Avariety of pure-breeding snapdragons produces a tall plant with green seeds (wwss). a second variety of pure-breeding snapdragons produce a short plant with white seeds (wwss). the two varieties are crossed, and the resulting seeds are collected. all of the seeds are green and when planted, they produce all tall plants (wwss). these tall f1 plants are allowed to self-fertilize. the results for seeds and plant height in the f2 generations are as follows. what is the х 2

The chi-square test is used to figure out if the difference between the observed numbers and the expected numbers are due to random chance or any other cause. In this example the X² = 6.13, and DF = 3.

Note: Please notice that the observed numbers used in this problem might differ from yours. Please check it out, and if they are different make your calculations using the same reasoning.

Available data:

1st cross: WWSS    x     wwss

F1) WwSs

2nd Cross: WwSs     x     WwSs

F2)

F2 Plant Phenotype                 Observed Number of individuals

Green Seed, Tall                                               565

White Seed, Tall                                                180

Green Seed, Short                                                180

White Seed, Short                                                  75                          

Total                                                                 1000

We know that when we cross two dihybrid individuals whose genes assort independently, they do not interact and express complete dominance, then the phenotypic ratio of the progeny is 9:3:3:1.

It might happen that the observed ratio is not exactly the same as the expected one. So we perform a chi-square test to see if the difference is by chance of any other reason.

So first, we need to get the expected numbers.

As we said before, we expect to see

Expected number of individuals

16 1000 individuals

9 X = 562.5 tall plants with green seeds

3 X = 187.5 tall plants with white seeds

3 X = 187.5 short plants with green seeds

1 X = 62.5 short plants with white seeds

Now, we need to get the chi-square value. We will use the following formula,

                   X² = Σ ((Observed - Expected)² / Expected)

                        Tall / Green       Tall / White     Short / Green     Short/White    

Observed             600                      175                   165                       60

Expected             562.5                  187.5                187.5                    62.5

(Obs-Exp)²/Exp     2.5                      0.83                  2.7                       0.1          

X² = Σ ((Obs-Exp)² / Exp) = 2.5 + 0.83 + 2.7 +0.1 = 6.13

X² = 6.13

Degree of freedom = n - 1 = 4 - 1 = 3  ⇒⇒  n = phenotypes

DF = 3

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Offspring resulting from F2 generation are -

WWSS-1

WWSs- 2

WwSS-2

WwSs- 4

WWss-1

Wwss-2

wwSS-1

wwSs-2

wwss-1

Explanation:

The offspring produced in the first generation after carrying a cross between WWSS and wwss are WwSs

F2 generation -

In this generation two parents having genotype WwSs are crossed. A punnet square showing a cross between them is shown below-

WS        Ws         wS         ws

WSWWSSWWSsWwSSWwSs

WsWWSsWWssWwSsWwss

wSWwSSWwSswwSSwwSs

wsWwSsWwsswwSswwss

The following genotype result from the cross

WWSS-1

WWSs- 2

WwSS-2

WwSs- 4

WWss-1

Wwss-2

wwSS-1

wwSs-2

wwss-1

the sunflower because each experiment got the same results

hope this !

-payshence

answer; i believe the correct answer is (anemia);