Probability Calculations for GFP Molecule Segregation in Cell

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Jun 11, 2024

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Name: ______________________________ Student’s perm #: ________________________ Student’s signature: ________________________ Grade =______________ /80 pts. MCDB 108C Spring 2024 MIDTERM EXAM #2 DO NOT OPEN THE EXAM UNTIL YOU ARE INSTRUCTED TO DO SO This exam should have 11 pages. Two blank pages are collated to the exam for you to draft your answers. Please put your name on every page You are allowed to use two pages of your own hand-written notes . While you may not use a computer or smart phone during the exam, you’re welcome to use a basic scientific calculator. In your answers, you should show how you arrived at your final conclusion(s) . Each point of this exam represents 1 point of the final grade of the class (total: 500 points). 1/11

Name: ______________________________ Part A: The following set of questions focuses on a problem introduced during the sections of Week 4. A “mother” cell contains 5 green fluorescence proteins (GFP). When the mother cell divides, the GFP molecules segregate at random and with equal probability between the two daughter cells, ⍺ and β . Note: In your computation of probabilities, you can leave numerical fractions without calculating their exact values. Question A1 [8 points]: What is the probability that all GFP molecules end in daughter cell ? ⍺ Briefly justify the probabilistic model you’re using, including its assumptions . 2/11

Name: ______________________________ Assumptions: the molecules segregate between cells ⍺ and β as a Bernoulli trial. This model is justified because (1) each molecule ends in either cell ⍺ or β (2 possible outcomes) and (2) the probability of each outcome is fixed and equal to 0.5. Finally, the application of the Bernoulli trial implies that each molecule segregates independently of each other. The outcome of a collection of ensemble of Bernoulli trials can be described by the Binomial distribution. If k denotes the number of GFP molecules in cell , ⍺ the probability that all 5 molecules are found in cell ⍺ is: P(k=5;n=5;p=0.5)=(1/2)^5=1/32=0.0312 Rubric: +4 pts for listing the 3 assumptions of a Bernoulli trial and the use of a Binomial distribution. In principle the students should explicitly mention the Bernoulli trial in their justification, but this might not be necessary if they justify the use of the Binomial distribution. +4 pts for computing the probability. Note that this value can be obtained without using the Binomial distribution. Question A2 [10 points]: What is the probability of observing 3 or 4 GFP molecules in cell ⍺ ? Let k be the stochastic variable representing the number of GFP molecules in cell . ⍺ The probability of observing 3 or 4 GFP molecules in cell ⍺ is computed by applying the addition rule of probabilities corresponding to mutually exclusive events. Thus the probability is: P(k=3 or k=4;n=5;p=0.5) = P(k=3;n=5;p=0.5) + P(k=4;n=5;p=0.5) = 5!/(3! * 2!) (1/2)^5 + 5!/(4!) (1/2)^5 = 10 (1/2)^5 + 5 (1/2)^2 = 0.4687 Rubric: +2 for associating the number of GFP molecules in cell alpha with a random variable +4 pts for applying and justifying the use of the additive rule for mutually exclusive events +4 pts for using the correct formula of the Binomial distribution for k=3 and k=4 Question A3 [8 points]: Consider the segregation of many GFP molecules between cells ⍺ and β . This time, the segregation is asymmetrical. The probability that a GFP molecule gets transmitted to cell β is very low: 10 -3 . The total number of GFP molecules is large: 2,000. What is 3/11

Name: ______________________________ the probability of observing at least 1 GFP molecule in the daughter cell β? Please briefly justify the probabilistic model you are using to answer this question. As discussed above the segregation of each GFP molecule can be viewed as a Bernoulli trial. In this problem, a successful outcome is the transmission of a given molecule to cell β. Since the number of Bernoulli trials is very large (2,000) and the probability that observing a successful outcome is very small (0.001), the Binomial distribution can be approximated by the Poisson distribution. Let k be a stochastic variable describing the final number of molecules in cell β. The expected (or average) number of GFP molecules in cell β is given by μ = n* p = 2. P(k≥1; μ=2) = 1-P(k=0; μ=2) = 1 – (2^0/0!) *e^(-2) = 1 – 1/e^2 = 0.8647 Rubric: +3 pts for justification of the application of the Poisson dist +2 pts for using the complementary probability (1-P) +3 pts for making proper use of the formula of the Poisson dist -3 pts if a student failed to use the (1-P) to calculate the final probability Part B: Imagine a small pond from which mosquitoes larvae hatch. During the middle of the night, 100 mosquitoes are released at the same time from the pond. Humans sleep in a house located near the pond. To estimate the likelihood that humans will get bitten by the mosquitoes, you are asked to model the dispersion of the mosquitoes as 1D random walk. Fact about the mosquitoes’ dispersion: Assume that all 100 mosquitoes are released from the same site x = 0 m at the same time. Positions to the right side of the pond are positive. During intervals of 1 hour , the mosquitoes move with equal probability either to the left or to the right by steps of 10 meters. Question B1 [8 points]: What is the fraction of mosquitoes found at position x = 0 m after 4 hours (equivalent to 4 time steps). Hint: Use the “law of motion.” 4/11

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