (495-500)^2/500 + (490-500)^2/500 + (505-500)^2/500
= 0.2+0.05+0.2+0.05
chi square value= 0.5

[ Note: Null hypothesis: That there is no significant difference between the observe and expected values.
Alternate hypothesis: There is difference between then observed and expected values.
The degree of freedom : no. of characters in the study = height and shape . so n = 2. Df = n-1 = 2-1 = 1.
The value in chi square table at df = 1 ; alpha = 0.05 is 3.841. The value we got 0.5 is less than 3.841. Hence we say that there is no significant difference in the observed and expected values. So we do not reject or fail to reject null hypothesis. ]
From these arguments following questions are answered.

C. critical value = 3.841

D. Fail to reject.