Post on 26-Dec-2015
Inheritance Patterns and Probability
July 2008
Pedigrees
1 2
21 3
1
I.
II.
III.
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
Dd, DD = normaldd = deaf
1 2
21 3
1
I.
II.
III.
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
Dddd
dd
1 2
21 3
1
I.
II.
III.
dd
dd
1 2
21 3
1
I.
II.
III.
dd
dd
Dd Dd
Dd or DD
Dd Dd
Dd or DDDd Dd
family 1family 2
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. Autosomal dominantB. Autosomal recessiveC. X-linked dominantD. X-linked recessiveE. 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 2
1
3 4
12 3 4 5
?
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 onlyC) I-1, I-3, II-1, and II-4 onlyD) I-3 and II-4 only E) I-3, II-4, and II-5 only
Diastrophic dysplasiaAutosomal recessive
“D” normal allele
“d” mutant allele
Matt Amy
Ron Peggy Gordon Pat
What is Matt’s genotype (Matt has diastrophic dysplasia)?
A. ddaa
B. ddAa
C. Ddaa
D. DdAa
E. ddAA
Achondroplasia Autosomal dominant “A” mutant allele
“a” normal allele
Diastrophic dysplasiaAutosomal recessive SLC26A2 gene
“D” normal allele
“d” mutant allele
Matt Amy
Ron Peggy Gordon Pat
ddaa
What is Ron’s genotype?
A. ddaa
B. Ddaa
C. DDaa
Achondroplasia Autosomal dominant FGFR3 gene
“A” mutant allele
“a” normal allele
DDAa (note AA embryos are not viable)
Ddaa Ddaa
Diastrophic dysplasiaAutosomal recessive SLC26A2 gene
“D” normal allele
“d” mutant allele
Matt Amy
Ron Peggy Gordon Pat
ddaa
What is Pat’s genotype?
A. DDAa
B. DDAA
C. Ddaa
D. None of the above
Achondroplasia Autosomal dominant FGFR3 gene
“A” mutant allele
“a” normal allele
DDAa (note AA embryos are not viable)
Ddaa Ddaa DDaa DDaa
Diastrophic dysplasiaAutosomal recessive SLC26A2 gene
“D” normal allele
“d” mutant alleleMatt Amy
Jeremy Zach Molly Jacob
ddaa
What is Zach’s genotype?
A. Ddaa
B. DdAa
C. DdAA
Achondroplasia Autosomal dominant FGFR3 gene
“A” mutant allele
“a” normal allele
DDAa (note AA embryos are not viable)
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) RRB) RrC) rrD) 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) RRB) RrC) rrD) 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) RRB) RrC) ½ Rr, ¼ RRD) 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 bornwill have blue cootie disease?A) 1/4B) 1/3C) 1/6D) 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.
What is the phenotype of the F1 generation?A) All red with antennae B) All red but half with antennae and half withoutC) 9 red antennae: 3 red no antennae: 3 blue antennae: 1 blue no antennaeD) ¼ red with antennae, ¼ red without antennae, ¼ blue with antennae, ¼ blue without antennae
XP
You allow the F1 generation to mate and produce offspring (F2 generation).
What is the probability that an F2 cootie will bered? A) 1/4B) 1/2C) 3/4D) 1
XP
RrAa
F1
Here is the F2 generation (RrAa X RrAa)
What is expected number of red cooties with antennae?A) 963B) 1700C) 1760D) 2063
1767
543
598
221
observed expected (O-E)2/E
Do the red and antenna gene follow rules of independent assortment?
Here is the F2 generation (RrAa X RrAa)
What is (O-E)2/E for the blue antennaless group?A) 625B) 25C) 3.2D) 0.13
1767 ---
543 586.7
598 586.7
221 196
observed expected (O-E)2/E
Do the red and antenna gene follow rules of independent assortment?
