Transcript Pedigrees - Talk Clickers

Inheritance Patterns and
Probability
July 2008
Pedigrees
I.
1
2
Dd, DD = normal
dd = deaf
II.
1
2
3
III.
1
This pedigree shows a family with a form of deafness that is inherited in
a recessive manner. Members of the family with filled symbols are
deaf. Which members of this family are definitely heterozygous
(Dd)?
A.
I-1 and I-2
B.
I-1, I-2, and II-1
C.
I-1, II-1, and II-3
D.
I-1, I-2, and II-3
E.
I-1, I-2, II-1, and II-3
I.
1
II.
1
III.
2
Dd
dd
2
dd
3
1
If II-2 and II-3 just had another baby boy. What is the chance that he is
deaf?
A.
1/8
B.
1/4
C.
1/2
D.
3/4
E.
1
family 2
family 1
I.
I.
Dd
Dd
1
II. Dd or
DD 1
Dd
1
2
Dd
dd
2
dd
III.
Dd
3
1
II.Dd or
DD
III.
1
2
dd
Dd
2
dd
3
1
What are the chances that II-1 from family 1 and II-1 from family 2 will have
a deaf child together?
A.
1/4
B.
1/9
C.
4/9
D.
1/16
Based on the pedigree above, which inheritance pattern can be
ruled out?
A. Autosomal dominant
B. Autosomal recessive
C. X-linked dominant
D. X-linked recessive
E. None of the above
Based on the pedigree above, which inheritance pattern can be
ruled out?
A.
B.
C.
D.
E.
Autosomal dominant
Autosomal recessive
X-linked dominant
X-linked recessive
None of the above
Based on the pedigree above, which inheritance pattern can be ruled out?
A.
X-linked dominant
B.
X-linked recessive
C.
neither of the above
?
Phenylketonuria (PKU) is an inherited disorder that can lead to mental
retardation if left untreated. PKU is inherited in an recessive manner.
What is the chance that the boy marked with a “?” in the pedigree
will have PKU?
A.
1/3
B.
1/4
C.
1/6
D.
1/8
I.
II.
III.
1
1
2
4
3
3
2
4
5
?
1
You would like to use mitochondrial DNA to try to determine if III-1 is a
member of the family shown in this pedigree. II-2 and II-3 are dead as
indicated with a slash and you are unable to collect mitochondrial DNA
from them.
If III-1 is a member of this family his mitochondrial DNA should match:
A) I-1 and II-1 only
B) I-1, I-2 and II-1 only
C) I-1, I-3, II-1, and II-4 only
D) I-3 and II-4 only
E) I-3, II-4, and II-5 only
Diastrophic dysplasia
Autosomal recessive
Achondroplasia
Autosomal dominant “A”
mutant allele
“D” normal allele
“d” mutant allele
Ron
Peggy
Gordon
Matt
Pat
“a” normal allele
Amy
What is Matt’s genotype (Matt has diastrophic dysplasia)?
A. ddaa
B. ddAa
C. Ddaa
D. DdAa
E. ddAA
Diastrophic dysplasia
Autosomal recessive
SLC26A2 gene
Achondroplasia
Autosomal dominant
FGFR3 gene
“D” normal allele
“A” mutant allele
“d” mutant allele
“a” normal allele
Ddaa
Ron
Peggy
Ddaa
Matt
ddaa
What is Ron’s genotype?
A. ddaa
B. Ddaa
C. DDaa
Gordon
Pat
Amy
DDAa (note AA embryos are not viable)
Diastrophic dysplasia
Autosomal recessive
SLC26A2 gene
Achondroplasia
Autosomal dominant
FGFR3 gene
“D” normal allele
“A” mutant allele
“d” mutant allele
“a” normal allele
Ddaa
Ron
Peggy
Ddaa
Matt
ddaa
What is Pat’s genotype?
A. DDAa
B. DDAA
C. Ddaa
D. None of the above
DDaa
Gordon
Pat
DDaa
Amy
DDAa (note AA embryos are not viable)
Diastrophic dysplasia
Autosomal recessive
SLC26A2 gene
Achondroplasia
Autosomal dominant
FGFR3 gene
“D” normal allele
“A” mutant allele
“d” mutant allele
Matt
ddaa
What is Zach’s
genotype?
