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Vika [28.1K]
3 years ago
11

Nucleotides can be radiolabeled before they are incorporated into newly synthesized DNA and, therefore, can be assayed to track

their incorporation. In a set of experiments, a student-faculty research team used labeled T nucleotides and introduced these into a culture of dividing human cells at specific times. Use the following information to answer the question below. The research team used their experiments to study the incorporation of labeled nucleotides into a culture of lymphocytes and found that the lymphocytes incorporated the labeled nucleotide at a significantly higher level after a pathogen was introduced into the culture. They concluded that __________.
a. their tissue culture methods needed to be relearned
b. infection causes lymphocyte cultures to skip some parts of the cell cycle
c. infection causes cell cultures in general to reproduce more rapidly
d. the presence of the pathogen made the experiments too contaminated to trust the results
e. infection causes lymphocytes to divide more rapidly
Biology
1 answer:
adoni [48]3 years ago
7 0

Answer:

e. infection causes lymphocytes to divide more rapidly

Explanation:

The cell cycle includes interphase and M phase which in turn together produce daughter cells from the existing parent cells. DNA replication occurs during the S phase of interphase to ensure that the daughter cells obtain the identical DNA present in the parent cell.

Lymphocytes are one of the types of white blood cells and are involved in cell-mediated and antibody-mediated immune responses. These cells are present in the blood as well as in lymphatic tissues. During the given experiment, lymphocytes exhibited a higher rate of incorporation of labeled nucleotide after the introduction of a pathogen in the culture. This suggests that the introduction of pathogen triggered the cell division in lymphocytes to produce more lymphocytes to fight the infection.

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2.86 A ball is dropped from rest from the top of a cliff that is 24 m high. From ground level, a second ball is thrown straight
Rashid [163]

Answer:

Distance Below the top = 6,02 m

Explanation:

To get the Final velocity of the first ball (that will be the intial velocity of the second) you need to solve the kinematic equation for Velocity:

(1) V_{1f} =V_{1O}  - gt

As the ball is dropped from rest V_{1O} = 0, so:

(2) V_{1f} = - gt

Note that the velocity is going to be negative as the ball is going down. To get the time it would take the ball to reach de base you can use the kinematic equation for position:

(3) h_{1} = h_{1O} + V_{1O} *t - \frac{1}{2} gt^{2}

We need the answer when h=0, and from the initial conditions V_{1O} = 0, h_{1O} = 24m, you get:

(4) 24m = \frac{1}{2} gt_{1f}^{2}

Solving for t, with 9,8\frac{m}{s^{2}} and :

(5) t_{1f} = \sqrt{\frac{24m*2}{g} } = 2,21 s

Replacing this time in (2), the final velocity is:

(6) V_{1f} = -21,66  \frac{m}{s}

So the initial velocity of ball 2 is equial to this but oppossite in direction so: V_{2O}= -V_{1f} = 21,66  \frac{m}{s}

The general position equation for ball 2 is (considering h_{2O} = 0 :

(7) h_{2} = V_{2O} *t - \frac{1}{2} gt^{2}

They cross paths when h_{1}=h_{2} so:

(8) h_{1O} - \frac{1}{2} gt^{2} = V_{2O} *t - \frac{1}{2} gt^{2}

Rearranging:

(9) t_{cross} =\frac{h_{1O}}{V_{2O} }

Replacing values:

t_{cross} = 1,108 s

To get the absolute position replace t_{cross} on equation (3) or (7):

h_{cross} = 17,98 m

To get it below the top of te cliff:

h_{cross, from above} = 24 m - 17,98 m

h_{cross, from above} = 6,02 m

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