IV. Predicting the outcome of various crosses
1. A female carrier of Tay-Sachs wants to have a child with a male carrier for
Tay-Sachs. Tay-Sachs is inherited in an autosomal recessive manner. Using the
letters “T” and “t” fill out the following table.
Female
Male
Genotype
Tt
Tt
Phenotype
Carrier = healthy
carrier
Genotypes of sperm/ova
T or t
T or t
2. Determine the probability of them having a child with Tay-Sachs using a
Punnett square.
T
t
T
TT
Tt
t
Tt
tt (25%)
3. You can also calculate the probability. This is especially convenient when
looking at more complex characters (imagine a Punnett square for a trihybrid
cross).
The probability (P) scale ranges form 0 to 1
The probabilities for all possible outcomes for an event must add up to 1.
Event
Probability
Tossing heads with a 2-headed coin
100%
Tossing tails with a 2-headed coin
0%
Tossing heads with a normal coin
50%
Tossing tails with a normal coin
50%
Rolling 3 on a 6-sided dice
1/6
Rolling a number other than 3
5/6
4.
You have just rolled three threes in a row. What is your chance of rolling
another 3?
1/6
5.
In the Tay-Sachs example above, the probability of receiving the “t”
allele from the mother is independent of the probability to receive an “t” allele
from the sperm. Apply the rule of multiplication :
P of egg receiving “t”
P of sperm receiving “t”
P of offspring "ff"
.5
.5
.25
6. A child heterozygous for Tay-Sachs can inherit the two alleles in two
independent ways: By receiving the “t” from the mother and the “T” from the
father, or the “t” from the father and the “T” from the mother. Apply the rule of
addition to calculate P of "Ff".
P of “t” in egg and
P of “T” in egg and P of "Ff"
“T” in sperm
.5 x.5 = .25
“t” in sperm
.5 x.5 = .25
.5
7. Now consider a trihybrid cross of garden peas, where
Character
Trait and genotype
Flower color
Purple: Pp, PP
White: pp
Seed color
Yellow: YY, Yy
Green: yy
Seed shape
Round: RR, Rr
Wrinkled: rr
You cross PpYyRr x Ppyyrr. List the genotypes and phenotypes of offspring
homozygous recessive for at least two of the characters.
Hint: you should find five genotypes.
9. Use the table below to calculate the probability for the offspring of PpYyRr x
Ppyyrr to be homozygous recessive for at least two traits.
Hint: Assume the alleles segregate independently from each other, use the rule
of multiplication.
10. Calculate the probability that the offspring would have any one of the five
genotypes above.
Hint: this increases the probability. The events are not independent. Use the
rule of addition.
Genotype with at least two
Probability of genotype
homozygous recessives
ppyyRr
.0625
ppYyrr
.0625
Ppyyrr
.125
PPyyrr
.0625
ppyyrr
.0625
Probability that offspring has any .3750
of these:
1. Flowers
cross 1: parental genotypes BB and WW
inheritance: incomplete dominant
cross 2: parental genotypes BW and BW
1. Blood type
a) cross 1: most likely AA x BB
cross 2: AB x AB
co-dominant
B) Ai x Bi A and B are completely dominant over i.
Ratio is off due to small sample size
2. Mice II
cross 1: The tail-less parent is Tt, the normal mouse is tt, you cannot figure out the
inheritance pattern without looking at cross 2
cross 2: Tt x Tt
TT dies (lethal combination)
3. Flies
cross 1: Xw+Xw x Xw+Y (x-linked recessive)
cross 2: Xw+ Xw+ x XwY
cross 3: Xw+ Xw x XwY
5. Pedigrees
a) both autosomal and sex-linked recessive fit the data
b) autosomal recessive
6. Mice III
a) cross 1: Dihybrid cross, blue and short-toothed are dominant over brown and longtoothed
cross 2: normal F1 cross yielding expected ratio of 9:3:3:1 for unlinked genes
b) cross 1
TtGG x ttgg (Tall and green are dominant, you can only figure that out if you look at
the second cross)
Cross 2: TTGg x ttgg
7. Pathways I
a) you would get colorless from any homozygous aa, red from any A_ and
homozygous for bb, and purple from any that is hetero- or homozygous dominant for
both traits.
This is epistasis.
b) you would get red from anyone with bb, colorless from anyone with B_ and aa,
and purple from anyone with B_A_
c) Sorry, ot call it colorless in the question and white in the cross is confusing.
Assume that it is the same.
aaBb -> blue
Aabb -> red
Aabb -> colorless
AaBb -> dead
Lethal, epistasis
8. Linkage
cross1: is your monohybrid cross with rough and hard being dominant
cross 2: It is is best to draw the chromosomes to understand how this works.
Assume that the genes for rough/smooth and hard/soft are on chromosome 1.
The rough, soft parent’s genotype, let’s call her mom, is RRss, the smooth, hard
parent’s genotype, let’s call him dad, is rrSS.
The resulting F1 has one chromosome from mom (RRss) and one from dad (rrSS)
9. Pedigree that illustrates multiple possible models
Hypothesis 1: autosomal recessive
Then 1, 3, 6 and 7 would be aa and 2, 4, 5 and 8 Aa
Reject: because if 14 is a legitimate child of 6 and 7 she should be aa.
Hypothesis 2: autosomal dominant
Then 1, 3, 6, 7 are Aa and 2,4,5,8 are aa
9-13 are A__ and 14 is aa.
Ratio in the third generation is off, but can be explained with small sample size.
Hypothesis 3: X-linked dominant:
Then 1 and 3 are XA Xa and 2, 4, and 5 are XaY
6 is XAXa and 7 is XAY
Reject: because 14 would have inherited the dominant allele from father
Hypothesis 4: X-linked recessive
Then 1 would be XaXa and 2 would be XAY
Reject: because 6 would have inherited a dominant allele from father