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love history [14]
3 years ago
15

A research scientist studied the effects of various treatments on the growth of poplar trees. He decided to treat a large sample

of saplings with a fertilizer, continual irrigation, irrigation and fertilizer and a control group (non-continual irrigation and no fertilization). He placed one quarter of all saplings into each of the groups and then measured their mass differential after 3 weeks. Did the treatment have an effect on the growth of the trees
Biology
1 answer:
STALIN [3.7K]3 years ago
4 0

Answer:

hello your question is incomplete attached below is the screenshot of the missing part

answer : The treatment had an effect on the growth of the trees

Explanation:

H0 : No significant effect

Ha : At least one treatment has a significant effect

To confirm the above hypothesis ( conduct one way ANOVA )

From the one way Anova

( X represents each group; from control to Fertilizer and irrigation )

∑ X1^2  = 0.2338 ,  ∑ X2^2 = 1.8284,   ∑X3^2 = 0.1982, ∑X4^2 = 11.8492

Also : ∑ N = 5 + 5 + 5 + 5 = 20  where N = number of  trees in each group

also K represents number of groups i.e. = 4

Next calculate the sum of squares ( calculated using online tools )

i) Between sum of squares:  value = 4.6824

ii) Within sum of squares : value = 4.3572

Next step : Calculate the degrees ( DF)

i) Between degrees of freedom

Dfb = k - 1  where K = 4   therefore Dfb = 3

ii) within degrees of freedom

Dfw = N - k , where N = 20  and K = 4 therefore Dfw = 16  

Next : calculate mean sum of squares

i) Between mean sum of squares

= 4.6824 / 3 = 1.5608

ii) within mean sum of squares  

= 4.3572 / 16 = 0.2723

Calculate the F-statistic value

F_{calculated } = Between mean sum of squares  / within mean sum of squares

= 1.5608 / 0.2723 = 5.73

determine the F critical value

Fcritical = F_{3,16,0.05} = 3.2389  ( obtained from F table )

since F_{calculated } > F_{critical}   hence we can say The treatment had an effect on the growth of the trees i.e. Reject H0

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2 years ago
If the frequency of the recessive allele for a gene is 0.3, calculate the expected frequency of heterozygotes in the next genera
Kisachek [45]
Q = recessive allele frequency = 0.3, and thus in H-W equilibrium there are ONLY two alleles, q (recessive) and
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So p, the dominant allele frequency, must be equal to 1 - q --> p = 1 - q
p = 1 - 0.3 = 0.7.
Since heterozygotes are a combination of the p and q, we must again look at the frequencies of each genotype: p + q = 1, then (p+q)^2 = 1^2
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Therefore if the population is in H-W equilibrium, then the expected frequency of heterozygous individuals = 2pq = 2(0.7)(0.3)
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What are the enzymes called when they are at low temperature?
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Answer:

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What is a process by which organisms with characteristic suited to the environment will survive and reproduce
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According to Charles Darwin's theory of Evolution by natural selection, organisms that possess heritable traits that enable them to better adapt to their environment compared with other members of their species will be more likely to survive, reproduce, and pass more of their genes on to the next generation.

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One word answer:Adaptation:

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2 years ago
Calculate the final concentration of BSA in the problems below using the formula
Rudiy27

Answer:

A. C_2=1.5\frac{mg}{mL}

B. C_2=0.075\frac{mg}{mL}

C. C_2=0.01\frac{mg}{mL}

D. C_2=0.001\frac{mg}{mL}

Explanation:

Hello.

In this case, we must compute the final concentration in all the cases so we solve for it in the given equation:

C_2=\frac{C_1V_1}{V_2}

Thus, we proceed as follows:

A. Here, the final diluted solution includes the 300 μL of the 5 mg/ml-BSA and the 700 μL of TBS.

C_2=\frac{300\mu L*5\frac{mg}{mL} }{(300+700)\mu L}\\\\C_2=1.5\frac{mg}{mL}

B. Here, the final diluted solution includes the 50 μL of the 1.5 mg/ml-BSA, the 450 μL of water and the 500 μL of TBS.

C_2=\frac{50\mu L*1.5\frac{mg}{mL} }{(50+450+500)\mu L}\\\\C_2=0.075\frac{mg}{mL}

C. Here, the final diluted solution includes the 10 μL of the 1 mg/ml-BSA and the 990 μL of TBS.

C_2=\frac{10\mu L*1\frac{mg}{mL} }{(10+990)\mu L}\\\\C_2=0.01\frac{mg}{mL}

D. Here, the final diluted solution includes the 10 μL of the 0.1 mg/ml-BSA and the 990 μL of TBS.

C_2=\frac{10\mu L*0.1\frac{mg}{mL} }{(10+990)\mu L}\\\\C_2=0.001\frac{mg}{mL}

Best regards.

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