A. Ddaa
B. DdAa
C. DdAA
Jeremy
“a” normal allele
Amy
DDAa (note AA embryos are not viable)
Zach
Molly
Jacob
Blue cootie disease is an autosomal recessive disorder that causes cooties
to be blue. Wild type cooties are red. R is the dominant color allele and r is
the recessive color allele.
?
?
What is the phenotype of the twins’ father?
A) RR
B) Rr
C) rr
D) red
Blue cootie disease is an autosomal recessive disorder that causes cooties
to be blue. Wild type cooties are red. R is the dominant color allele and r is
the recessive color allele.
?
?
What is the genotype of the twins’ father?
A) RR
B) Rr
C) rr
D) 1/2 Rr, 1/2 RR
Blue cootie disease is an autosomal recessive disorder that causes cooties
to be blue. Wild type cooties are red. R is the dominant color allele and r is
the recessive color allele.
?
?
What is the genotype of the twins' mother?
A) RR
B) Rr
C) ½ Rr, ¼ RR
D) 2/3 Rr, 1/3 RR
Blue cootie disease is an autosomal recessive disorder that causes cooties
to be blue. Wild type cooties are red. R is the dominant color allele and r is
the recessive color allele.
?
?
What is the probability that the first twin born
will have blue cootie disease?
A) 1/4
B) 1/3
C) 1/6
D) 0
The next few questions are not about pedigrees, but follow the cootie
example
Antennaless is an autosomal recessive disorder that leads to cooties
without antenna. A is the dominant WT allele and a is the mutant recessive
allele.
A true-breeding WT red cootie mates with a true-breeding blue antennaless
cootie.
P
X
What is the phenotype of the F1 generation?
A) All red with antennae
B) All red but half with antennae and half without
C) 9 red antennae: 3 red no antennae: 3 blue antennae:
1 blue no antennae
D) ¼ red with antennae, ¼ red without antennae,
¼ blue with antennae, ¼ blue without antennae
You allow the F1 generation to mate and produce offspring (F2 generation).
P
X
F1
RrAa
What is the probability that an F2 cootie will be
red?
A) 1/4
B) 1/2
C) 3/4
D) 1
Here is the F2 generation (RrAa X RrAa)
observed
expected
(O-E)2/E
1767
543
598
221
Do the red and antenna gene follow rules of independent assortment?
What is expected number of red cooties with
antennae?
A) 963
B) 1700
C) 1760
D) 2063
Here is the F2 generation (RrAa X RrAa)
observed
expected
1767
---
543
586.7
598
586.7
221
196
(O-E)2/E
Do the red and antenna gene follow rules of independent assortment?
What is (O-E)2/E for the blue antennaless group?
A) 625
B) 25
C) 3.2
D) 0.13
Here is the F2 generation (RrAa X RrAa) 3138 total
observed
expected
1767
---
0.02
543
586.7
3.3
598
586.7
0.21
221
196
---
(O-E)2/E
Do the red and antenna gene follow rules of independent assortment?
A) Yes, accept hypothesis – differences are likely due to chance
B) Yes, accept hypothesis – differences are not likely due to chance
C) No, reject hypothesis – differences are likely due to chance
D) No, reject hypothesis – differences are not likely due to chance
Calculating probability of inheritance
(monhybrid, dihybrid crosses)
Results of the F1 cross Yy X Yy
What is the phenotype of the circled green pea?
A) YY
B) Yy
C) yy
D) green
E) need more information
Results of the F1 cross Yy X Yy
What is the genotype of the circled yellow pea?
A) YY
B) Yy
C) yy
D) yellow
E) need more information
Plant 1: Yellow,
round peas
X
Plant 2: Green,
wrinkled peas
P:
F1:
1/2 Yellow, round
peas
Y - Yellow
y - Green
R - Round
r - wrinkled
1/2 Yellow, wrinkled
peas
What is the genotype of the yellow, round parent?
A:
B:
C:
D:
E:
YYRR
YyRR
YYRr
YyRr
Cannot be determined
Use Mendel’s Dihybrid cross results:
P
X
F1
F2
315
101
108
32
Given this data, what do you think the ratio of offspring is?
