In the flowering plant, white flowers (B) are dominant over red flowers (b), and short plants (E) are dominant over tall flower (e). An F2 generation was created by crossing two F1 individuals (each BbEe). The data below are the counted number of flowers from each phenotype in the F2 generation.

White short= 206
Red, short= 83
White, tall= 85
Red, tall= 30

Required:
a. What is your null hypothesis?
b. What is your expected phenotypic ratio based on Mendelian inheritance?

(c)Calculate the expected number of flowers you should have gotten based on the Mendelian inheritance. Then calculate a chi-square value, degrees of freedom, and a p-value.

Chi-square statistic: _____
Degrees of freedom (# phenotypes -1):
P-value:

(d)Interpret your results. Do you reject it or fail to reject your null hypothesis (please restate the null)?

Answer:

(a)[tex]H_0:$The given data fit the predicted phenotype[/tex]

(b)9:3:3:1

(c)

Chi-square statistic: 3.8914
Degrees of freedom (# phenotypes -1) =3
P-value: 0.2734

(d)We fail to reject the null hypothesis.

Step-by-step explanation:

In the flowering plant, white flowers (B) are dominant over red flowers (b), and short plants (E) are dominant over tall flowers (e). An F2 generation was created by crossing two F1 individuals (each BbEe).

(a)The null hypothesis is:

[tex]H_0:$The given data fit the predicted phenotype[/tex]

(b)The gametes are BE, Be, bE and be.

The offsprings are presented in the table below:

[tex]\left|\begin{array}{c|cccc}&BE&Be&bE&be\\--&--&--&--&--\\BE&BE&BE&BE&BE\\Be&BE&Be&BE&Be\\bE&BE&BE&bE&bE\\be&BE&Be&bE&be\end{array}\right|[/tex]

The expected phenotypic ratio based on Mendelian inheritance

BE:Be:bE:be=9:3:3:1

(c)

[tex]\left|\begin{array}{c|c|c|c|c|c}$Phenotype&Observed&$Expected&O-E&(O-E)^2&\dfrac{(O-E)^2}{E} \\-----&--&--&--&--&--\\$White short(BE)&206&\frac{9}{16}*404 \approx 227 &-21&441&1.9427\\$Red, short(bE)&83&\frac{3}{16}*404 \approx 78 &5&25&0.3205\\$White, tall(Be)&85&\frac{3}{16}*404 \approx 78 &7&49&0.6282\\$Red, tall(be)&30&\frac{1}{16}*404 \approx 25 &5&25&1\\-----&--&--&--&--&--\\$Total&404&--&--&--&3.8914\end{array}\right|[/tex]

Therefore:

Chi-square statistic: 3.8914
Degrees of freedom (# phenotypes -1): 4-1 =3
P-value: 0.2734

(d) Our null hypothesis is:

[tex]H_0:$The given data fit the predicted phenotype[/tex]

Since p>0.05, the given data fit the predicted phenotypic ratio.

We, therefore, fail to reject the null hypothesis.

The difference in the observed and expected are sosmall that they can be attributed to random chance.