Probabilites and Binomial Expansion 3-2
Introduction
Overview of methods to predict outcomes of genetic crosses in biology.
Focus will be on the Punnett square and the rules of probability.
Punnett Square
Definition: A diagram used to predict the genotype and phenotype combinations of a genetic cross.
Named after Reginald Punnett, an English geneticist.
Steps in Creating a Punnett Square
Determine the Gametes:
Identify the alleles for each parent.
Example traits: tall (T) and short (t) plants.
Tall plant is heterozygous (Tt) for height.
Short plant is homozygous recessive (tt).
Draw the Punnett square:
Create a grid to show combinations of alleles.
Example:
For the short plant: both gametes will be (t, t).
For the tall plant: gametes can be (T, t).
Fill the grid:
Cross gametes: T with t and t with t.
Resulting genotypes: TT, Tt, Tt, tt.
Determine Genotype and Phenotype Ratios:
Genotype ratio from example: 1 TT : 2 Tt : 1 tt = 1:1 for Tt (heterozygous) and tt (homozygous recessive).
Phenotype ratio:
Two tall plants (TT and Tt), two short plants (tt).
Resulting ratio: 2 tall : 2 short = 1:1.
Example Problem with Cats
Given that black fur (n) is recessive to brown fur (N).
Problem:
Cross a black cat (nn) with a heterozygous brown cat (Nn).
Use a Punnett square:
Brown cat gametes: (N, n).
Black cat produces only n gametes.
Resulting combinations lead to:
Genotypes: 50% brown (Nn) and 50% black (nn).
Probability of a kitten being black = 50%.
Rules of Probability
Definition: The likelihood of a specific event occurring.
Calculating probability: [ P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} ]
Examples in Probability
Probability of an Ace in a Deck of Cards:
Favorable outcomes (Aces): 4.
Total outcomes (cards): 52.
Probability: ( P(Ace) = \frac{4}{52} = \frac{1}{13} )
Ace of Spades Probability:
Two conditions: must be a spade and an ace.
Probability of drawing an Ace: ( \frac{4}{52} ) = 0.077.
Probability of drawing a spade: ( \frac{13}{52} = \frac{1}{4} ).
Combined Probability for an Ace of Spades:
[ P(Ace\, of\, Spades) = P(spade) \times P(ace) = \frac{1}{4} \times \frac{1}{13} = \frac{1}{52} ]
Rolling Dice:
Probability of rolling a four on two rolls (using “and”): 1/6 for each roll.
Overall probability: ( P(four) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36} ).
Probability Rules
Multiplicative Rule
Used for independent events occurring together (conditioned by AND).
Probability that two independent events occur together: [ P(A\, and\, B) = P(A) \times P(B) ]
Addition Rule
Used to find the probability of one or the other occurring (conditioned by OR).
Example: Probability of getting a three or a four on a die:
Probability of three: ( \frac{1}{6} )
Probability of four: ( \frac{1}{6} )
Combined Probability: [ P(three\, or\, four) = P(three) + P(four) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3} ]
Complex Probability Scenarios
Events with Multiple Outcomes
Probability calculations for scenarios with specific outcomes:
E.g., in a family with two parents, what’s the probability of three children having albinism?
Use Punnett square for complexity when two alleles are involved (recessive and dominant).
Example of Three Children with Albinism
If both parents are heterozygous (Aa), then using Punnett square:
P(albinism) = 1/4 for each child.
Probability of three children with albinism: [ P(all\, three\, albino) = \left( \frac{1}{4} \right)^3 = \frac{1}{64} ]
Multiple Children and Outcomes
Searching for combinations of affected and unaffected scenarios can be complex:
Probability of 3 children with albinism and 1 not: multiple combinations (3 choose 1).
Use expansions or factorial methods (combinations to determine scenarios).