Probability
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Lab 2- Probability 

Introduction 

What determines the numbers that are selected in the lottery or whether heads or tails results after a coin flip? We can ask similar questions about biological events. What determines whether chromosome 14 will be pulled to the left side of the cell or the right side during anaphase I of meiosis? The answer is chance.  

Despite common understandings of chance as something completely unpredictable, in math and science chance is expressed through probability in precise mathematical terms that place predictable limits on chance events. Specifically, probability is defined as the likelihood of a particular chance event occurring among the total number of equally likely chance events. Mathematically,                                                   

                        probability (P) =   

 For a coin toss, we can calculate the probability that heads will result from one toss. If heads is the number of particular chance events of interest, then the numerator is simply 1. The total number of equally likely events is 2 because tails is just as likely as heads. Thus, the probability is or 50 percent. 

We can now ask, What is the probability of obtaining heads if we flip the coin a second time? The First Law of Probability states that the results of one chance event have no effect on the results of subsequent chance events. Thus, the probability of obtaining heads the second time you flip it remains at . Even if you obtained five heads in a row, the odds of heads resulting from a sixth flip remain at . 

What if we ask a different question, What is the probability of flipping a coin twice and obtaining two heads (or, equivalently, flipping two coins at the same time)? Notice that this is different from the previous question; no coins have been flipped yet. In this case, we turn to the Second Law of Probability, which states that the probability of independent chance events occurring together is the product of the probabilities of the separate events. Thus, if the probability of one coin coming up heads is , and the independent likelihood of the second coin coming up heads is , then the likelihood that both will come up heads is x = .  

The Third Law of Probability considers what happens with chance events that are mutually exclusive, which means that they are not independent. In other words, if one event occurs then another event cannot occur. This frequently applies to combinations. For example, we could ask, What is the probability that in three coin flips, two heads and one tail will result? There are three possibilities for how this might occur: 

Possibility 1: head-head-tail

Possibility 2: head-tail-head

Possibility 3: tail-head-head 

If sequence 2 comes up, neither 1 nor 3 can happen, although in advance you have no way of knowing which might happen. All three sequences are equally likely, however. To calculate this, use the Third Law which states that the probability that any of a number of mutually exclusive events will occur is the sum of their independent probabilities. For the example above: 

P(1) = x x = 1/8

P(2) = x x = 1/8

P(3) = x x = 1/8

                                                                         P(total) =    3/8 

 

Thus, the probability of obtaining two heads and one tail in three separate coin flips is 3/8. These same rules of probability allow us to calculate the odds of parents conceiving particular numbers of girls or boys or of predicting the likelihood that specific chromosomes will segregate together into the same gamete. All are chance events. 

 

Procedure 

Part I:  Simple Probability 

1.         Assume that two sides of a coin represent two alleles, Q and q, for a single gene in the cells that    give rise to sperm. During meiosis these alleles will segregate at anaphase I, and only one allele will end up in the sperm that ultimately fertilizes an egg. If meiosis and fertilization are followed in 30 heterozygous males, Qq, how many times would you predict the Q allele to segregate into the successful sperm? The q allele? Place your answers under Expected in Table 1.1.  

2.         What is the probability that the Q and q alleles will be found in these sperm? 

3.         Designate sides of the coin as Q and q and toss the single coin 30 times. Record the number of times the Q and q alleles are found in successful sperm. Place your answers in Table 1.1 under. Observed and calculate and record the difference between Observed and Expected. (Use negative  numbers as appropriate.) 

4.         Now consider a mating between two heterozygous individuals, Qq. Meiosis must occur in each, resulting in the formation of egg cells and sperm cells. Those gametes then can fuse to form a zygote, or fertilized egg. Using the second law of probability, what is the chance that the fertilized egg will be QQ? What is the probability the fertilized egg will be qq? 

5.         Using the third law of probability, what is the chance that the fertilized egg will be Qq?  

6.         Apply the probabilities calculated above to 40 matings, and predict the number of heterozygotes and homozygotes of each type. Record these under Expected in Table 1.2. 

7.         Designate the sides of two coins as Q and q and toss them together 40 times, recording the allele combination each time in Table 1.2. 

8.         Complete Table 1.2 to compare your predictions (Expected Number) with the actual results (Observed Numbers). 

9.         Now lets examine a couple that is planning a large family. They plan to have three children, and  they would like to know the probability that all three will be boys; all three girls; two boys/one girl; etc. Determine all the possible combinations and probabilities and record them in Table 1.3. Now predict the expected numbers of each class if 32 couples have three children each. 

10.        Designate the sides of three coins as boy and girl and toss them together 32 times, recording the offspring combination each time. Calculate the difference between observed and expected. 

 

Part II:  Binomial Expansion 

Assume that a couple plans to have five children. In this case, it is somewhat tedious to outline and then calculate all the possible combinations because the number of independent events has increased beyond three or four.  

Fortunately, if one knows the number of events and the individual probabilities of each alternative event, the probabilities of various combinations can be calculated directly using binomial expansion. For a hypothetical couple that plans to have one child, boy is one possible outcome (and we call that a) and girl is the other alternative (and we call that b).  

