  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 ½ 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 ½. Even if you obtained five heads in a row, the odds of heads resulting from a sixth flip remain at ½.

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 ½, and the independent likelihood of the second coin coming up heads is ½, then the likelihood that both will come up heads is ½ x ½ = ¼.

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:

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) = ½ x ½ x ½ = 1/8

P(2) = ½ x ½ x ½ = 1/8

P(3) = ½ x ½ x ½ = 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 let’s 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 ½), 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 = (½)2 = ¼

Probability of one boy, one girl: 2ab = 2(½)(½) = 2/4 = ½

Probability of two girls: b2 = (½)2 = ¼

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.         Let’s 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. Let’s assume that you observed 13 Q and 17 q segregating into successful sperm. To calculate chi square, use the equation:

å   (observed value – expected value)2

c2 =                        expected value

This equation sums (å) 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                 ½ x ½ x ½                                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