Download Homework I. Mendelian genetics

Homework I. Mendelian genetics
KEY (Sec 28)
1- An organism with genotype AAEeDd can make how many different gametes (list
each one)? (1 pt)
It could produce four different types: AED, AeD, AEd, Aed
2- You hypothesized that you have ONE autosomal dominant trait. You crossed the
following true breeding parents: wild type female and mutant male. In the F2
generation you observe the following phenotypes: 35 dominants females, 35
dominants males, 20 recessive males and 10 recessive females. What is your Chisquare statistical result, and does this fail to reject the stated hypothesis? (2 pts).
As you are hypothesizing that the trait is autosomal dominant (then NO SEXLINKED), you sum males and females with same phenotype together to get the
observed values for each phenotype. Remember that for a monohybrid cross the
expected phenotypic ratio is 3 dominants : 1 recessive, so out of 100 observations you
would expect to observe 75 dominant and 25 recessive.
Dominant
Recessive
Total
Observed
(O)
70
30
100
Expected
(E)
75
25
100
(O-E)
(O-E)2
-5
5
25
25
X2
25 / 75 = 0.33
25 / 25 = 1
1.33
You have only 2 phenotypes, dominant and recessive, so the degrees of freedom are
2 -1 = 1. The tabulated Chi-square for 1 degree of freedom is equal to 3.841. Your
calculated Chi-square is less than the theoretical value, which means that you hold
your hypothesis (the trait is autosomal and dominant).
3- You are studying ONE mutant trait that is sex-linked dominant. You crossed the
following true breeding parents: mutant female and wild type male. What are
the phenotypes and phenotypic ratios for the F2 generation? (1 pt)
Let’s define A = mutant trait, a = wild type
XA
XA
Xa
XAXa
XAXa
Y
XAY
XAY
The phenotypic proportions in the F1 are 1 mutant female : 1 mutant male
Now, to get the F2, you have to breed one female and one male from the F1:
A
X
Xa
XA
XAXA
XAXa
Y
XAY
Xa Y
Remember that the mutant trait is dominant. Then, the phenotypes and phenotypic
ratio observed in the F2 are:
2 mutant females : 1 mutant male : 1 wild type male
4- Both Mrs. Smith and Mrs. Jones had babies the same day in the same hospital.
Mrs. Smith took home a bay girl, whom she named Sharon. Mrs. Jones took
home a girl, whom she named Jane . Mrs. Jones began to suspect, however, that
the child had been accidentally switched with Mrs. Smith baby in the nursery.
Blood test were made; Mr. Smith was type A, Mrs. Smith was type B, Mr. Jones
was type A, Mrs. Jones was type A. Sharon was type O, and Jane was type B.
Had a mix-up occurred? Show your work! (1 pts)
You know the phenotype of each parent but not their genotype. Then, Mr. Smith could
be either IA IA or IA I, and the same thing for Mrs. Smith, Mr. Jones and Mrs. Jones.
We are going to use the following notation:
Mr. Smith
IA __
Mrs. Smith
IB __
Mr. Jones
IA __
Mrs. Jones
IA __
If you do the Punnett square, you will see that:
The Smith baby could be: IA IB, IA i, IB i,or i i
The Jones baby could be: IA IA, IA i, or i i
Therefore, the parents of Jane can only be the Smith; there was a mix -up!
5- What is the genotypic ratio of the progeny produced from this mating?
Parental cross: female aaBb crossed with a male AABb. (1 pt)
Fill in the Punnett square:
? \?
aB
ab
AB
AaBB
AaBb
The genotypic ratio is 1 AaBB : 2 AaBb : 1 Aabb
Ab
AaBb
Aabb
6- You have found a blue tomato growing in your garden and want to find its
genetic basis. You make a cross of a true breeding blue tomato plant with true
breeding red tomato plant. Out of 1000 F2 plants you find that 712 are red, and
288 are blue. What is the likely genetic basis for the blue tomato (which trait is
dominant?). Test your hypothesis with a Chi-square test. (2 pts).
The red trait is most likely dominant, and this is probably a monohybrid cross. So,
you have two phenotypic classes: red tomatoes and blue tomatoes. If this trait is
inherited as expected by Mendel’s First Law, you expect to observe 750 red tomatoes
and 250 blue ones in your F2.
Red
Blue
Total
Observed
(O)
712
Expected
(E)
750
288
1000
250
1000
(O-E)
(O-E)2
X2
-38
1444
1.92
38
1444
5.77
7.69
You have only 2 phenotypes, red (dominant) and blue (recessive), so the degrees of
freedom are 2 -1 = 1. The tabulated Chi-square for 1 degree of freedom is equal to
3.841. Your calculated Chi-square is bigger than the theoretical value, which means
that you reject your hypothesis (tomato color is not inherited as expected by first
Mendel’s Laws).