• Two basic rules of probability are helpful in solving genetics problems: the rule of multiplication (or the rule of and) and the rule of addition (or the rule of or).
• Rule of multiplication is that the probability that independent events will occur simultaneously is the product of their individual probabilities. For example:
  • Question:
  • In a Mendelian cross between pea plants that are heterozygous for flower color (Pp), what is the probability that the offspring will be homozygous recessive?
  • Answer:
  • Probability that an egg from the F₁ (Pp) will receive a p allele = 1/2.
  • Probability that a sperm from the F₁ will receive a p allele = 1/2.
  • The overall probability that two recessive alleles will unite, one from the egg and one from the sperm, simultaneously, at fertilization is: 1/2 X 1/2 = 1/4.

• Rule of addition is that 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. For example:
  • Question:
  • In a Mendelian cross between pea plants that are heterozygous for flower color (Pp), what is the probability of the offspring being a heterozygote?
  • Answer:
  • There are two ways in which a heterozygote may be produced: the dominant allele (P) may be in the egg and the recessive allele (p) in the sperm, or the dominant allele may be in the sperm and the recessive in the egg. Consequently, the probability that the offspring will be heterozygous is the sum of the probabilities of those two possible ways:
  • Probability that the dominant allele will be in the egg with the recessive in the sperm is 1/2 X 1/2 = 1/4.
  • Probability that the dominant allele will be in the sperm and the recessive in the egg is 1/2 X 1/2 = 1/4.
  • Therefore, the probability that a heterozygous offspring will be produced is 1/4 + 1/4 = 1/2.

  The Statistical Nature of Inheritance

• If a seed is planted from the F₂ generation of a monohybrid cross, we cannot predict with certainty that the plant will grow to produce white flower (pp).
• We can say that there is a 1/4 chance that the plant will have white flowers.
• Stated in statistical terms: among a large sample of F₂ plants, approximately 25% will have white flowers.
• The larger the sample size, the more likely it will be that the percentage of the results will approximate the prediction.

• Probability (p) is the chance that an event will occur. Probability equals the case (a) divided by the total number of cases (n) or p = a/n, assuming all cases are equally likely events.
  • p for heads? heads/heads plus tails. 1/2
  • p for tails? tails/heads and tails. 1/2
  • p for heads or tails? 1/2 + 1/2 = 1
  • p for heads and tails? These are mutally exclusive. But heads and then tails is possible. 1/2 X 1/2 = 1/4

• The sum rule: if either/or then add the probabilities.
  • Die: What is the probability that the die will fall on either 1 or 2 spots? = 1/6 + 1/6.

Product rule: the and rule, multiply the probabilities.

• What is the probability that die will fall on 1 and next throw, 2? 1/6 X 1/6.

Binomial Data: Tossing two coins, heads or tails, discrete outcomes will occur. Whole number outcomes like 1 heads and 1 tails, or two heads, but not between.

Proportions verses percentages? Proportions are a specific chance over total chances, then reduced so that total chances equal 1. For percentages, you would multiply this number by 100.

Permutation? How many different arrangements in a row.

Permutations are the number of different ways a certain number of things can be arranged in a row.

It might seem tangental to our discussion, but in fact it is a crucial concept.

The .swf images below provide an opportunity to play with the concept.

The above image will demonstrate the concept of permutations. Use the image below and attempt to determine how many are possible with three.

The above image will again demonstrate the concept of permutations. Use the image below and attempt to determine how many are possible with four.

The above image will again demonstrate the concept of permutations, plus illustrate the rapid increase in complexity that occurs when more individuals are added. How would you like to use a manual system such as this on five individuals?

A Mathematical formula would be useful. The formula for individuals is simply n!, where n is the number of individuals.

For five, the answer is 5 X 4 X 3 X 2 X 1, or 120.

When calculating permutations for categories, instead of individuals, the number of possible orders is fewer. Use the image below and determine the number of orders.

The above image demonstrates the concept of permutations for categories. Use the image below and attempt to determine how many orders are possible with two men and two women.

For categories you might want to know how many ways 1 boy and 2 girls can be arranged.

The formula is n!/(s!t!...u!), where s,t...,and u are the number of individuals in each discreet category present.

The above example would be 4!/(2!2!), which equals six.

Genetics textbooks usually offer the student a choice between two approaches for solving problems involving permutations: the Binomial Theorem and the method outlined below.

I like to break up the formula into two parts, and de-mystify the math logic.

Following the dialog and examples should simplify the subject matter for you.

