Download BIO152 Genetics problems Tutorial 8 outline

BIO152 Genetics problems
Tutorial 8
November 10
11/9/2006
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Tutorial 8 outline
11/9/2006
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Lec 17-Question 3
A B/a b x ?? for testcross
(what is the testcross_________).
If the two loci are 10 m.u. apart, what
proportion of progeny will be A B/a b?
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3- answer
You perform the following cross and are told that
the two genes are 10 m.u. apart.
A B/a b x a b/a b
Among their progeny, 10 percent should be
recombinant (A b/a b and a B/a b) and 90
percent should be parental (A B/a b & a b/a b).
Therefore, A B/a b should represent 1/2 of the
parentals or 45 percent.
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Lecture 17-Question 4
R and S are loci 35 m.u. apart. If the plant
of genotype R S/r s is selfed, what
progeny phenotypes will be seen and in
what proportions?
(once you know the genotype you can
determine the phenotype proportions)
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4- answer
P
R S/r s x R S/r s
gametes
genotypes & expected
proportions:
1/2 (1 – 0.35)R S parental
1/2 (1 – 0.35)r s
parental
1/2 (0.35) R s recombinant
1/2 (0.35) r S recombinant
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4-Answer continued F1
genotypes
0.1056 R S/R S
0.1056 r s/r s
0.2113 R S/r s
0.1138 R S/r S
0.1138 R S/R s
Phenotypes
0.6058 R S
0.1056 r s
0.1444 R s
0.1444 r S
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0.1138 r s/r S
0.1138 r s/R s
0.0306 R s/R s
0.0306 r S/r S
0.0613 R s/r S
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Pedigree from Lec 16
I
II
III
IV
If dominant what is the chance that III6 will have
affected children?
If recessive who are obligate carriers
(heterozygous)
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Who are at risk of being carriers?
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I
II
III
IV
If dominant what is the chance that III6 will have affected children?
zero
If recessive who are obligate carriers (heterozygous)
I2,
II1, II5, II6, II8,
III1, III2, III4, III5, III6, III7,
IV1
Who are at risk of being carriers?
III11, III12 all with chance 50%
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III10,
10
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Construct a pedigree (Lec 16)
Alice and Bob have a two year old son, Charles, who is showing mental
retardation, short stature, micropenis, and cryptorchidism.
Alice has two living, unaffected, brothers but her eldest brother died at
age 9 and a second brother died aged 10 months. Both had similar
problems to Charles.
Alice's father, David, who was symptomless, has a sister, Ethel, who
has an unaffected boy and girl, and a brother, Fred, who also has
two unaffected children.
Alice's mother, Gertrude, has two living sisters and had a brother who
had died in childhood and who, she remembers, had been mentally
retarded.
Bob has two brothers, Henry and Ignatius, who are still unmarried. His
parents, John and Kate, had tragic lives, both were adopted and
never knew their biological parents and both died as the result of a
road accident.
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Nuclear family in question
A = Alice mom
B = Bob dad
C = affected son
Charles
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• Step 1
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Alice’s (A) family
Alice & her 4 brothers are
connected vertically to a
horizontal line which is, in turn,
connected to the line drawn
between her parents David
and Gertrude.
Her two dead brothers (whom we
presume died of the same
genetic disease - though this
can sometimes be a foolish
assumption without medical
evidence) are shaded in (to
show that they suffered from
the disease) and are crossed
through (to show that they are
dead).
Step 2
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Construct a pedigree-Step 3
• Now add (Alice’s mother) Gertrude's
siblings to the pedigree.
• And (Alice’s father) David's siblings and
his nephews & nieces
• Finally add Bob's side of the family
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7
What is the most likely pattern?
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(Lec 16) Inheritance pattern?
#1
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8
answer
#1autosomal dominant
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(Lec 16) Inheritance pattern?
#2
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9
Answer
#2 autosomal recessive
I3 & I4 must be carriers
to have an affected
son
Affects both sexes
equally
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#3
www.langara.bc.ca/.../mario/Assets/pedigr
ee3.gif
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#3 autosomal recessive
www.langara.bc.ca/.../mario/Assets/pedigr
ee3.gif
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What is the most likely pattern of
inheritance?
Autosomal recessive ?
Autosomal dominant ?
X-linked recessive ?
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11
answer
Autosomal dominant:
• roughly half of the offspring of affected persons
are themselves affected,
• that every affected person has an affected
parent
• X linkage is ruled out by male to male
transmission.
• the other two answers were possible but
unlikely.
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Probability of Autosomal recessive
• three people half shaded would all need to be
carriers marrying into the family. This is possible
but how likely it is depends on the frequency of
carriers in the population.
• disease affects one birth in 10,000 how frequently
will random members of the population turn out to
be carriers?
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12
To calculate the frequency of carriers of a condition with a
frequency of homozygotes of one in 10,000 you need to
use the Hardy Weinberg formulae:
P2 +2pq + q2 =1 p+q =1
q2 = 1/10,000 = .0001 q= .01 p= .99 2pq=.02
So in that case the likelihood of three
carriers is approximately (1/50)3 which is 1
in 125,000 which is unlikely
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Probability of X linked recessive?
Which of these two facts argues more strongly
against X linkage?
