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geneticspg5

# geneticspg5 - Observation 1 PROBABILITY Understanding...

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Unformatted text preview: Observation 1: PROBABILITY Understanding genetics relies on being able to calculate probabilities of certain events occurring. To calculate the probability of an event occurring, we need to know how many different options we have. In a coin toss, there are two options; heads or tails. The chance of heads is 1 out of 2 different options or: 1/2 (as a decimal = .5 or as a percent = 50%). The chance of getting tails is the same (1 chance out of two different options). Example: What is the probability of drawing the Queen of Hearts from a deck of cards? The probability of all of your different options must add up to one. The probability of getting heads and of getting tails is 1/2 + 1/2 = l. The probability of getting any card in the deck (1/52), since there are 52 different possibilities they all add up to 1. 7. What is the probability of ﬂipping a coin and getting heads? l /2 8. What is the probability of rolling a ”3 ” twice on a six—sided die (pl dice. 7 ”Le x i Z l— (9 (0 3b If you need to calculate the probability of two independent events occurring, you compute the probability for each independent event, then multiply them together For example the chance of getting heads twice is: 1/2 x 1/2— — 1/4 9. What IS the probability of picking an Ace of Hearts from the deck, HL y ‘i w _L putting it back, and then a Jack of Diamonds? 15'; 62‘ ' (32> 2 10 What is the probability of ﬂipping a coin 3 times and getting heads each time? J’Xer‘t‘li “"54 GAMETES AND PROBABILITY . '2. Z #— 8 11. What type of gametes will be produced ifan individual has the genotype Aa? A (if CL 12. What is the probability of getting an ”A ” sperm from a parent with the genotype Aa? I Z 13. What type of gametes will be produced if an individual has the genotype AaBb? Aabb? Garnete 1 Gamete 2 Garnete 3 Garnete 4 .690 14. What is the probability of getting an: a. "AB “ sperm from a parent with the genotype AaBb? llq _ b. “Ab ” sperm from a parent with the genotype AaBb? ll “i a "Ab ” sperm ﬁom a parent with the genotype Aabb? ll; 15. Under conditions of normal meiosis, it is not possible to get a ":10 sperm Explain sect: .Xheiiliﬁxe / kt; to :3 s, .' .-__y_ ,4 .. n A -:’. =5 16. Use probability to determine the number of oﬁfspring that will haw the .4033 get-1013p? ﬁom parental cross of A (11% x AaBb. “l, {7:24 ‘1 10.09 26 ...
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