Because boy or girl is equally likely (and equal to ), the total probability of having a boy or girl is a + b = 1. For a couple interested in two children, we simply calculate (a + b)(a + b) = 1, or (a + b)2 = 1. Expanding this gives: a2 + 2ab + b2 = 1.

 

                                                Probability of two boys: a2 = ()2 =

                                                Probability of one boy, one girl: 2ab = 2()() = 2/4 =

                                                Probability of two girls: b2 = ()2 =

                                                                                                            Total: 1

 

Generally, the formula for binomial expansion is (a + b)n, where n equals the number of independent events. (Your text, P-108, shows how to set up a simple table for determining the coefficients.) For example, if four children were planned, the expansion would appear thus:  

(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4 

This can be interpreted as indicating that there are 4 ways to get three boys and one girl, 4a3b (and, likewise, to get one boy and 3 girls, 4ab3), 6 ways to get 2 boys and 2 girls (6a2b2), etc. The terms in this binomial expansion show directly that the probability of 4 boys is:  

a4 = (1/2)4 = 1/16 

Similarly, the probability of 3 boys and 1 girl is:  

4a3b = 4(1/2)3(1/2) = 4/16 

1.         Complete the rest of the combinations to be sure you understand this. 

2.         Lets return to the couple interested in five children. Use binomial expansion to calculate the probability of having five boys; four boys and one girl; and all the other combinations. Complete your work on the back of the Probability Worksheet and hand this in before you leave lab. 

 

Part III:  Chi Square 

One problem with tests of probability is that they rarely come out perfectly. For example, you may have expected the Q allele to end up in the successful sperm 15 times out of 30 trials (Table 1.1). (In other words, your hypothesis was that Q segregates randomly.) But how would you interpret an observed experimental result that indicates 13 times, not 15?

Was the coin you flipped not a fair coin (or was the meiotic segregation not random) or was the deviation simply within the scope of chance deviation? Chi square analysis allows us to answer these and similar questions. In the segregation results you recorded in Table 1.1, there were only two outcome classes, Q and q. Lets assume that you observed 13 Q and 17 q segregating into successful sperm. To calculate chi square, use the equation:

 

                                                   (observed value expected value)2

                                    c2 =                        expected value 

This equation sums () the square of the differences between observed and expected and divides by expected. The expected, or predicted, numbers for this case of 50 percent probability with 30 trials are 30(.50) = 15 Q and 30(.50) = 15 q. Construct a table of these values: 

 

                        Observed         Expected         (O-E)                 (O-E)2                (O-E)2/E

                            13                    15                     -2                    4                        0.266

                            17                     15                      2                    4                        0.266

            Total         30                     30                      0                                              0.532 = c2 

 

Generally speaking, when the difference between the expected and observed outcomes is small, c2 is small, and when the difference is large, c2 is large. For c2 to be truly meaningful, however, we must compare it against a table of probabilities, such as the one shown below. 

This table shows the probabilities that a particular value of c2 is due only to chance variation (and not, in our example, to an unfair coin or nonrandom segregation). To interpret our value of c2, we must know the degrees of freedom (df) for the experiment. In our example, if Q resulted, then q could not, and vice versa.  

Thus, there was only one degree of freedom, which is defined as one less than the total number of outcome classes. From the table, we can see that 0.532 and df = 1 falls between 0.455 and 1.642, which correspond to probabilities of 0.50 and 0.20, respectively. In other words, if we were to repeat this same experiment many times (or observe many instances of meiotic segregations), we would expect this much difference between observed and expected values between 20 and 50 percent of the time due to chance deviation alone.  

Statisticians have set 5 percent as the lower limit for accepting a hypothesis, which our example easily exceeds. In this case, we would accept the hypothesis that Q segregates randomly and is found in successful sperm in half of all cases. You can check your calculations using an online Chi Square calculation located at "http://faculty.vassar.edu/lowry/csfit.html". 

 

                                                    Chi Square Probability Values                                           Reject            Reject

DF

 

P= .99

 

.95

.90

.75

.50

.25

.10

.05

 

.01

 

1

 

0.00016

 

0.00393

0.01579

0.10153

0.45494

1.32330

2.70554

3.84146

 

6.63490

 

2

 

0.02010

 

0.10259

0.21072

0.57536

1.38629

2.77259

4.60517

5.99146

 

9.21034

 

3

 

0.11483

 

0.35185

0.58437

1.21253

2.36597

4.10834

6.25139

7.81473

 

11.3448

 

4

 

0.29711

 

0.71072

1.06362

1.92256

3.35669

5.38527

7.77944

9.48773

 

13.2767

 

5

 

0.55430

 

1.14548

1.61031

2.67460

4.35146

6.62568

9.23636

11.0705

 

15.0862

 

 

6

 

0.87209

 

1.63538

2.20413

3.45460

5.34812

7.84080

10.6446

12.5915

 

16.8118

 

7

 

1.23904

 