A: 3:1
B: 1:2:1
C: 9:3:3:1
D: 2:1
Results of the F1 cross Yy X Yy
1
2
3
4
The test cross that would most clearly distinguish the
genotype of the circled yellow pea is:
A) Yellow pea 1 X Yellow pea 2
B) Yellow pea 2 X Yellow pea 3
C) Yellow pea 2 X Green pea 4
D) You would need to do all of the above crosses
Genotype and phenotype
Genotypes
You cross a yellow with a green
and see a 50:50 ratio of green
and yellow progeny.
What is the genotype of the
original yellow pea?
A)
B)
C)
D)
YY
Yy
yy
Need more information
Phenotypes
Dihybrid cross
Mating between individuals that differ in two traits
Round, Yellow
P
Wrinkled, Green
X
RRYY
rryy
F1
100% Round, Yellow
RrYy
What are the possible gametes produced by the
F1 peas?
A) rryy, RrYy, RRYY
B) R, r, Y, y
C) Rr, Yy, RR, rr, YY, yy
D) RY, Ry, rY, ry
Dihybrid cross
F1
RrYy
X
RY
Ry
rY
ry
RY
RRYY RYRy RrYY RrYy
Ry
RRYy RRyy RrYy Rryy
RrYy
rY
ry
RrYY RrYy
rrYY
rrYy
RrYy
rrYy
rryy
Rryy
F2 Generation
Question 6: What fraction of the
F2 generation is green?
A) 1/16
B) 1/2
C) 1/9
D) 1/4
What is the phenotype ratio of progeny in F1
generation of the following cross?
Round, yellow
Wrinkled, green
X
RrYy
rryy
A
B
C
D
Round, Yellow
9
3
1
1
Wrinkled, Yellow
3
1
1
3
Round, Green
3
3
1
3
1
1
1
9
Wrinkled, Green
Can you use the outcome to
deduce the parental genotype?
• Suppose you cross a yellow and green and get
50% yellow and 50% green?
What are the parental genotypes?
A) YY X yy
B) Yy x yy
C) yy x yy
D) Yy x Yy
Monohybrid cross probability
• Consider Yy X Yy cross
What is the probability of
getting a Y from parent 1?
A)
B)
C)
D)
1/4
1/2
1
1/16
Monohybrid cross probability
• Consider Yy X Yy cross
What is the probability of
getting a Y from one
parent *AND* Y from the
other parent (i.e. YY)?
A)
B)
C)
D)
1/4
1/2
1
1/16
Monohybrid cross probability
• Consider Yy X Yy cross
What is the probability of
being Yellow (i.e. YY OR
Yy)?
A)
B)
C)
D)
1/4
1/2
3/4
1
Consider the following cross:
AaBBCcddEe X aabbCCDdEe
What is the probability their first offspring will be
aaBbCCDdee?
A)1/8
B)1/16
C)1/32
D)1/64
E)Cannot be determined
What is the probability of rolling a two OR a three
with one role of a six-sided die?
A)1/3
B)1/2
C)1/6
D)1/36
E)1/64
A male smurf has an dominant X-linked disorderthat causes red skin.
He marries smurfette (who is normal blue).
What are the possible phenotypes of their male children?
A) all blue skin
B) all red skin
C) patches of red and blue skin
D) more than one of the above
A male smurf has an dominant X-linked disorder that causes red skin.
He marries smurfette (who is normal blue).
What are the possible phenotypes of their female children?
A) all blue skin
B) all red skin
C) patches of red and blue skin
D) more than one of the above
Statistical Analysis of Crosses
Use Mendel’s Dihybrid cross results:
P
X
F1
F2
315
101
108
32
Total
seeds
observed =
556
3. Calculate Expected (e) numbers for each class if hypothesis correct
How many seeds would you expect to be green and round (to the nearest whole
number)?
A: 100
B: 104
C: 105
D: 108
3. Calculate Expected (e) numbers for each class if hypothesis correct
4. Calculate X2 = ∑ (o -e)2/e
Always use real numbers, not % or fraction
∑ means ’Sum of all classes’
Use a table:
Observed expected
(o-e)2/e
315
312
(315-312)2/312=0.029
101
104
0.087
108
104
0.154
32
35
0.257
X2 = 0.527
5. Calculate Degree of Freedom
A: 1
B: 4
C: 3
6. Look up probability (p) for X2 at a given df in the table
A: Accept the hypothesisB: Reject the hypothesis
How many degrees of freedom are there in the F2 generation
of the following cross?
P
X
F1
F2
315 101 108 32
A) 1
B) 2
C) 3
D) 4
E) 5
What if his results had been 5120 yellow and 2903 green?
