Download presentation

Presentation is loading. Please wait.

Published byWilliam Emmerson Modified over 2 years ago

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?

Similar presentations

OK

Chi-Square Test A fundamental problem in genetics is determining whether the experimentally determined data fits the results expected from theory (i.e.

Chi-Square Test A fundamental problem in genetics is determining whether the experimentally determined data fits the results expected from theory (i.e.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on carl friedrich gauss facts Ppt on work power and energy class 11 Ppt on aircraft landing gear systems Edit ppt on iphone Ppt on combination of resistances vs health Ppt on glorious past of india Ppt on different types of computer softwares and hardwares Ppt on ms-excel functions Ppt on preservation of public property tax Ppt on pi in maths cheating