Mathematics, 16.06.2020 22:57 wolfsaway

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?

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Answer from: anitadefrances

Question Completion

(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)?

(a) H_0:$The given data fit the predicted phenotype

(b)9:3:3:1

(c)

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

H_0:$The given data fit the predicted phenotype

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

The offsprings are presented in the table below:

$\left|\begin{array}{c|cccc}&BE&Be&bE&be\\--&--&--&--&--\\BE&BE&BE&BE&BE\\Be&BE&Be&BE&Be\\bE&BE&BE&bE&bE\\be&BE&Be&bE&be\end{array}\right|$

The expected phenotypic ratio based on Mendelian inheritance

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

(c)

$\left|\begin{array}{c|c|c|c|c|c}$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|$

Therefore:

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

(d) Our null hypothesis is:

H_0:$The given data fit the predicted phenotype

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.

Answer from: Quest

solution:

the given function is ,

$f(x) = 4 x^3 - 72 x^2 + 320 x,$

= $4x( x^2- 18 x + 80)\\\\ 4 x(x^2-10 x -8 x + 80) \\\\ 4 x[(x-10)(x-8)]$

the three roots of this quadratic function is , 0 ,8 and 10.all the roots are real and rational.

if a quadratic equation is of the form ,

$f(x) = a x^3+ b x^2 + c x+d,$

sum of the roots = a + b +c= 0+8+10=18= $\frac{-b}{a}$

product of roots taken two at a time = a b + b c+ c a = 0+80+0=80 = $\frac{c}{a}$

product of roots = a b c = 0= $\frac{d}{a}$

there are 11 doors in a hall starting from 0 to 11.the entry in the hall is permitted through gate number 0 and exit is allowed through only gate number 8 and 10.so, three roots(0,8,10) of quadratic function $f(x) = 4 x^3 - 72 x^2 + 320 x,$ is given by entry and exit route of this hall.

Answer from: Quest

9.8? i'm not really sure

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