Homework 12-KEY
BIO 2
1.
Fill in the table below. The first row is done for you as an example.
Use the rule I showed you in class. Each heterozygous locus is assigned the number
2 and each homozygous locus is assigned the number 1. Then multiply these
numbers together to calculate the number of gametes.
Genotype
# of gametes
Gametes
AAbb
1
Ab
4
AB, ab, Ab, aB
4
ABC, Abc, ABc, AbC
1
ABC
8
ABC, ABc, AbC, Abc, aBC, aBc, abC, abc
(1)(1)
AaBb
(2)(2)
AABbCc
(1)(2)(2)
AABBCC
(1)(1)(1)
AaBbCc
(2)(2)(2)
2.
Use a Punnett Square to predict the genotypes and phenotypes of all of the offspring produced
by a cross between a pea plant with the genotype PpYyTt x PPyyTt, where P = purple
flowers, p = white flowers; Y = yellow seeds, y = green seeds; T = tall, t = dwarf. In your
answer, indicate the relative proportion of the offpspring with each genotype and phenotype.
Bio 2
PYT
PYt
PyT
Pyt
pYT
pYt
pyT
pyt
PyT
PPYyTT
PPYyTt
PPyyTT
PPyyTt
PpYyTT
PpYyTt
PpyyTT
PpyyTt
Pyt
PPYyTt
PPYytt
PPyyTt
PPyytt
PpYyTt
PpYytt
PpyyTt
Ppyytt
Genotypes:
Phenotypes:
PPYyTT: 1/16
Purple flowers, yellow seeds, tall
PPYyTt: 2/16
Purple flowers, yellow seeds, tall
PpYyTT: 1/16
Purple flowers, yellow seeds, tall
PpYyTt: 2/16
Purple flowers, yellow seeds, tall
PPyyTT: 1/16
Purple flowers, green seeds, tall
PPyyTt: 2/16
Purple flowers, green seeds, tall
PpyyTT: 1/16
Purple flowers, green seeds, tall
PpyyTt: 2/16
Purple flowers, green seeds, tall
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PPYytt: 1/16
Purple flowers, yellow seeds, short
PpYytt: 1/16
Purple flowers, yellow seeds, short
PPyytt: 1/16
Purple flowers, green seeds, short
Ppyytt: 1/16
Purple flowers, green seeds, short
2
2
The phenotypes can be quickly confirmed using the multiplication rule:
Purple flowers, yellow seeds, tall (P_Y_T_) = (1)(1/2)(3/4) = 6/16
Purple flowers, green seeds, tall (P_yyT_) = (1)(1/2)(3/4) = 6/16
Purple flowers, yellow seeds, short (P_Y_tt) = (1)(1/2)(1/4) = 2/16
Purple flowers, green seeds, short (P_yytt) = (1)(1/2)(1/4) = 2/16
3.
Use the multiplication rule to calculate the fraction of offspring from a cross between
AaBbCcDd and AaBbCcDd that would have the same phenotype as the two parents.
A_B_C_D_ = (3/4)(3/4)(3/4)(3/4) = 81/256
4.
Use the multiplication and addition rules to calculate the probability that a couple who plans to
have 4 children will have two boys and two girls (any order). Assume that the couple will
succeed in having exactly four children.
There are 6 possible orders. Each order has a probability of 1/16 (as shown for the
first possibility) since the chance of either sex in any given pregnancy is ½ and the
pregnancies are independent events.
GGBB = (1/2)(1/2)(1/2)(1/2) = 1/16
BBGG = (1/2)(1/2)(1/2)(1/2) = 1/16
GBGB = (1/2)(1/2)(1/2)(1/2) = 1/16
BGBG = (1/2)(1/2)(1/2)(1/2) = 1/16
BGGB = (1/2)(1/2)(1/2)(1/2) = 1/16
GBBG = (1/2)(1/2)(1/2)(1/2) = 1/16
The total probability therefore = (1/16 x 6 ) = 6/16
5.
About 1 in 25 Ashkenazai Jews are carriers (heterozygotes) for the autosomal recessive allele
that causes Tay Sachs Disease.
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a.
If two Ashkenazi Jews who have no family history of the disease decide to have
children, what is the probability that their first child will have sickle cell anemia?
Indicate how you arrived at your answer.
(1/25) = probability man is a carrier
(1/25) = probability woman is a carrier
(1/4) = probability two carriers’ first child will be affected
All of these events are independent of each other and all must be true for the
child to be affected. Therefore, the combined probability is:
(1/25)(1/25)(1/4) = 1/2500
b.
If the couple’s first child is affected with the disease, what is the probability that their
second child will also be affected? Indicate how you arrived at your answer.
(1) = probability man is a carrier
(1) = probability woman is a carrier
(1/4) = probability two carriers’ first child will be affected
(1)(1)(1/4) = 1/4
c.
If, after having one diseased child, the couple decides to have 3 more children, what is
the probability that at least one of the additional 3 children will also have Tay Sachs?
