Cell Cycle and Chi Square

Cell Cycle Regulation

CDK-cyclin complexes CDK is turned on or off due to additional transcription factors Cyclins vary in type and concentration depending upon the cell

Super Complex System of protein checks!

P53 = master guardian gene

p53 will initiate p21 gene to halt cell cycle if DNA is damaged

Rb gene Rb protein binds to E2F halting cell cycle at G1 The CDK-cyclinA complex inactivates Rb so E2F activates leading to S-phase

Tumor Suppressor Genes Rb and p53 – they act to halt cell cycle If one allele for a TSG is mutated then… If both alleles for TSG are mutated then…

Proto-oncogenes RAS signal transduction pathway

Mutation in RAS leads to cellular division even if no ligand present

Mutations are…and can be caused by…

AP LAB BOOK – page 87 Double blind study? Null Hypothesis? Root tips A Root tips B

Total # of cells counted Hypothetical Data Jar A – slides A Jar B – slides B A slides B slides Interphase # of cells 543 493 Mitosis 251 298 Total # of cells counted 794 791

Hypothetical Data Jar A – control Jar B – experimental X2 = ∑ (observed – expected)2 expected A slides B slides Interphase # of cells 543 493 Mitosis 251 298 Total # of cells counted 794 791 Observed = experimental values Expected = determine % expected from the control values

What does Chi-square analysis tell us? If the difference between your observed results and expected results are due to random chance or not. EX –flip a coin enough and you would expect to get heads 50% of time and tails 50% of time. If this number is way off even after 1,000 tosses then something else besides randomness is occurring

The Chi Squared Equation X2 = ∑ (observed – expected)2 expected # of Possible Outcomes Observed Expected X2 Allium root tip squashes lab: Null hypothesis? # of possible outcomes for the onion root tip cells? Observed = experimental Expected = use values from control to determine

Hypothetical Data Jar A – control Jar B – experimental X2 = ∑ (observed – expected)2 expected A slides B slides Interphase # of cells 543 493 Mitosis 251 298 Total # of cells counted 794 791 Observed = experimental values Expected = determine % expected from the control values

Calculations X2 = ∑ (observed – expected)2 expected Observed Interphase Expected Interphase Observed Mitosis Expected Mitosis 493 538 (.68 x 791) 298 250 (.32 x 791) A slides B slides Interphase # of cells 543 493 Mitosis 251 298 Total # of cells counted 794 791

How it works continued Determine degrees of freedom – it’s one less than the number of possible outcomes Ex – expect 4 different phenotypes, degrees of freedom is 3 (as in a punett square) Ex 2 – expect two different phases, degrees of freedom is 1

How it works Once you have your degrees of freedom and Chi square sum, go to the distribution table Biology p value 0.05, find the critical value

NULL HYPOTHESIS – Reject or Accept REJECT Null Hypothesis if your chi-square value is greater than or equal to the critical value (3.84 in this case) ACCEPT Null Hypothesis if chi-square value is less than critical value

Why? p is probability of an event occurring due to randomness If p = .50 then the difference between observed and expected results is due to random events 50% of time (random not due to biological process or experimental treatment)

You explain If p = .05 then what does that mean? the difference between observed and expected results is due to random events occurring only 5% of time

if critical value is higher than critical value at if critical value is higher than critical value at .05 your null hypothesis is REJECTED meaning something is going on other than randomness if your critical factor is equal to or less than value at .05 = your NULL hypothesis is supported, difference is simply due to randomness

APPLY THIS NEW DATA TO YOUR ALLIUM ROOT TIP LAB Jar A – control Jar B – experimental A slides B slides Interphase # of cells 255 498 Mitosis 136 212 Total # of cells counted

Chi- square and Mendelian Genetics! P = Purple p = Yellow S = Smooth PpSs x PpSs s = Shrunken What are the expected results? Hypothesis? 9:3:3:1 due to independent assort of alleles

Results analysis Observed results: An ear of corn has a total of 381 grains, including 216 Purple & Smooth, 79 Purple & Shrunken, 65 Yellow & Smooth, and 21 Yellow & Shrunken. Expected?

Table, Chi Square and Degrees of Freedom Possible Outcomes Observed Expected Chi Square X2 = ∑ (observed – expected)2 expected Purple/ smooth 216 Purple/ shrunken 79 Yellow/ smooth 65 Yellow/ shrunken 21 Df: SUM of Chi Squares:

Hypothesis supported?

Try this one Same hypothesis: independent assort therefore 9:3:3:1 outcome Observed: An ear of corn has a total of 389 grains, including 219 Purple & Smooth, 70 Yellow & Smooth, and 100 white.

Possible Outcomes Observed Expected Purple/ smooth Purple/ shrunken Chi Square X2 = ∑ (observed – expected)2 expected Purple/ smooth Purple/ shrunken Yellow/ smooth Yellow/ shrunken

FRUIT FLY GENETICS Cross two heterozygous wild type fruit flies for the traits below. Predict expected outcomes of offspring using a punnett square. W= normal wings w=vestigial B = normal body b = black Go to the next slide for actual results then perform Chi Square to determine if there is a significant statistical difference between the expected and observed results. If yes – what? If no – why not?

Actual results of the cross 800 Wild/Wild 55 wild-type wings/ black body 50 vestigial wings/ wild type body 902 vestigial/ black body significant statistical difference between the expected and observed results? If yes – what? If no – why not?