**PROBABILITIES OF GENETIC TRAITS**

**Introduction**

Before we investigate the probabilities of genetic traits, let's look at probability in general. The probabilities involved in rolling a pair of dice will be used to begin the investigation. When we roll each die, it lands with one side up out of a possible six sides. So the probability for one die to land on any specific number is . To calculate the probability of both dice landing with a four showing, we multiply the two probabilities together. That is:

*

so the probability of rolling two fours is . The probability of rolling a four __then__ a one is
, but the probability
of rolling a four then a one __or__ two is .

The probability of rolling a four and a two is a little more complicated, because the four could be rolled on the first die with the two on the second die or visa-versa. To calculate the probability of a four and a two showing on the dice, we add the individual double events. That is:

* *.*

So the probability of rolling a four and a two is arrived at by calculating the probability of rolling a four then a two, and adding it to the probability of rolling a two then a four. Therefore the probability is .

**Exercises Involving Dice**

When two dice are thrown, find the probability of rolling:

1. two five's 2. a five then a two 3. a five and a two

4. no fours 5. all even number 6. an odd and an even

7. two sevens 8. two values each 9. two values with a

less than five sum of a) eleven

b) ten

When we flip two coins, they have the possibility of landing heads/heads, heads/tails, tails/heads, or tails/tails. Each of these is equally possible to happen, so each represents one-fourth of all the possible occurrences. These values are also the probabilities of the events occurring. The probability of flipping two coins landing:

heads/heads_{ },
tails/tails_{ },
and heads with tails in any order_{ }.

The genetic probability of traits is similar to the probability of flipping a coin. In human genetics, the egg and sperm each supply the baby one of two pieces of genetic information per trait from its parents. These pieces of genetic information are called alleles. Alleles are either dominant or recessive. Dominant alleles are indicated by capital letters, whereas recessive allels are indicated by lower case letters. Let's look at the probability of a baby inheriting either ridges or no ridges on its fingernails. We'll use the letter "R" to represent the dominant allele, and "r" to represent the recessive allele. Suppose the genetic make-up of both parents is Rr. Each allele has an equal chance of being passed to the child, so the probability for , and the probability for (just as in flipping a coin). Now lets see how the probability of genetics works.

*Note: The sum of all the probabilities for allele combinations (as above) is one.

**Checking for Understanding**

Find a baby's probability of inheriting ridged fingernails if the genetic make-up of the parents is as follows:

1. father = Rr, and mother = rr

Solution: Since the mother does not have an "R" allele, the baby can't inherit "RR", so . The possible inheritance for the baby is "Rr" (mother's first r), "Rr" (mother's second r), "rr", and "rr" again.

2. father = RR, and mother = Rr

3. father = rr, and mother = RR

4. father = rr, and mother = Rr

Each allele of the mother and father has an equal chance of being passed down
to the next generation. Because of this equal chance, there is another way to calculate
genetic probability. We can look at the ratios for the allele combinations. Let's
rework problem 4 above. We will indicate the two lower case "r" alleles
of the father as .
The genetic make-up of the parents is: father_{ }, and mother_{ }. The only possible genetic
make-up for the offspring is . There are just as many "rr" combinations as there are "Rr"
combinations. That means the probability of inheriting two recessive alleles is the
same as the probability of inheriting one dominant and one recessive allele, therefore
. Since the
sum of these probabilities is one, we conclude that .

**More About Probability of Genetics**

When finding the probabilities involved for inheriting two (or more) genetic traits, we apply what we have learned so far. The steps for the calculations are:

* find all the possible genetic combinations.

* find the probabilities for each of the inherited traits.

* multiply the probabilities together.

For example, suppose the genetic make-up of the father is AaBb and the genetic make-up of the mother is AaBB.

Step 1) Find the possible "A trait" genetic combinations for the child.
*They are AA, Aa, aA, and aa*.

Find the possible "B trait" genetic combinations for the child.

*They are BB, BB, bB, and bB.*

Step 2) Calculate the probabilities for the "A trait".

, and .

Calculate the probabilities for the "B trait".

, and .

Step 3) Calculate each of the __double allele probabilities__:

*Note: if there where three traits, we would multiply three probabilities together, etc.

For the offspring to inherit the dominant trait, it has to receive only one dominant allele from one parent. With this information, let's find the probability that the child will inherit both the dominant "A trait" and the dominant "B trait".

.

**Genetics Probability Exercises**

Given the genetic make-up of the parents, determine the indicated probabilities.

1. If the make-up of the father = EeGg, and the mother = EEGg, then find the probability that the child will be:

a. EEGg b. EEgg c. Eegg

d. EeGG e. EeGg f. EEGG

2. If the make-up of the father = Ttqq, and the mother = TtQQ, then find the probability that the child will be:

a. TtQq b. Ttqq c. ttQq

d. TTQq e. have the f. have both dominant Q-trait dominant traits

3. If the make-up of the father = RrHh, and the mother = RrHh, then find the probability that the child will be:

a. RRHH b. RRHh c. Rrhh

d. RrHH e. RrHh f. RRhh

g. rrHH h. rrHh i. rrhh

j. have the dominant k. have both traits

R-trait dominant

4. If the make-up of the father = Ddtt, and the mother = DdTt, then find the probability that the child will be:

a. the same as the father b. the same as the mother c. DDtt

d. all dominant traits e. dominant T-trait and f. all traits

recessive d-trait recessive

5. If the make-up of the father = AaBb, and the mother = AaBb, then find the probability that the child will have:

a. all recessive alleles b. all recessive traits c. AaBb

d. all dominant alleles e. all dominant traits f. Aabb

g. dominant A-trait h. recessive a-trait i. aaBb

j. dominant A-trait & k. recessive a-trait &

recessive b-trait dominant B-trait

6. If the make-up of the father = EeFfGg, and the mother = eeFFGg, then find the probability that the child will be:

a. EEFFGG b. eeffgg c. EeFfGg

d. only the E-trait e. only the e-trait f. all 3 traits

is dominant is recessive recessive

g. 3 dominant traits h. 3 recessive traits i. 2 recessive

traits