Problem

The following problem extends the Hardy-Weinberg …

Problem 119 Easy Difficulty

A flask of nutrient broth, buffered to maintain pH, is inoculated with a strain of E. coli. The flask is placed in a constant temperature environment where it is aerated by shaking.
A. Predict the effect of a change in energy availability over time.
B. Represent the change graphically in terms of the number of cells as a function of time.
C. In your graph as time progresses there is a change in the growth rate of the population. Add annotation to your graph to describe the time interval during which the growth rate is increasing linearly in proportion to the number of cells. Add annotation to your graph to describe another time interval during which the growth rate is decreasing in proportion to the square of the number of cells. Add a third annotation to describe an interval of time where the rate of growth is zero.
D. Select and justify two measurements of the E. coli population that could be made at two different points in time during growth that would be sufficient to answer questions about the population size at any time.
E. Describe the population of E. coli if the environment was continuously supplement by additional nutrient broth.

Answer

ANSWER NOT AVAILABLE

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Biology

Biology for AP Courses

Chapter 36

Population and Community Ecology

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Video Transcript

In the problem we have part a so dash t is given us 200 over t plus 1 to the power r. So this is 0 to 4200. Upon t plus 1 to the power r d t is equals to 200 point integration, 0 to 4 t plus 1 to the power minus 2 d. This is equal to 200 or it is minus 200 into 1 upon t plus 10 to 4. This equals to 1 ud 60 point. So this is the answer for part a of the problem. Now we have part b where we have integration 0 to 6200 over t plus 1 to the power 3 d t and now here this is 2 and in part b we have rituals 3. So this is 200 point: integration, 0 to 6 t plus 1 to the power minus 3 d. This is equal to 200 into t plus 1 to the power minus 2 upon minus 20 and 6 is equals to minus 100 into 1, upo t plus 1 whole square 0 and 6. This is equal to minus 100 into 1 upon fourty, 9 minus 1. So after calculating we have this equals 4 in double 0 over fourty 9, and that responds to 98 point in part c of the problem we have delta p will decrease, which the pathetics as r increases delta p will decrease. Now we have part d where this is through 50, that is, 200 integration, 0 to 10 t plus 1 to the power minus r t t here. Integration result is 315 and want to find the value of r. So this is the 50 upon 200 point. That is integration 0 to 10 t plus 1 to the power minus r d. T already is 7 upon 4 equals 1 upon r plus 111 to the power minus r, plus 1 minus 1 to the power minus r plus 1 point now for the resolving it and simplifying we have the value of r equals 1.3. So this is the answer for part d of the problem. Now we have the last part, which is part e here it is 200 integration, 0 to infinity t plus 1 to the power minus 3 d. The value of i is 3 here am i come to find the integration from 0 to infinity. Therefore, it is 200 into t plus 1 to the power minus 2 over minus 20 to infinity, so it is 100 into 1 upon t plus 1 whole square 0 to infinity, so it is minus 100 into 1 upon infinity minus 1 upon 1. So this is 0 that is equal to 100 point. So this is the answer to the given father.

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