Here is the F2 generation (RrAa X RrAa) 3138 total
Do the red and antenna gene follow rules of independent assortment?A) Yes, accept hypothesis – differences are likely due to chanceB) Yes, accept hypothesis – differences are not likely due to chanceC) No, reject hypothesis – differences are likely due to chanceD) No, reject hypothesis – differences are not likely due to chance
1767 --- 0.02543 586.7 3.3598 586.7 0.21221 196 ---
observed expected (O-E)2/E
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) YYB) YyC) yyD) greenE) need more information
Results of the F1 cross Yy X Yy
What is the genotype of the circled yellow pea?A) YYB) YyC) yyD) yellowE) need more information
What is the genotype of the yellow, round parent?Y - Yellowy - Green
R - Roundr - wrinkled A: YYRR
B: YyRRC: YYRrD: YyRrE: Cannot be determined
Plant 1: Yellow, round peas
Plant 2: Green, wrinkled peasX
F1:
1/2 Yellow, round peas
1/2 Yellow, wrinkled peas
P:
Use Mendel’s Dihybrid cross results:
XP
F1
F2
315 101 108 32
Given this data, what do you think the ratio of offspring is?
A: 3:1B: 1:2:1C: 9:3:3:1D: 2:1
Results of the F1 cross Yy X Yy
The test cross that would most clearly distinguish the genotype of the circled yellow pea is:A) Yellow pea 1 X Yellow pea 2B) Yellow pea 2 X Yellow pea 3C) Yellow pea 2 X Green pea 4D) You would need to do all of the above crosses
1 2 3 4
Genotype and phenotype
PhenotypesGenotypes
You cross a yellow with a green and see a 50:50 ratio of greenand yellow progeny. What is the genotype of theoriginal yellow pea?
A) YYB) YyC) yyD) Need more information
Dihybrid cross
Mating between individuals that differ in two traits
X
Round, Yellow Wrinkled, Green
RRYY rryy
What are the possible gametes produced by the F1 peas?A) rryy, RrYy, RRYYB) R, r, Y, yC) Rr, Yy, RR, rr, YY, yyD) RY, Ry, rY, ry
P
F1
RrYy
100% Round, Yellow
Dihybrid cross
F1
X
RrYy
RrYy
RY Ry rY ry
RY
Ry
rY
ry
F2 Generation
RRYY RYRy RrYY RrYy
RRYy RRyy RrYy Rryy
RrYY RrYy rrYY rrYy
RrYy Rryy rrYy rryy
Question 6: What fraction of theF2 generation is green?
A) 1/16B) 1/2C) 1/9D) 1/4
What is the phenotype ratio of progeny in F1 generation of the following cross?
RrYy rryy
X
Round, yellow Wrinkled, green
Round, Yellow
Wrinkled, Yellow
Round, Green
Wrinkled, Green
9
3
3
1
3
1
3
1
1
1
1
1
1
3
3
9
A B C D
Can you use the outcome todeduce 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 yyB) Yy x yyC) yy x yyD) Yy x Yy
Monohybrid cross probability
• Consider Yy X Yy cross
What is the probability ofgetting a Y from parent 1?
A) 1/4B) 1/2C) 1D) 1/16
Monohybrid cross probability
• Consider Yy X Yy cross
What is the probability ofgetting a Y from oneparent *AND* Y from theother parent (i.e. YY)?
A) 1/4B) 1/2C) 1D) 1/16
Monohybrid cross probability
• Consider Yy X Yy cross
What is the probability ofbeing Yellow (i.e. YY ORYy)?
A) 1/4B) 1/2C) 3/4D) 1
Consider the following cross:AaBBCcddEe X aabbCCDdEeWhat is the probability their first offspring will be aaBbCCDdee?
A)1/8B)1/16C)1/32D)1/64E)Cannot be determined
What is the probability of rolling a two OR a three with one role of a six-sided die?
A)1/3B)1/2C)1/6D)1/36E)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 skinB) all red skinC) patches of red and blue skinD) 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 skinB) all red skinC) patches of red and blue skinD) more than one of the above
Statistical Analysis of Crosses
Use Mendel’s Dihybrid cross results:
XP
F1
F2
315 101 108 32
3. Calculate Expected (e) numbers for each class if hypothesis correctHow many seeds would you expect to be green and round (to the nearest whole
number)?A: 100B: 104C: 105D: 108
Total seeds observed = 556
Observed expected (o-e)2/e
5. Calculate Degree of Freedom
3. Calculate Expected (e) numbers for each class if hypothesis correct
4. Calculate X2 = ∑ (o -e)2/eAlways use real numbers, not % or fraction
∑ means ’Sum of all classes’
Use a table:
X2 =
315 312 (315-312)2/312=0.029101 104 0.087108 104 0.15432 35 0.257
0.527
A: 1B: 4C: 3
6. Look up probability (p) for X2 at a given df in the table
A: Accept the hypothesis B: Reject the hypothesis
XP
F1
F2
315 101 108 32
How many degrees of freedom are there in the F2 generation of the following cross?