Could Mendel still accept his hypothesis?
X2 = 535
A) Accept the hypothesis
B) Reject the hypothesis
What does P = 0.005 mean for the 28:20
ratio?
A) 28:20 is likely to be 3:1
B) 28:20 is NOT likely to be 3:1
C) 28:20 is not statistically “significant” and so
cannot be used to assess 3:1 ratio
D) This experiment is totally flawed and cannot
be interpretted
Exceptions to Mendel’s Laws
(maternal,
cytoplasmic/mitochondrial, sexlimited, co-dominance,
incomplete dominance, lethal,
epistatsis, heterozygous
advantage, imprinting)
Maternal
A maternal effect gene exists in a dominant N (normal) allele and a
recessive n (mutant) allele. What would be the ratios of genotypes and phenotypes
for the following cross?
nn female X NN male
A) all Nn, all normal
B) all Nn, all mutant
C) all nn, all mutant
D) all NN, all normal
A maternal effect gene exists in a dominant N (normal) allele and a
recessive n (mutant) allele. What would be the ratios of genotypes and phenotypes
for the following cross?
NN female X nn male
A) all Nn, all normal
B) all Nn, all mutant
C) all nn, all mutant
D) all NN, all normal
Worm Mel2 gene products are deposited into the egg by the mother and
are required from embryonic development. Mutations in the mel2 gene
are recessive and cause maternal effect embryonic lethality.
In a cross between mel2 heterozygotes, what percent of embryos will
die?
A) 100%
B) 50%
C) 25%
D) 0%
Zebrafish Ack15 gene products are deposited into the egg by the mother
and are required from embryonic development. Mutations in the ack15
gene are recessive and cause maternal effect embryonic lethality.
In a cross between ack15 homozygous mutant female and a
heterozygous male, what percent of embryos will die?
A) 100%
B) 50%
C) 25%
D) 0%
You are studying Drosophila mulleri and you discover a maternal effect
gene you call nanu. nanu mRNA is localized to the anterior end of the
embryo and promotes the formation of anterior structures. Mutations
in the nanu gene are recessive.
In a cross between two nanu heterozygotes, how many of the embryos will
have defects in their anterior structures?
A.
100%
B.
50%
C.
25%
D.
0%
You are studying Drosophila mulleri and you discover a maternal effect
gene you call nanu. nanu mRNA is localized to the anterior end of the
embryo and promotes the formation of anterior structures. Mutations
in the nanu gene are recessive.
In a cross between a nanu/nanu mutant female and a +/+ male, how many
of the embryos will have defects in their anterior structures?
A.
100%
B.
50%
C.
25%
D.
0%
Cytoplasmic/mitochondrial
Sex Limited
Co-dominance/incomplete
dominance
Variable Expression
Conditional
You cross a true-breeding white buffalo to a true breeding black buffalo.
All the F1 are brown.
P
F1
An F1 brown buffalo is crossed to the white parent. If they
have 4 offspring, how many do you predict will be white?
A. 0
B. 1
C. 2
D. 4
E. Not enough information
A true-breeding albino buffalo is crossed to a true-breeding
black buffalo and all of the progeny are brown. Crossing the
brown buffalos to each other yields an approximate ratio of 1
albino: 2 brown: 1 black.
The alleles for buffalo color show:
A) complete dominance
B) partial dominance
C) co-dominance
CU has asked the MCDB2150 class to help it with the breeding of Ralphie
buffalo. You do the following cross:
X
Long haired Ralphie
buffalo
Short haired Ralphie
buffalo
Ralphie buffalo with a mix of
short and long hairs
You try to establish a true breeding herd of Ralphie buffalo with a mix of short
and long hair using the F1 Ralphies but you are unsuccessful. Which mode of
interaction between alleles is a possible reason for your lack of success?
A. Codominance
B. Incomplete dominance
C. Complete dominance
D. Recessive epistasis
The blood type alleles in humans show
example(s) of:
A) co-dominance
B) complete dominance
C) multiple alleles
D) two of the above
E) all of the above
A man with blood type A and a woman with blood type B have a child
with blood type O. This couple can also have children with which blood
types?