1. The transmission of the condition from
male to male, (I3 to II7)
2. The transmission of the condition to
females
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1. The transmission of the condition from
male to male, (I3 to II7)
it is almost impossible for a father to pass
his X chromosome to a son, and
certainly not to a fertile son.
[two possible exceptions to this rule, XX
males and XXY Klinefelter's syndrome.
Both are rare and both are infertile.
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The transmission of the condition to
females
frequency of this genetic disease is1 in
10,000, how likely is it that a female
marrying into this family is a carrier of this
condition if it is X linked?
the disease frequency is 1 person in 10,000
is almost the same as saying 1 male in
5,000
use Hardy-Weinberg to estimate the
frequency of carrier females
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The gene frequency 1 in 5000 (q in the
Hardy Weinberg formula)
In the case of X linked genes, males come
in two types, normal and mutant, at
frequencies p and q. Females, who have
two X chromosomes, come in three types,
homozygous normal (frequency p2),
heterozygotes (frequency 2pq) and
homozygous mutant (frequency q2).
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if this gene is X linked then the frequency of carrier
females would be about 0.0004 or 1 in 2500
The two people half shaded must have been
heterozygotes. This is possible but not likely.
Chance would be about 1/2500 × 1/2500 which
is 1 in 6.25 million.
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Genetic questions posted for Tutorial 6
1.
A sexually reproducing organism is heterozygous for
two genes located on different chromosomes, one for
ear shape and one for toe length. Its genotype is
AaBb. Which of the following genotypes is most
probable in a gamete from this organism?
a.
b.
c.
d.
e.
AB
AaBb
Aa
Bb
A
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1-Answer
1.
A sexually reproducing organism is heterozygous for
two genes located on different chromosomes, one for
ear shape and one for toe length. Its genotype is
AaBb. Which of the following genotypes is most
probable in a gamete from this organism?
a.
b.
c.
d.
e.
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AB
AaBb
Aa
Bb
A
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#5
#5. Two mice are heterozygous for albinism (Aa).
The dominant allele (A) codes for normal
pigmentation, and the recessive allele (a)
codes for no pigmentation. What percentage of
their offspring will show albinism?
a. 25 percent
b. 50 percent
c. 75 percent
d. 100 percent
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5-Answer
#5. Two mice are heterozygous for albinism (Aa).
The dominant allele (A) codes for normal
pigmentation, and the recessive allele (a)
codes for no pigmentation. What percentage of
their offspring will show albinism?
a. 25 percent
b. 50 percent
c. 75 percent
d. 100 percent
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#20
#20 A man and woman are both of normal
pigmentation, but both have one parent who is
albino (without melanin pigmentation). Albinism
is an autosomal (not sex-linked) recessive trait.
What is the probability that their first child will be
an albino?
A.
B.
C.
D.
E.
0%
1/8
1/2
1/4
100%
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Answer
#20 A man and woman are both of normal pigmentation,
but both have one parent who is albino (without
melanin pigmentation). Albinism is an autosomal (not
sex-linked) recessive trait. What is the probability that
their first child will be an albino?
A. 0%
B. 1/8
C. 1/2
D. ¼ (what about the 1st child albino AND a girl?)
E. 100%
What is the probability of their second child being an
albino?
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#7
#7 If the first five seeds (offspring) grown from a
cross between two heterozygous parent peas
with the genotype Rr are all round, what is the
probability that the next offspring will be
wrinkled?
a. 0%
b. 100%
c. 25%
d. Cannot determine from the information given
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7-Answer
#7 If the first five seeds (offspring) grown from a
cross between two heterozygous parent peas
with the genotype Rr are all round, what is the
probability that the next offspring will be
wrinkled?
a. 0%
b. 100%
c. 25%
d. Cannot determine from the information given
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#16 & #17
If an organism is heterozygous for two traits that are
linked, how many different genotypes are possible in
the gametes?
#16. Assume no crossing over occurs.
#17. if even a single crossing over event occurs?
a.
b.
c.
d.
1
2
4
8
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16&17-ANSWER
If an organism is heterozygous for two traits
that are linked, how many different
genotypes are possible in the gametes?
#16. Assume no crossing over occurs.
b. 2
#17. if even a single crossing over event
occurs?
C. 4
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Another problem…
Suppose two AaBbCc individuals are mated.
Assuming that the genes are not linked, what
fraction of the offspring are expected to be
homozygous recessive for the three traits?
A. 1/4
B. 1/8
C.1/16
D.1/64
E. 1/256
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Another problem…
Suppose two AaBbCc individuals are mated.
Assuming that the genes are not linked, what
fraction of the offspring are expected to be
homozygous recessive for the three traits?
A. 1/4
B. 1/8
C.1/16
D.1/64 [ ¼ x ¼ x ¼ ]
E. 1/256
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How many unique gametes could be produced
through independent assortment by an
individual with the genotype AaBbCCDdEE?
A. 32
B. 4
C. 16
D. 8
E. 1/64
[hint: apply the formula (2n), where n represents
the number of heterozygous gene pairs.]
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Hint
apply the formula (2n), where n represents
the number of heterozygous gene pairs.
Aa BbCCDdEE = 23 =8 possible unique
gametes
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22