2.16735

2.83311

4.25485

6.34581

9.03715

12.0170

14.0671

 

18.4753

 

8

 

1.64650

 

2.73264

3.48954

5.07064

7.34412

10.2188

13.3615

15.5073

 

20.0902

 

9

 

2.08790

 

3.32511

4.16816

5.89883

8.34283

11.3887

14.6836

16.9189

 

21.6659

 

10

 

2.55821

 

3.94030

4.86518

6.73720

9.34182

12.5488

15.9871

18.3070

 

23.2092

 

 

11

 

3.05348

 

4.57481

5.57778

7.58414

10.3410

13.7006

17.2750

19.6751

 

24.7249

 

12

 

3.57057

 

5.22603

6.30380

8.43842

11.3403

14.8454

18.5493

21.0260

 

26.2169

 

13

 

4.10692

 

5.89186

7.04150

9.29907

12.3397

15.9839

19.8119

22.3620

 

27.6882

 

14

 

4.66043

 

6.57063

7.78953

10.1653

13.3392

17.1169

21.0641

23.6847

 

29.1412

 

15

 

5.22935

 

7.26094

8.54676

11.0365

14.3388

18.2450

22.3071

24.9957

 

30.5779

 

 

16

 

5.81221

 

7.96165

9.31224

11.9122

15.3385

19.3688

23.5418

26.2962

 

31.9999

 

17

 

6.40776

 

8.67176

10.0851

12.7919

16.3381

20.4886

24.7690

27.5871

 

33.4086

 

18

 

7.01491

 

9.39046

10.8649

13.6752

17.3379

21.6048

25.9894

28.8693

 

34.8053

 

19

 

7.63273

 

10.1170

11.6509

14.5620

18.3376

22.7178

27.2035

30.1435

 

36.1908

 

20

 

8.26040

 

10.8508

12.4426

15.4517

19.3374

23.8276

28.4119

31.4104

 

37.5662

 

 

21

 

8.89720

 

11.5913

13.2396

16.3443

20.3372

24.9347

29.6150

32.6705

 

38.9321

 

22

 

9.54249

 

12.3380

14.0414

17.2396

21.3370

26.0392

30.8132

33.9244

 

40.2893

 

23

 

10.1957

 

13.0905

14.8479

18.1373

22.3368

27.1413

32.0069

35.1724

 

41.6384

 

24

 

10.8563

 

13.8484

15.6586

19.0372

23.3367

28.2411

33.1962

36.4150

 

42.9798

 

25

 

11.5239

 

14.6114

16.4734

19.9393

24.3365

29.3388

34.3815

37.6524

 

44.3141

 

 

26

 

12.1981

 

15.3791

17.2918

20.8434

25.3364

30.4345

35.5631

38.8851

 

45.6416

 

27

 

12.8785

 

16.1514

18.1139

21.7494

26.3363

31.5284

36.7412

40.1132

 

46.9629

 

28

 

13.5647

 

16.9278

18.9392

22.6571

27.3362

32.6204

37.9159

41.3371

 

48.2782

 

29

 

14.2564

 

17.7083

19.7677

23.5665

28.3361

33.7109

39.0874

42.5569

 

49.5878

 

30

 

14.9534

 

18.49266

20.59923

24.47761

29.33603

34.79974

40.25602

43.77297

 

50.8921

 

 

 

Assignment: 

1.         Write the hypotheses that you were testing (perhaps unknowingly) in the experiments that produced the data in tables 1.2 and 1.3. 

2.         On a blank sheet of paper, construct appropriate tables to calculate c2 for the data in Tables 1.2 and 1.3. 

3.         Interpret your c2 values and comment on whether you should support or reject your hypothesis.

                                                                                                                        Probability Worksheet                         Name:

 Table 1.1 Segregation of Alleles at Meiosis 

Outcome Classes         Observed (O)                Expected (E)                Difference (O-E) 

 Q (in successful sperm) 

 q (in successful sperm) 

 Totals                               30 

 

 

Table 1.2 Calculating Probabilities of Zygote Genotypes 

Outcome                      Combinations              Observed         Expected         Difference

 Classes                                                                (O)                    (E)                    (O-E) 

Q on both coins                  QQ

            (homo. zyg.)

Q on one coin,                    Qq or qQ

    q on the other

            (hetero. zyg.)

q on both coins                   qq

            (homo. zyg.)

 

 Totals                               4                                 40 

 

Table 1.3 Probabilities of Different Offspring Combinations

 

Outcome          Combinations  Probability       Observed         Expected         Difference

 Classes                                                                (O)                    (E)                    (O-E)

 

3 boys              BBB                 x x                                 32 x 1/8

                                                     = 1/8                                       = 4 

2 boys, 1 girl      

 

1 boy, 2 girls 

 

3 girls

 

 Totals

  

Materials:

Calculators

PC's with interconnection

pile of pennies

 

References:

Chi Square Table Values: http://www.statsoft.com/textbook/sttable.html#chi

Chi Square Calculator: http://faculty.vassar.edu/lowry/csfit.html