A: O only
B: AB and O
C: A, B and O
D: A, B, AB and O
Charlie Chaplin (multiple alleles)
• Charlie was blood type O
• Girlfriend was blood type A
• Her (out-of-wedlock child) B
FACTS:
I locus (blood group) has 3 alleles
A = genotype IAIA or Iai -> Ab to B
B = genotype IBIB or Ibi -> Ab to A
AB = genotype IAIB
-> No Ab
O = genotype ii
-> Ab to both A and B
Mr. Chaplin’s case
•
•
A)
B)
C)
Given IA and IB are dominant to I
His girlfriend sued for paternity who won?
Girlfriend won - baby COULD be his
Chaplin won - baby COULD NOT be his
Hung jury, can’t tell from the facts
Only 66% of women with a heterozygous BRCA1
mutation get breast cancer by age 55 and most do
so in only one breast.
BRCA1 mutant allele…
A: shows incomplete penetrance
B: shows variable expressivity
C: both
Lethal
Lethal Alleles
All the sneetches want their children to have
stars on their bellies. The combination of
alleles that makes a black, starless sneetch is
lethal.
If a true breeding black star bellied sneetch
mates with a true breading yellow sneetch,
what is the probability that their first child will
have a star?
A) 1
B) 1/2
C) 1/4
D) 3/16
YYss X yySS
YY, Yy = yellow
yy = black
SS, Ss = star
ss = no star
Lethal Alleles
All the sneetches want their children to have
stars on their bellies. The combination of alleles that
makes a black, starless sneetch is lethal.
If two heterozygous yellow star-bellied sneetches mate,
what is the likelihood that their first child will not have a
star?
YySs X YySs
A) 1
B) 1/4
C) 1/5
D) 3/16
YS
Ys
yS
YS
YYSS
YYSs
YySS
Ys
YYSs
YYss
YySs
yS
YySs
YySs
yySS
ys
YySs
Yyss
yySs
ys
YySs
Yyss
yySs
yyss
Sue
SLC26A
gene
DdAa
DA
Zach
DdAa
SLC26A
gene
FGFR3
gene
ttp://images.icnetwork.co.uk/upl/icnewcastle/aug2005/4/8
/00015218-E719-12F1-88F40C01AC1BF814.jpg
FGFR3
gene
Da
dA
da
DA
DDAA
DDAa
DdAA
DdAa
Da
DDAa
DDaa
DdAa
Ddaa
dA
DdAA
DdAa
ddAA
ddAa
da
DdAa
Ddaa
ddAa
ddaa
What is the chance that Zach and Sue will have a child with
diastrophic dysplasia and achondroplasia?
A.
1/6
B.
1/8
C.
3/16
D.
9/16
E.
None of the above
Sue
SLC26A
gene
DdAa
DA
Zach
DdAa
SLC26A
gene
FGFR3
gene
ttp://images.icnetwork.co.uk/upl/icnewcastle/aug2005/4/8
/00015218-E719-12F1-88F40C01AC1BF814.jpg
FGFR3
gene
Da
dA
da
DA
DDAA
DDAa
DdAA
DdAa
Da
DDAa
DDaa
DdAa
Ddaa
dA
DdAA
DdAa
ddAA
ddAa
da
DdAa
Ddaa
ddAa
ddaa
What is the chance that Zach and Sue will have a child with only
diastrophic dysplasia?
A.
1/3
B.
1/6
C.
1/12
D.
1/16
Epistasis
P Brown X Yellow
F1
Black
F2 Brown: Black: Yellow
Partial Dominance Model
1: BB -> Brown
2: Bb -> Black
1: bb -> Yellow
Recessive epistasis Model
9 B_E_ -> Black
3 bbE_ -> Brown
4 _ _ee -> Yellow
Crossing F2 yellows to brown parent gave a mix of brown,
yellow and black. Which model does this support?
A: partial dominance
B: recessive epistasis
Epistasis and Labrador retriever coat color
The B locus determines if pigment can be produced
“B” codes for black pigment
“b” codes for brown pigment
The E locus determines if the pigment can be deposited in the
hair shaft
“E” allows dark pigment (black or brown) to be deposited
“e” prevents dark pigment from being
deposited
(the dogs are yellow)
What is the phenotype of a BbEe lab?
A.
Black
B.
Brown
C.
Yellow
The B locus determines if pigment can be produced
“B” codes for black pigment
“b” codes for brown pigment
The E locus determines if the pigment can be deposited in the
hair shaft
“E” allows dark pigment (black or brown) to be deposited
“e” prevents dark pigment from being
deposited
(the dogs are yellow)
What is the phenotype of a bbEe lab?
