1. AP Biology

2.  Segregation of the alleles into gametes is like a coin toss (heads or tails = equal probability)  Rule of Multiplication  Probability that independent events will occur simultaneously is the product of their individual probabilities

3.  You have 2 coins. What is the probability that you will flip two heads?  Coin #1 = 1 in 2 chance heads = ½  Coin #2 = 1 in 2 chance heads = ½  ½ x ½ = ¼  Answer = 1 in 4 (or, 25%) chance

4.  What is the probability that offspring of an F 1 generation cross will be homozygous recessive? (Pp x Pp  pp)  Mom (Pp) = 1 in 2 chance for p = ½  Dad (Pp) = 1 in 2 chance for p = ½  ½ x ½ = ¼  Answer = 1 in 4 (or, 25%) chance

5.  Rule of Addition  The probability of an event that can occur in two or more independent ways is the sum of the separate probabilities of the different ways.

6.  You have 2 coins. What is the probability that you will flip a heads and a tails? Possibility #1Possibility #2 Coin 1 = ½ headsCoin 1 = ½ tails Coin 2 = ½ tailsCoin 2 = ½ heads ½ x ½ = ¼ ANSWER: ¼ + ¼ = ½

7.  What is the probability that two heterozygous parents will produce heterozygous offspring? (Pp x Pp  Pp) Possibility #1Possibility #2 Mom = ½ PMom = ½ p Dad = ½ pDad = ½ P ½ x ½ = ¼ ANSWER: ¼ + ¼ = ½

8.  What is the probability that two parents heterozygous for both height and flower color will produce tall offspring with purple flowers?  Probability of Tt:  Probability of Pp = ½ (previous example)  Answer = ½ Tt x ½ Pp = ¼ chance of TtPp Possibility #1Possibility #2 Mom = ½ TMom = ½ t Dad = ½ tDad = ½ T ½ x ½ = ¼ ANSWER: ¼ + ¼ = ½

9.  Statistics can be used to determine if differences among groups are significant or simply the result of predictable error  Chi-square test is used to determine if differences in experimental data (what is observed) and expected results is due to chance or some other circumstance

10.  A Punnett square of the F1 cross Gg x Gg would predict the expected portion of green: albino seedlings would be 3:1  Complete the Expected (e) column and the (o-e) column PhenotypeGenotype# Observed (o) # Expected (e) (o-e) GreenGG or Gg72 AlbinoGg12 Total:84

11.  There is a small difference between the observed and expected results  Are these data close enough that the difference can be explained by random chance or variation in the sample?  To determine if the observed data fall within acceptable limits, a chi-square analysis is performed

12.  Null hypothesis  There is no statistically significant difference between the observed and expected data  Alternative hypothesis  Accepted if the chi-square test indicates that the data vary too much from the expected value (in this case, 3:1)

13.  Formula:  o = observed number of individuals  e = expected number of individuals  Σ = sum of the values (in this case, the difference between observed and expected, squared, divided by the number expected

14.  This statistical test will examine the null hypothesis, which predicts that the data from the experimental cross from the example will be expected to fit the 3:1 ratio  Complete the rest of your data table

15.  X 2 = 5.14 is now compared to the critical values table to determine if it is an acceptable value Phenotype# Observed (o) # Expected (e) (o-e)(o-e) 2 e Green72639811.29 Albino1221-9813.85 X 2 = Σ (o-e) 2 e 5.14

16.  The chi-square critical values table provides two values that you need to calculate chi-square:  Degrees of freedom. This number is one less than the total number variables (df = n -1, where n is the # of variables)  Example: In a monohybrid cross, such as our example, there are two classes of offspring (green or albino peas). Therefore, there is just one degree of freedom.  Probability. The probability value (p) is the probability that a deviation as great as or greater than each chi-square value would occur simply by chance.  Many biologists agree that deviations having a chance probability greater than 0.05 (5%) are not statistically significant. Therefore, when you calculate chi-square you should consult the table for the p value in the 0.05 row.

17. ProbabilityDegrees of Freedom (df) 12345 0.053.845.997.829.4911.1 0.016.649.2111.313.215.1 0.00110.813.816.318.520.5 If your chi-squared value is below the value for 0.05, you can accept the null hypothesis If your chi-squared value is above the value for 0.05, you should reject the null hypothesis

18.  We reject our null hypothesis… what does this mean?  Chance alone cannot explain the deviations we observed and there is, therefore, reason to doubt our original hypothesis (or to question our data collection accuracy)  Probability p = 0.05 means that only 5% of the time would you expect to see similar data if the null hypothesis was correct  You are 95% sure the data do not fit a 3:1 ratio  Consider why…  Additional experiments necessary  Sample too small? Errors in data collection?