Think Ahead
Probability and genes 1
To prepare for the next week of class, you should fully understand and be able to answer the
following questions. You can get help with this work from:


your instructor during office hours
http://biology.clc.uc.edu/courses/bio105/geneprob.htm
One of the easiest ways to think about probabilities is to use the classic coin toss. When you toss
1 coin, there are only 2 possible outcomes: heads or tails. So each time you toss a coin, the
“probability” (the measure of the chance that the event will happen) of tossing heads can be
written:
P(heads) = ½
And the “probability” of tossing tails can be written:
P(tails) = ½
Of course, ½ could be also be written as 0.5 or as 50%, since these are all equivalent expressions.
Mathematically, a probability is expressed with a number between 0 and 1. A probability of 0
means the event is not possible. A probability of 1 means an event is certain to occur.
A more complex situation occurs when you are tossing 2 coins. Since the first coin could come
up either heads (h) or tails (t) and the second coin also come up either heads (h) or tails (t), there
are 4 possible outcomes of tossing 2 coins: h/h, h/t, t/h, t/t.
One easy way to figure out all the possible outcomes is by using a table:
h
t
first coin
h
t
second coin
Now fill in the open squares with the combinations:
h
t
h h/h h/t
t t/h t/t
first coin
second coin
1. There are _4_ possible outcomes.
2. What is the probability of tossing heads on both coins? Express this number as a
fraction, a decimal, and a percent. P(both heads) = ¼, 0.25, 25%
3. What is the probability of tossing 2 tails? Express this number as a fraction, a decimal,
and a percent. P(2 tails) = ¼, 0.25, 25%
4. What is the probability of tossing 1 head and 1 tail? Express this number as a fraction, a
decimal, and a percent. P(1 head, 1 tail) = 2/4 (1/2), 0.50, 50%
You can use the same approach to look at some genetic outcomes. For simplicity’s sake we’ll
consider genes with only two possible forms (or alleles): “dominant” (A) and “recessive” (a).
Every person has 2 forms of the gene, or 2 alleles, one on each of a pair of chromosomes. When
2 people have a child, each of them passes 1 gene to the child. So the inner boxes in our table
represent all of the possible genetic outcomes for children. (In genetics this is called a Punnett
Square.) This is quite similar to the 2 coin situation – we have 2 objects (2 coins, 2 genes) and
each object has 2 possible forms: the coin can land either heads or tails, and the gene can be in
either its dominant or recessive form.
Let’s consider an imaginary scenario in which the ability to appreciate art is genetic and is
determined by a single gene. People who have at least one A (the dominant form of the gene)
can appreciate art. People who don’t have an A cannot appreciate art. Fill in the table below and
answer the questions:
A a
father
a
a
mother
5. Can the mother appreciate art? Explain your answer. No, because she does not have a
dominant gene (A) that allows the appreciation of art.
6. Can the father appreciate art? Explain your answer. Yes, because he has one A.
7. What is the probability that a child will have the ability to appreciate art? Express this
number as a fraction, a decimal, and a percent.
A a
a Aa aa
a Aa aa
8. What is the probability that a child will not have the ability to appreciate art? Express this
number as a fraction, a decimal, and a percent. P = 2/4 (1/2), 0.50, 50%
9. What is the probability that a child will have 2 recessive alleles? Will this child be able to
appreciate art? P = 2/4 (1/2), 0.50, 50%
10. What is the probability that a child will have 2 dominant alleles? P = 0
11. What is the probability that a child will have 1 dominant and 1 recessive allele?
P = 2/4 (1/2), 0.50, 50%
12. What is the probability that a child will be able to appreciate art?
P = 2/4 (1/2), 0.50, 50%
13. Set up a Punnett Square and answer the same questions for a family in which the father
is homozygous recessive and the mother is homozygous dominant.
a
a
A Aa Aa
A Aa Aa
father
mother
5. Can the mother appreciate art? Explain your answer.
Yes, because she has at least one A.
6. Can the father appreciate art? Explain your answer. No, because he has no A.
7. What is the probability that a child will have the ability to appreciate art? Express this
number as a fraction, a decimal, and a percent. P = 4/4 (1), 1, 100%
8. What is the probability that a child will not have the ability to appreciate art? Express this
number as a fraction, a decimal, and a percent. P = 0
9. What is the probability that a child will have 2 recessive alleles? Will this child be able to
appreciate art? P = 0
10. What is the probability that a child will have 2 dominant alleles? P = 0
11. What is the probability that a child will have 1 dominant and 1 recessive allele?
P = 4/4 (1), 1, 100%
12. What is the probability that a child will be able to appreciate art?
P = 4/4 (1), 1, 100%