A) 1B) 2C) 3D) 4E) 5
X2 = 535
A) Accept the hypothesisB) Reject the hypothesis
What if his results had been 5120 yellow and 2903 green?Could Mendel still accept his hypothesis?
What does P = 0.005mean for the 28:20 ratio?
A) 28:20 is likely to be 3:1B) 28:20 is NOT likely to be 3:1C) 28:20 is not statistically “significant” and so
cannot be used to assess 3:1 ratioD) This experiment is totally flawed and cannot
be interpretted
Exceptions to Mendel’s Laws (maternal,
cytoplasmic/mitochondrial, sex-limited, 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 phenotypesfor the following cross?
nn female X NN male
A) all Nn, all normalB) all Nn, all mutantC) all nn, all mutantD) 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 phenotypesfor the following cross?
NN female X nn male
A) all Nn, all normalB) all Nn, all mutantC) all nn, all mutantD) 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 ExpressionConditional
You cross a true-breeding white buffalo to a true breeding black buffalo. All the F1 are brown.
An F1 brown buffalo is crossed to the white parent. If they have 4 offspring, how many do you predict will be white?
A. 0B. 1C. 2D. 4E. Not enough information
P
F1
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 dominanceB) partial dominanceC) co-dominance
CU has asked the MCDB2150 class to help it with the breeding of Ralphie buffalo. You do the following cross:
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. CodominanceB. Incomplete dominanceC. Complete dominance D. Recessive epistasis
X
Long haired Ralphie buffalo
Short haired Ralphie buffalo
Ralphie buffalo with a mix of short and long hairs
The blood type alleles in humans show example(s) of: A) co-dominanceB) complete dominanceC) multiple allelesD) two of the aboveE) 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 onlyB: AB and OC: A, B and OD: A, B, AB and O
Charlie Chaplin (multiple alleles)
• Charlie was blood type O• Girlfriend was blood type A• Her (out-of-wedlock child) BFACTS:I locus (blood group) has 3 allelesA = genotype IAIA or Iai -> Ab to BB = genotype IBIB or Ibi -> Ab to AAB = genotype IAIB -> No AbO = genotype ii -> Ab to both A and B
Mr. Chaplin’s case
• Given IA and IB are dominant to I• His girlfriend sued for paternity who won?A) Girlfriend won - baby COULD be hisB) Chaplin won - baby COULD NOT be hisC) 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 penetranceB: shows variable expressivityC: both
Lethal
Lethal AllelesAll the sneetches want their children to havestars 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?
YYss X yySSA) 1B) 1/2C) 1/4D) 3/16
YY, Yy = yellowyy = blackSS, Ss = starss = no star
Lethal AllelesAll the sneetches want their children to havestars 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) 1B) 1/4C) 1/5D) 3/16
YS Ys yS ysYS YYSS YYSs YySs YySsYs YYSs YYss YySs YyssyS YySS YySs yySS yySsys YySs Yyss yySs yyss
DdAa
ttp://images.icnetwork.co.uk/upl/icnewcastle/aug2005/4/8/00015218-E719-12F1-88F40C01AC1BF814.jpg
Zach
Sue
DdAa
DA
Da
dA
da
DA Da dA da
DDAA
DDAa
DdAA
DdAa
DDAa
DDaa
DdAa
Ddaa
DdAA
DdAa
ddAA
ddAa
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
SLC26A gene
FGFR3 gene
SLC26A gene
FGFR3 gene
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
DdAa
ttp://images.icnetwork.co.uk/upl/icnewcastle/aug2005/4/8/00015218-E719-12F1-88F40C01AC1BF814.jpg
Zach
Sue
DdAa
DA
Da
dA
da
DA Da dA da
DDAA
DDAa
DdAA
DdAa
DDAa
DDaa
DdAa
Ddaa
DdAA
DdAa
ddAA
ddAa
DdAa
Ddaa
ddAa
ddaaSLC26A gene
FGFR3 gene
SLC26A gene
FGFR3 gene
Epistasis
Crossing F2 yellows to brown parent gave a mix of brown, yellow and black. Which model does this support?
