Download Segregation, Independent Assortment, and Probability

More fun with Genetics:
Segregation, Independent Assortment, and Probability
In order to answer the questions like the ones that follow, you have to be able to work with
punnet squares and calculate probability from your results. Here are a few "helpful" hints to
guide you in calculating probability and help you to answer the questions on the take home quiz!
1. When you have dihybrid, trihybrid, etc. crosses with many possible combinations, you may
calculate the proportion of individuals with a specific genotype by calculating the proportion
for each combination separately. For example:
S = smooth
s = wrinkled
Y = yellow
y = green
T = tall t = short
You cross two individuals with the following genotypes: SsyyTt x ssYyTT
Question:
What proportion of the resulting individuals will have the following genotype: SsYyTT
Answer:
You could do a great big huge punnet square and drive us both crazy, or you can think
about each of the traits individually, because they assort independently.
So: Ss x ss
Start with the first trait
yy x Yy
Tt x TT
S
s
s
s
Ss
Ss
ss
ss
therefore, 1/2 have the
genotype of interest=Ss
Using the same method, you can determine that 1/2 will have the Yy genotype and
1/2 will have the genotype TT. So, you can multiply these together to get the
proportion of individuals with the genotype you are looking for: SsYyTT = 1/2 x 1/2
x 1/2 = 1/8
Question:
What proportion of the individuals would be dominant for all traits?
Answer:
Here we are asking about phenotype. Therefore, 1/2 of the individuals would
smooth, 1/2 would be yellow, but all (1) would be tall.
Your answer would be 1/2 x 1/2 x 1 = 1/4
be
Now let’s look at probability.
2. There are a few simple formulas to remember. They can look very ominous, but if you
cancel, they are not as formidable as you might first guess.
To determine the number of permutations (how many possible combinations) of an event
n!
s! x r!
where n = the total number of individuals, and s and r refer to different
combinations of events.
Here is an example: If you had only two types of children, good or bad, in a group of 9
children, how many different ways could you have 6 good children and 3 bad children in any
order?
n = 9, s = 6, r = 3
9! =
6! x 3!
9x8x7x6x5x4x3x2x1
6x5x4x3x2x1 x 3x2x1
= 9x8x7x6x5x4x3x2x1
6x5x4x3x2x1 x 3x2x1
9x8x7 = 504 = 84 different ways
3x2x1
6
So lets say that good (G) is dominant over bad (g). One homozygous dominant individual (GG)
marries a homozygous recessive individual (gg). Out of 9 children, what is the probability of
having four bad ones?
You know that GG x gg will give you four Gg offspring. Since a bad child would have to be gg,
the probability is 0.
If you have two heterozygous individuals (Gg) marry and produce nine children, what would be
the probability of 4 bad ones?
Answer: Do a punnet square showing the probable offspring of 2 Gg individuals. You will see
that 3/4 of the children will be GG or Gg, while 1/4 of the children they produce will be gg. The
probability of having 4 of the 9 children be bad would be:
9! (3/4)5(1/4)4 = 11.7%
5! x 4!
If no number of children were stipulated and you were asked the probability of having four of the
children from this marriage be good, you would not need the permutation and your answer
would be (3/4)4 = 31.6%
This can actually be a lot of fun! So let's try a few as an "all or none" take home. If you get
stuck, give me a call and I will try to help you get back on the right track!
Name ______________________
Date _____________________________
Please show all of your work on a separate sheet of paper.
1. In sesame, both the number of seed pods per leaf axil (3-pod or 1-pod) and the shape of the
leaf (wrinkled or smooth) are determined by pairs of alleles at two different genetic loci, and they
show independent assortment. The result of six crosses gave the results below. Fill in the full
genotypes for all parents and the generalized genotypes for the four categories of offspring.
Number of progeny
Parents
1-pod smooth x 3 pod wrinkled
__________
___________
___________
________
211
0
205
0
78
90
84
88
447
158
146
52
318
98
323
104
110
113
33
38
362
118
0
0
______________
1-pod smooth x 3-pod smooth
____________
__________
______________
1-pod smooth x 1-pod wrinkled
___________
3-pod
wrinkled
______________
1-pod smooth x 3-pod smooth
__________
3-pod
smooth
_____________
1-pod smooth x 1-pod smooth
___________
1-pod
wrinkled
_____________
1-pod wrinkled x 3-pod smooth
__________
1-pod
smooth
______________
2. In humans, two traits, widow’s peak and free-hanging earlobes, depend on separate dominant
genes located on different chromosomes. A man with a widow’s peak and attached earlobes
(whose father has free-hanging earlobes) married a woman without a widow’s peak but with
free-hanging earlobes (whose father had attached earlobes).
a. What are the genotypes of the two parents?
b. What is the probability that their first child will not have either a widow’s peak or
attached earlobes?
c. What is the probability that their first child will have both a widow’s peak and freehanging earlobes?
d. What is the probability that their first child will have a widow’s peak or attached
earlobes?
e. What is the probability that their first child will not have a widow’s peak or freehanging earlobes?
SHOW ALL YOUR WORK ON A SEPARATE SHEET OF PAPER!