Question

Fluorescent proteins played an important role in studies of protein expression copy number regulation in cells. For this problem, consider a fluorescent protein that can be in either an 'on' or 'off' state, with equal probability. In real fluorescent proteins, the 'on' state refers to when they are actually fluorescent and the 'off state is dark. For this problem they are just two states, like heads or tails on a coin toss.

(a) If a cell has 10 copies of the fluorescent protein, what is the probability that exactly 5 of them are in the 'on' state?
(b) For the same cell, what is the probability that 7 or more of the proteins are in the 'on' state?
(c) Make a plot of the probability distribution for the fraction of proteins in the 'on' state for a cell with 10 proteins and compare it to the same plot for a cell with 100 proteins. Since we are asking about the fraction of proteins in the 'on' state, your x-axis on this plot should run from 0 to 1 for both plots. You don't have to calculate every point for these plots, just check a few points to make the sketch and explain.

Answer 1

The question is about calculating the probabilities of fluorescent proteins being in the 'on' state in a cell. For 10 proteins, the probability that exactly 5 are 'on' is approximately 0.246. The probability that 7 or more are 'on' is approximately 0.172, and the distributions for 10 and 100 proteins show a bell-shaped curve centered at 0.5.

Step-by-step explanation:

The question deals with the calculation of probabilities regarding the state of fluorescent proteins, which can be either in an 'on' or 'off' state with equal probability.

(a) To calculate the probability that exactly 5 out of 10 fluorescent proteins are in the 'on' state, we use the binomial probability formula P(X=k) = C(n, k) * p^k * (1-p)^(n-k), where C(n, k) is the combination of n items taken k at a time, p is the probability of a single event occurring, and k refers to the number of events.

In this case:
C(10, 5) = 10! / (5! * 5!) = 252,
p = 0.5 (since the protein has an equal chance to be 'on' or 'off'),
and k = 5.

The probability of exactly 5 'on' states is thus 252 * (0.5)^5 * (0.5)^(10-5) = 252 * (0.5)^10 ≈ 0.246.

(b) The probability that 7 or more proteins are in the 'on' state requires us to calculate the probabilities for 7, 8, 9, and 10 'on' states and sum them up. This yields a probability of approximately 0.172.

(c) For a cell with 10 proteins, the probability distribution of the 'on' state is bell-shaped, with the peak at 0.5 'on' state. As the number of proteins increases to 100, the peak becomes narrower, but still centered around 0.5, indicating a more precise average as the sample size increases.