A: partial dominanceB: recessive epistasis
Partial Dominance Model1: BB -> Brown2: Bb -> Black1: bb -> Yellow
Recessive epistasis Model9 B_E_ -> Black3 bbE_ -> Brown4 _ _ee -> Yellow
P Brown X Yellow
F1 Black
F2 Brown: Black: Yellow
Epistasis and Labrador retriever coat colorThe 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. hairlessB. curly hairC. straight hair
hairless mouse aaBB X curly hair AAbb
F1s have straight hair AaBb
F2 9 A- B-4 aa -- 3 A- bb
What is the order of function?
A. A, then BB. B, then AC. A and B act simultaneouslyD. Not enough data
18 straight hair mice A-B-
8 hairless mice aaB- & aabb
6 curly hair mice A-bb
What is the order of function?
A. B, then EB. E, then BC. E and B act simultaneouslyD. Not enough data
Recessive epistasis Model9 B_E_ -> Black3 bbE_ -> Brown4 _ _ee -> Yellow
If gene A is epistatic to gene B which intermediate will build upin a AB double mutant?
A) intermediate 1B) intermediate 2C) intermediate 3
Mutant strain A: intermediate 2 builds up
Mutant strain B: intermediate 3 builds up
Mutant strain C: intermediate 1 builds up
Epistasis AA – fluffy hairaa – baldBB – red hair pigmentbb – no red pigment (blue hair)
X
AaBb AaBb
What are the possible gametes produced by each chuzzel?A) Aa, BbB) AaBbC) A, a, B, bD) 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
91 47 33
3. You cross two of the F1 gold buffalo together and in the F2 generation get:
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
2. You find out that all of the F1 buffalos are gold
91 47 33
3. You cross two of the F1 gold buffalo together and in the F2 generation get:
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
AAttaaTT
AaTt
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:AaTt
91 47 33
AT At aT at
AT AATT AATt AaTT AaTt
At AATt AAtt AaTt Aatt
aT AaTT AaTt aaTT aaTt
at 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
AaTt
X
Punnett square for cross of two gold buffalo AaTt x AaTt:
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:AaTt
91 47 33
AT At aT at
AT AATT AATt AaTT AaTt
At AATt AAtt AaTt Aatt
aT AaTT AaTt aaTT aaTt
at 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
AaTt
X
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:AaTt
91 47 33
AT At aT at
AT AATT AATt AaTT AaTt
At AATt AAtt AaTt Aatt
aT AaTT AaTt aaTT aaTt
at 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
AaTt
X
3. You cross two of the F1 gold buffalo together and in the F2 generation get:
91 47 33
AT At aT at
AT AATT AATt AaTT AaTt
At AATt AAtt AaTt Aatt
aT AaTT AaTt aaTT aaTt
at AaTt Aatt aaTt aatt
Approximately how many true-breeding tan buffalo are in the F2 generation in your herd?
A. 1
B. 3
C. 11
D. 22
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. FatherB. MotherC. SonD. Father and the sonE. Everyone
Imprinting
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.
Aa A’a’
A’a Aa’ aa’ AA’
1 2 3 4
Which of the offspring will be affected?A) 1 and 3B) 2 and 3C) 3 onlyD) 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
Aa A’a’
A’a AA’ aa’
Which disorder does the mother have?A) NoneB) PWSC) ASD) need more information
1 2 3
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
Aa A’a’
A’a AA’ aa’
Which disorder does the offspring 1 have?A) NoneB) PWSC) ASD) need more information
1 2 3
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
Aa A’a’
A’a AA’ aa’
Which disorder does the offspring 3 have?A) both syndromesB) PWSC) ASD) need more information
1 2 3
An individual with AS married a normal individual and produced anoffspring with PWS. What is the gender of the parent with AS?
A) maleB) femaleC) 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.
What is the gender of the child with PWS?A) maleB) femaleC) need more information
Dominance vs. recessive
•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 greenchuzzles 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) redB) green
Chuzzle Population