A.
Black
B.
Brown
C.
Yellow
The B locus determines if pigment can be produced
“B” codes for black pigment
“b” codes for brown pigment
The E locus determines if the pigment can be deposited in the
hair shaft
“E” allows dark pigment (black or brown) to be deposited
“e” prevents dark pigment from being
deposited
(the dogs are yellow)
What is the phenotype of a bbee lab?
A.
Black
B.
Brown
C.
Yellow
The B locus determines if pigment can be produced
“B” codes for black pigment
“b” codes for brown pigment
The E locus determines if the pigment can be deposited in the
hair shaft
“E” allows dark pigment (black or brown) to be deposited
“e” prevents dark pigment from being
deposited
(the dogs are yellow)
What is the phenotype of a BBee lab?
A.
Black
B.
Brown
C.
Yellow
You cross a hairless mouse aaBB to a mouse with curly
hair AAbb. All of the F1s have straight hair. You cross two
of the F1 mice together. In the F2 generation: 18 mice
have straight hair, 8 mice are hairless, and 6 have curly
hair.
What is the phenotype of the aabb mice in the F2
generation?
A. hairless
B. curly hair
C. straight hair
hairless mouse aaBB X curly hair AAbb
F1s have straight hair AaBb
F2
9 A- B4 aa -3 A- bb
18 straight hair mice A-B8 hairless mice aaB- & aabb
6 curly hair mice A-bb
What is the order of function?
A. A, then B
B. B, then A
C. A and B act simultaneously
D. Not enough data
Recessive epistasis Model
9 B_E_ -> Black
3 bbE_ -> Brown
4 _ _ee -> Yellow
What is the order of function?
A. B, then E
B. E, then B
C. E and B act simultaneously
D. Not enough data
Mutant strain A: intermediate 2 builds up
Mutant strain B: intermediate 3 builds up
Mutant strain C: intermediate 1 builds up
If gene A is epistatic to gene B which intermediate will build up
in a AB double mutant?
A) intermediate 1
B) intermediate 2
C) intermediate 3
Epistasis
AA – fluffy hair
aa – bald
BB – red hair pigment
bb – no red pigment
(blue hair)
X
AaBb
AaBb
What are the possible gametes produced by each chuzzel?
A) Aa, Bb
B) AaBb
C) A, a, B, b
D) AB, Ab, aB, ab
1.You cross a true-breeding white buffalo to a true-breeding tan buffalo in the P generation
2. You find out that all of the F1 buffalos are gold
3. You cross two of the F1 gold buffalo together and in the F2 generation get:
91
47
33
The T locus determines if pigment can be produced:
“T” codes for gold pigment
“t” codes for tan pigment
The A locus determines if pigment can be deposited into the hair shaft:
“A” allows pigment (gold or tan) to be deposited into the hair shaft
“a” prevents pigment from being deposited into the hair shaft (the buffalo are white)
What is the genotype of the tan buffalo in the P generation?
A.
AAtt
B.
Aatt
C.
Either AAtt or Aatt
1.You cross a true-breeding white buffalo to a true-breeding tan buffalo in the P generation
aaTT
AAtt
2. You find out that all of the F1 buffalos are gold
AaTt
3. You cross two of the F1 gold buffalo together and in the F2 generation get:
91
47
33
The T locus determines if pigment can be produced:
“T” codes for gold pigment
“t” codes for tan pigment
The A locus determines if pigment can be deposited into the hair shaft:
“A” allows pigment (gold or tan) to be deposited into the hair shaft
“a” prevents pigment from being deposited into the hair shaft (the buffalo are white)
What is the genotype of the white buffalo in the P generation?
A.
aatt
B.
aaTT
C.
aaTt
X
2. You find out that all of the F1 buffalos are gold
AaTt
AaTt
3. You cross two of the F1 gold buffalo together and in the F2 generation get:
91
47
33
Punnett square for cross of two gold buffalo AaTt x AaTt:
AT
At
aT
at
AT
At
aT
at
AATT
AATt
AaTT
AaTt
AATt
AaTT
AaTt
AAtt
AaTt
Aatt
AaTt
aaTT
aaTt
Aatt
aaTt
aatt
How many of the
genotypes in the
Punnett square will
result in a gold
buffalo?
