. To calculate
the probability of both dice landing with a four showing, we multiply the two probabilities
together. That is: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