Credit worth 30 points

Probability in Genetics.

Introduction. In order to understand genetics you need to have some basic concepts concerning probability. The notion of probability and chance is used in every are of genetics from basic considerations of the outcome of meiosis and Mendel’s so called laws of inheritance to gene sequencing problems and data base searches. The goal of this exercise is to introduce the important probability concepts and illustrate their use with some elementary examples from genetics.

Learning goals:

Define probability in a simple sense.

Define Sample space and be able to describe the sample space for a random experiment

Distinguish between ordered Vs unordered events

Define Probability

Apply the concepts of sample space and ordered Vs unordered events to describe simple genetics problems.

State the basic rules for combining Independent and mutually exclusive events

Apply the rules for independent and mutually exclusive events to simple genetics problems

Define conditional probability apply this concept to simple problems.

Demonstrate the use of tree or branch diagrams as an aid to doing probability problems

Basic Definitions:

Sample space.

All possible outcomes for a random experiment. For example. If you flip a coin once the sample space is the set {H,T}. Suppose you have two coins the sample space when both coins are flipped is {HH, HT, TH, TT}. An event is a subset of the sample space.

Activity 1

Suppose you have two dice each with faces 1 through 6. List the sample space (the set of all possible outcomes for a single toss of both dice):

Activity 2

You have four balls in an urn. A red ball, a green ball, a blue ball and an orange ball. Suppose you remove the balls from the urn in random order. You might for instance pull the green ball out first, the blue one next etc. List the sample space for this experiment. Hint you are listing all possible arrangements of the four balls.

Ordered Vs unordered event.

If the sequence of the joint occurrence of several events is important then this is ordered. For example, maybe it is important to distinguish between {HT} and {TH} when two coins are flipped. This is an ordered event. Another example would be the sample space for the experiment in activity 2. Suppose for the coin example if you define your event as being ‘exactly one head’ when two coins are flipped, then this can happen in two ways {HT} and {TH}. So the event ‘exactly one head’ is an unordered event because you don’t care which head comes first! An unordered events if you think about it carefully describes a subset of ordered events in your original sample space.

Activity 3

Returning to the dice example from activity 1. List the outcomes in the sample space that correspond to the unordered event: ‘the sum of both faces when the dice are tossed once is equal to 6’:

Some counting rules: Often times we won’t want to list every possible of event but instead need to count how many there are and there are some useful rules.

Counting rule for the number of ordered events: In general if for a single event you can have one of X possible outcomes, then if you have Y joint events then the total number of possible combinations of Y joint events is going to be X^Y. Thus for a single die there are six possible outcomes. If you have 10 dice, there are 6^10 possible outcomes if all 10 dice are tossed at once, 6 outcomes for the first die times 6 for the second die etc.

A powerful rule for unordered events: In general we can ask for N experiments with two possible outcomes for each experiment about the number of events involving M of the first outcome and N-M of the second outcome. For example if you flip 10 coins you can ask about the number of ways you can get 6 heads in 10 flips. This is going to be all the possible ordered combinations of 6 Heads in 10 flips{TTTTHHHHHH, THTTTHHHHH, etc.}.Useful rule for the number of unordered joint events where there are N possible independent events and M events of one particular type: The number of Number of possible unordered joint events = N!/[(M!)(N-M)!]Where N! is N factorial or 1*2*3* …*N.

For our coin example, the number of ways you can get 6 heads in 10 tosses is:10!/(6! 4!) = 10*9*8*7/(4*3*2*1)Make sure you understand why this is!. Note that for this example the total number of possible ordered events is 2^N or 2 raised to the Nth power.

In statistics the expression

N!/[(M!)(N-M)!] is often called N choose M which represents the number of ways of selecting M things out of a group of N things.

Applications of these ideas to genetics.

Activity 4

A. The cells in your body receive half their chromosomes from your father and one half of their chromosomes from your mother. So for each pair of homologous chromosomes one will be a maternal chromosome and one will be a paternal chromosome.. During meiosis when the haploid gametes are formed, each member of the pair but not both ends up in a gamete. This is the principle of segregation.1. For humans (N =23) how many possible combinations of maternal and paternal chromosomes are there? Show your reasoning.