Indicate how you arrived at your answer. (Use the space at the top of the next
page for your answer.)
For at least one child to be affected, the following probabilities exist:
AAA = (1/4)(1/4)(1/4) = 1/64 (all children affected)
AAN = (1/4)(1/4)(3/4) = 3/64 (first two affected, third normal)
ANA = (1/4)(3/4)(1/4) = 3/64 (first and third affected, second normal)
NAA = (3/4)(1/4)(1/4) = 3/64 (first normal, second and third affected
ANN = (1/4)(3/4)(3/4) = 9/64 (first affected, second and third normal)
NAN = (3/4)(1/4)(3/4) = 9/64 (first and third normal, second affected)
NNA = (3/4)(3/4)(1/4) = 9/64 (first two normal, third affected)
The probability of any one of these mutually exclusive events occurring is the
sum of their individual probabilities = 37/64
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Note that the opposite of having at least one child affected is having no
children affected. So an easier way to approach this problem is to calculate
the probability that none of the 3 children will be affected and subtract this
number from 1.
NNN = (3/4)(3/4)(3/4) = 27/64
64/64 – 27/64 = 37/64
5.
X-linked dominant disease mutations are rare in humans, but some have been reported. One
such disease, hypertrichosis, causes excessive hair growth (all over the body, especially on
the face and hands). If a man with hypertrichosis has 6 girls and 2 boys with his unaffected
wife, how many of the children will be affected? What will the sex of the affected children be?
How can you tell? Indicate how you arrived at your answer, using a Punnett Square.
XH= hypertrichosis (dominant); Xh= normal
Xh
XH
Y
XHXh
XhY
All of the daughters will be affected because the disease is dominant and they will
all inherit their father’s affected X chromosome. None of the sons will be affected
because they inherit Y, not X, from their fathers.
6.
Jane has type A blood and is married to Andy, who has type B blood. Their first child, Steven,
has type A blood and their second child, Maury, has type B blood.
a.
What are the genotypes of Jane and Andy at the ABO blood locus? (Assume that Andy
is the biological father of both children and indicate how you arrived at your answer.)
Jane must be IAi and Andy must be IBi. Jane passed her IA allele to Steven
and Andy passed his i allele, which explains why Steven has Type A blood.
Jane passed her i allele to Maury and Andy passed his IB allele, which
explains why Maury has Type B blood. If Jane was IAIA she could not have
had a child with Type B blood. If Andy was IBIB he could not have had a child
with Type A blood.
b.
If the couple has additional children, could one of the children have a blood type that
would prove that Andy was not the father? Explain.
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In this family, no. All genotypes are possible:
7.
IA
i
IB
IAIB (Type AB)
IBi (Type B)
i
IAi (Type A)
ii (Type O)
Gabriella is 29 and her mother died of Huntington Disease, an autosomal dominant
neurological genetic disorder that has a late onset (about age 45). Gabriella does not want
genetic testing to see if she has the disease allele. However, she has two young daughters.
What is the probability that one (but not both) of the girls carry the disease allele? Indicate
how you arrived at your answer. Assume that Gabriella’s husband does not carry the allele.
Gabriella has a ½ chance of having the disease allele. If she is a carrier, there are
two ways that one (but not both) of her daughters could have the allele: AN and NA.
Each of these has a ¼ chance, and they are mutually exclusive, with a combined
probability of ½.
Probability Gabriella has the allele = ½
Probability one (but not both) of her daughters got the allele if she carries it: ½
Overall probability that one (but not both) of her daughters has the allele:
(1/2(1/2) = 1/4
8.
Gerard and Kate are both carriers for a recessive disease allele that causes Fanconi’s anemia,
a severe form of anemia that is usually fatal in childhood. The couple wants to start a family
with 4 children and have asked their genetic counselor to predict the probabilities of each of
the following outcomes.
a.
All four children are affected.
(1/4)(1/4)(1/4)(1/4) = 1/256
b.
Only one of the four children is affected.
Possible orders: NNNA, NNAN, NANN, ANNN, each with a probability of
(1/4)(3/4)3 = 27/256. Total probability is (4)(27/256) = 108/256
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c.
Two of the four children are affected.
Possible orders: AANN, NNAA, ANAN, NANA, NAAN, ANNA, each with a
probability of (1/4)2(3/4)2 = 9/256. Total probability is (6)(9/256) =
54/256
d.
Three of the four children are affected.
Possible orders: AAAN, AANA, ANAA, NAAA, with each order having a
probability of (1/4)3(3/4) = 3/256. Total probability is (4)(3/256) =
12/256
e.
None of the children is affected.
NNNN = (3/4)4 =81/256
f.
Add up all the probabilities you calculated for a-e. Do they add up to 1?
1/256 + 108/256 + 54/256 + 12/256 + 81/256 = 256/256 = 1
Note that the most probable outcome for this couple is that 1 in 4 of their
children will be affected. This makes sense since the statistical probability is
1 in 4 for each pregnancy.
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