A.
9
B.
4
C.
3
D.
2
E.
1
X
2. You find out that all of the F1 buffalos are gold
AaTt
AaTt
3. You cross two of the F1 gold buffalo together and in the F2 generation get:
91
AT
At
aT
at
47
33
AT
At
aT
at
AATT
AATt
AaTT
AaTt
AATt
AaTT
AaTt
AAtt
AaTt
Aatt
AaTt
aaTT
aaTt
Aatt
aaTt
aatt
How many of the
genotypes in the
Punnett square will
result in a white
buffalo?
A.
9
B.
4
C.
3
D.
2
E.
1
X
2. You find out that all of the F1 buffalos are gold
AaTt
AaTt
3. You cross two of the F1 gold buffalo together and in the F2 generation get:
91
AT
At
aT
at
47
33
AT
At
aT
at
AATT
AATt
AaTT
AaTt
AATt
AaTT
AaTt
AAtt
AaTt
Aatt
AaTt
aaTT
aaTt
Aatt
aaTt
aatt
How many of the
genotypes in the
Punnett square will
result in a tan
buffalo?
A.
9
B.
4
C.
3
D.
2
E.
1
3. You cross two of the F1 gold buffalo together and in the F2 generation get:
91
AT
At
47
33
AT
At
aT
at
AATT
AATt
AaTT
AaTt
AATt
AAtt
AaTt
Aatt
Approximately how
many true-breeding
tan buffalo are in
the F2 generation in
your herd?
A. 1
B. 3
aT
AaTT
AaTt
aaTT
aaTt
C. 11
D. 22
at
AaTt
Aatt
aaTt
aatt
E. 33
Heterozygous Advantage
A man who does not have sickle cell anemia and has no history
of it in his family (assume he is not a carrier) marries a woman
who has sickle cell anemia. They have a son.
This family is planning to travel to the Solomon Islands. Which
family member(s) should take Lariam, a very expensive drug that
prevents malaria?
A. Father
B. Mother
C. Son
D. Father and the son
E. Everyone
Imprinting
A’a’
Aa
1
2
3
4
A’a
Aa’
aa’
AA’
A mutation (a) occurs on an imprinted gene (A). The maternal copy of the gene is
methylated and not expressed. ‘ denotes the alleles inherited from the father.
Which of the offspring will be affected?
A) 1 and 3
B) 2 and 3
C) 3 only
D) none of the offspring will be affected
Prader-Willi if the deletion is on the chromosome inherited from mom.
Angelman syndrome if the deletion is on the chromosome inherited from
dad.
A is the normal chromosome 15 a contains the deletion
A is from mom A’ is from dad
A’a’
Aa
1
2
3
A’a
AA’
aa’
Which disorder does the mother have?
A) None
B) PWS
C) AS
D) need more information
Prader-Willi if the deletion is on the chromosome inherited from mom.
Angelman syndrome if the deletion is on the chromosome inherited from
dad.
A is the normal chromosome 15 a contains the deletion
A is from mom A’ is from dad
A’a’
Aa
1
2
3
A’a
AA’
aa’
Which disorder does the offspring 1 have?
A) None
B) PWS
C) AS
D) need more information
Prader-Willi if the deletion is on the chromosome inherited from mom.
Angelman syndrome if the deletion is on the chromosome inherited from
dad.
A is the normal chromosome 15 a contains the deletion
A is from mom A’ is from dad
A’a’
Aa
1
2
3
A’a
AA’
aa’
Which disorder does the offspring 3 have?
A) both syndromes
B) PWS
C) AS
D) need more information
Prader-Willi if the deletion is on the chromosome inherited from mom.
Angelman syndrome if the deletion is on the chromosome inherited from
dad.
An individual with AS married a normal individual and produced an
offspring with PWS. What is the gender of the parent with AS?
A) male
B) female
C) need more information
What is the gender of the child with PWS?
A) male
B) female
C) need more information
Dominance vs. recessive
Chuzzle Population
•A chuzzle populations contains 640 red chuzzels and 320 green
chuzzles.
•Chuzzles are not choosy about their mates. Either color will mate
with the other at equal frequencies.
•When red chuzzles mate all the pups are red. When red and green
chuzzles mate some pups are red and some are green.
•There is no advantage (for mating or survival) based on color.
Which trait is dominant?
A) red
B) green