1. Cell Cycle and Chi Square

2. Cell Cycle Regulation

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

4. Super Complex System of protein checks!

5. P53 = master guardian gene

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

7. 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

8. 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…

9. Proto-oncogenes
RAS signal transduction pathway

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

11. Mutations are…and can be caused by…

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

14. 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

15. 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

16. 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

17. 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

18. 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

19. 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

20. 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

21. 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

22. 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

23. 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)

24. 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

26. 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

27. 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

28. 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

29. 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?

30. 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:

31. Hypothesis supported?

32. 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.

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

34. 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?

35. 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?