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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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AfilCa [17]

Options are not provided inthe question. Th ecomplete question is as following:

In the classic Meselson and Stahl experiment, Coli are first grown in 15N-enriched media (0th generation) and subsequently switched to 14N media. A DNA sample isolated during the first generation of E. Coli grown in 14N media is subjected to a centrifugation step in a dense CsCl gradient. Assuming that replication occurs in a semiconservative fashion, the DNA sample will _______.

1) run as two bands (of equal intensity), one at the same position as a uniformly 15N-labeled DNA, and one at the same position as a uniformly 14N-labeled DNA.

2) run as one band, at the same position as a uniformly 15N-labeled DNA.

3) run as one band, at the same position as a uniformly 14N-labeled DNA. run as one band, at a position expected for a DNA containing a 14N:15N ratio equal to 7.

4) run as one band, at a position expected for a DNA containing a 14N:15N ratio equal to 1.

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1)

Explanation:

In 1957, Meselson and Stahl gave an experimental evidence for semi-conservative replication  of DNA, means that each DNA strand act as template for synthesis of new DNA.

For the experiment, they use  E. coli grown in heavy isotope of 15N nitrogen and switch to 14N nitrogen undergoing one, two, or three generations and the DNA samples are collected. The DNA samples are then mix with cesium chloride and separated as light and heavy DNA.

As the replication occurs in a semiconservative fashion, the double DNA sample will run as two bands with intermediate density, having mixtures of 15N and 14N DNA, either of which would have appeared as DNA.

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