Question: What is the probability of a woman who will have 5 children giving birth to four girls and 1 boy in that order?

Answer: The probability of giving birth to the first girl is 1/2, as is the second and third girl. [1/2 X 1/2 X 1/2 X 1/2 or (1/2)⁴] The probability of the last child being a boy is also 1/2. [(1/2)⁴ X 1/2 = 1/32]

Each individual outcome is multiplied by the next because the nature of the question is that of an and question, not an or question. "What is the probability of having a girl and then a boy..."

In terms of how many ways or orders are involved, because only one order is requested by the question, permutations are an unnecessary consideration. The calculated probability is for just one possible order, the one given in the question.

Question: What is the probability of a woman who will have 5 children giving birth to four girls and 1 boy?

Answer: In this question, the order is not specified. One order would be the way the first question was stated. Its probability was 1/32. Would you expect the probability of any order to be greater than one order? I would. How much greater? The product of the number of different orders and the probability of one order. How many orders could there be for a woman giving birth to four girls and 1 boy? [5!/(4!1!) or 5] The correct answer to the question is 1/32 X 5 or 5/32.

The steps are to calculate the probability for one order, then calculate how many orders are possible, and then multiply the probability of one order by the number of possible orders.

What is the probability that brown eyed heterozygous parents will have 2 brown eyed girls, 2 brown eyed boys and 1 blue eyed girl?

Probability of getting a brown eyed boy is (3/4)(1/2) = 3/8.

Probability of getting a brown eyed girl is (3/4)(1/2) = 3/8.

Probability of getting a blue eyed girl is (1/4)(1/2) = 1/8.

• The answer is (3/8)²(3/8)²1/8 X (5!/(2!2!1!)) =

STATISTICS  

• It is rare that the actual observed numbers will fit exactly the numbers predicted. How do we test whether data support a hypothesis or not?
• We can determine the reliability or confidence limits of our conclusion. We test the hypothesis using certain statistical methods.
• If the expected ratio of offspring is 3:1, tall to short, then what would be the probability that of 4 progeny, 3 will be tall and 1 short?
• p = (3/4)³(1/4) X (4!/3!1!) = 27/64
• So even though you expect to get a 3:1 ratio, the probability, due to random chance, says you will only expect to see it, in the above described case, 42% of the time!
•  
• How do we decide that two things are likely to be different?
• Null hypothesis states they are the same; a statistical test will help you decide whether to reject or accept the null hypothesis.
• If we cannot reject the null hypothesis, does that mean that they are the same? No.
• It only means we cannot support that they are different. We cannot reject the null hypothesis.
• For example, assume that there exists a 10% difference between 2 populations. If we have few observations, there will be no significant difference between the 2 populations. If there are many observations, then we will find the two populations differ significantly.
• What if we have few observations, but observe a large difference, say 95%? We are more likely going to be able to reject the null hypothesis.
• The larger the number of observations the more sensitive the statistical test.

Chi-Square  

• The Chi-Square analysis is a statistical test useful on data with discreet outcomes.
• It is a simple to use and interpret test.
• Data are tested in two steps. First, a chi-square value is calculated. Next, a table is used to determine the probability. If the probability is below a certain value, the null hypothesis is rejected.

• Chi-square (x²) equals the sum of all the squared values for the differences of the observered and expected values over the expected for each.
• The x² value must be converted into a probability using a conversion table.
• Degrees of freedom must be calculated in order to use the conversion table.
• Degrees of freedom is the number of different categories minus 1.
• Degrees of freedom is a count of the independent categories.
• Remember that when using statistics you will be testing your hypothesis. Therefore, the number of categories will be determined from your hypothesis, not from the number of categories you observe.
• A horse with four socks only has three degrees of freedom of how to put the socks on. He can choose on the first three socks but the next sock must go on the last foot and so is not a free choice.
• After calculation of x² and degrees of freedom, use a chi-square table and find location that has the value that most closely matches.
• For 1 degree of freedom and a value of 3.842, the null hypothesis is rejected. The expected fails to likely to equal the observed. It is possible that they are equal, but unlikely.
• The value .05 or less on the table is the area outside the 95% of the normal observations, assuming that the two populations are the same or equal. [1 - .05 = .95] [.95 X 100 = 95%]. See Figures 3.11 and 3.12.

• 333.3 = _x² Lookituponthetable.... Where does it fit? What is the p value?
• The greater the chi square value the more different the values are from the expected, and the less likely the null hypothesis is true.