Suppose two heterozygous genes (Mm) are being crossed. The possible outcomes are

Question

Suppose two heterozygous genes (Mm) are being crossed. The possible outcomes are: MM, Mm, Mm, mm If you wanted to run a simulation to look at these outcomes, you could generate four random numbers (1 – 4). You could arbitrarily assign a number to each outcome, so that a 1 represents MM, a 2 or 3 represents Mm, and a 4 represents mm. (a) Suppose you want to simulate something with a 20% (1 in 5) chance of success. What numbers would you generate (from ___ to ___)? What number(s) would indicate “success?” (b) Suppose you want to simulate something with a 30% (3 in 10) chance of success. What numbers would you generate? What number(s) would indicate “success?” (c) Suppose you want to simulate something with a 2 in 13 chance of success. What numbers would you generate? What number(s) would indicate “success?”

Explanation / Answer

(a) To simulate something wth a 20% chance of success, I will generate 2-digit random numbers in the range 00 to 99. Thenumbers divisible by 5, namely 00, 05, 10,...,95 will be considered a success Since there are 20 such numbers divisible by 5 and the total population of 2-digit random numbers (starting from 00 to 99), is 100, the chance of success is 20/100 or 20%.

(b) To simulate something wth a 30% chance of success, I will generate 2-digit random numbers in the range 00 to 99. Thenumbers <= 29, namely 00, 01, 02, ..., 29 will be considered a success Since there are 30 such numbers <=29 and the total population of 2-digit random numbers (starting from 00 to 99), is 100, the chance of success is 30/100 or 30%.

(c) To simulate something wth a 2 in 13 chance of success, I will generate 3-digit random numbers in the range 000 to 999 and ignore any number > 129. Thenumbers <= 19, namely 00, 01, 02, ..., 19 will be considered a success. Since there are 20 such numbers <=19 and the total population of 3-digit random numbers <= 129 (starting from 000 to 129), is 130, the chance of success is 20/130 or 2 in 13..

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