B, How many of the combinations from part A involve 10 maternal chromosomes and 13 paternal chromosomes? Hint are we dealing with an ordered or unordered event here?

Activity 5

Recall that DNA is made from a sequence of four nucleotides that differ in terms of which nitrogen base A,T,G,C is present in each nucleotide. Suppose you have a region of DNA which is 100 nucleotides long. How many possible nucleotide sequences are there for this region of DNA?

Probability

Typically we repeat experiment many times in genetics and other areas of science and we are often interested in how often a certain outcome might be expected to happen. The expected frequency of a particular event when an experiment is repeated an infinite number of times is the probability of the event. For a single coin toss, the probability of a head on a single toss is 1/2. Probabilities are always assumed to be real numbers between 0 and 1. Probabilities in genetics are often predicted based on certain hypotheses and then the predictions are used to test the hypothesis using real data. We will refer to the probability of an event as P(event) or sometimes Pr(event). So for a coin flipped once and only once for the sample space {T,H} P(H) = 1/2. Notice that in the absence of other information, we will often assume that the probability of elementary events such as the result of a single toss of a coin are equally likely. The problem is for us is going to be when elementary events are combined.

Independent and mutually exclusive events:

Two events A and B are said to be independent events if the probabilities of both events happening jointly.

Useful rule: multiplication rule for the joint probability of two independent events A,B:

P(A and B) = P(A)*P(B)

So for example if we flip two coins P(H on the first coin and H on the second coin) = P(H on the first coin)*P(H on the second coin) = 1/4

Two events are said to be mutually exclusive if P(A and B) = 0

For example for a single flip of a coin the event H and the event T are mutually exclusive since they cannot happen at the same time.

Note that if for all the mutually exclusive events in a sample space the probabilities must some to 1. Thus P(A or B) = 1

For example a die has 6 different faces which are mutually exclusive. If each one is equally likely then the probability of any on occurring on a single toss is 1/6. So all the probabilities have to sum to 1 for our sample space{1,2,3,4,5,6}

Useful addition rules for mutually exclusive events:

Rule 1: Given N mutually exclusive events {A1,A2, … ,AN} then

P(A1) + P(A2) + … + P(AN) = 1.0

Rule 2: P(A particular mutually exclusive event) = 1 – P(all the others) This rule is very useful!

For example:

Suppose I toss a coin 10 times (or 10 coins once) and ask what is the probability of getting at least one head in 10 tosses. We could do this in two ways one is to sum probabilities P(1 at least head in 10) + P(2 heads in 10) + P(10 heads in 10) or we can simply go:

P(1 at least head in 10 tosses) = 1 – P(no heads in 10 tosses) = 1 – 1/(2^10)

Application to genetics:

Activity 6

For maternal and paternal chromosomes in the human gamete example, what is the probability of a gamete having at least one maternal chromosome?

Activity 7

Many types of color blindness are what are called X linked, that is determined by genes on the X chromosome. Suppose a woman is carrying one X chromosome with the gene for a particular type of color blindness; her other X chromosome does not have this gene. If she is married to a man who does not have this gene on his X chromosome. You may remember that color blindness is X linked recessive.

A. What is the probability that her first child will carry the X chromosome with the gene associated with color blindness?

B. Suppose amniocentesis reveals that the child is male. What is the probability that the child is color blind. Hint: this involves conditional probability

Activity 8

Albinism in humans is controlled by a recessive gene (c). Suppose two heterozygous individuals (genotype Cc) marry. Assume they have six children.

A. What is the probability that at least one child will be albino?

B. What is the probability that three will be albino and three will have pigmentation?

C. A challenge: What is the probability that three of the children will be boys AND all three boys will be albino? Hint assume gender is independent of whether or not a child turns out to be albino. Not too tricky if you think carefully!

Activity 9

Flip 5 coins 128 times.You can use actual coins or use. You should have 6 expected classes. Get your expected ratio from Pascal’striangle

http://shazam.econ.ubc.ca/flip/

Keep a tally sheet. Do a chi square. Do your data fit expections?

Class Expected Observed O – E (O – E)2 (O – E)2/E

Total

Degrees of Freedom: ______________Probability of the X2: ____________________________________

Conclusion:

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