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3 years ago

Help please, 30 points.

Biology

1 answer:

Kisachek [45]3 years ago

4 0

Answer:

Option 1 i think it's A

Explanation:

hope it's helpful

Answer of 2nd question

option A 

glucose is a unit of energy

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22. A plant geneticist is investigating the inheritance of genes for bitter taste (Su) and explosive rind (e) in watermelon (Cit

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Answer:

See the answer below

Explanation:

Using the formula for calculating Chi square ( $X^2$ ):

$X^2$ = $\frac{(O - E)^2}{E}$ where O = observed frequency and E = expected frequency.

The observed frequencies for the four phenotypes are 88, 62, 62, and 81 respectively.

For the expected frequency, the phenotypes are supposed to assort in 9:3:3:1 according to Mendel ratio.

Hence, expected frequencies are calculated as:

phenotype (1) = 9/16 x 293 = 164.81

phenotype (2) = 3/16 x 293 = 54.94

phenotype (3) = 3/16 x 293 = 54.94

phenotype (4) = 1/16 x 293 = 18.31

The $X^2$ is calculated thus:

Phenotype O E $X^2$

1 88 164.81 $\frac{(88 - 164.81)^2}{164.81}$ = 35.80

2 62 54.94 $\frac{(62 - 54.94)^2}{54.94}$ = 0.91

3 62 54.94 $\frac{(62 - 54.94)^2}{54.94}$ = 0.91

4 81 18.31 $\frac{(81 - 18.31)^2}{18.31}$ = 214.64

Total $X^2$ = 35.80 + 0.91 + 0.91 + 214.64 = 252.26

To the nearest tenth = 252

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

$X^2$ tabulated = 7.815

The calculated  $X^2$  exceeds the critical value, hence, the null hypothesis is rejected. The two genes did not assort independently. 

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