1. Inheritance Patterns and Probability
July 2008

2. Pedigrees

3. 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)?
I-1 and I-2
I-1, I-2, and II-1
I-1, II-1, and II-3
I-1, I-2, and II-3
I-1, I-2, II-1, and II-3
Spring ’08 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Tin Tin Su and Dr. Michelle Smith
77% correct

4. I. 1 2
II. dd Dd 1 2 3
III. dd 1
If II-2 and II-3 just had another baby boy. What is the chance that he is deaf?
1/8
1/4
1/2
3/4 1
Spring ’08 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Tin Tin Su and Dr. Michelle Smith
96% correct

5. family 2 family 1 I. Dd Dd Dd I. Dd 1 2 1 2 Dd or DD II. II. Dd or DD3 1 2 3
III. dd
III. dd 1 1
What are the chances that II-1 from family 1 and II-1 from family 2 will have a deaf child together?
1/4
1/9
4/9
1/16
Spring ’08 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Tin Tin Su and Dr. Michelle Smith
70% correct

6. Based on the pedigree above, which inheritance pattern can be ruled out?
Autosomal dominant
Autosomal recessive
X-linked dominant
X-linked recessive
None of the above
Spring ’08 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Tin Tin Su and Dr. Michelle Smith

7. 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
Spring ’08 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Tin Tin Su and Dr. Michelle Smith

8. Based on the pedigree above, which inheritance pattern can be ruled out?
X-linked dominant
X-linked recessive
neither of the above
Spring ’08 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Tin Tin Su and Dr. Michelle Smith

9. ?
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?
1/3
1/4
1/6
1/8
Spring ’08 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Tin Tin Su and Dr. Michelle Smith
18% correct

10. I. 1 2 3 4
II. 1 3 2 4 5 ?
III. 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
Spring ’08 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Tin Tin Su and Dr. Michelle Smith
81% correct

11. What is Matt’s genotype (Matt has diastrophic dysplasia)? A. ddaa
Autosomal recessive
“D” normal allele
“d” mutant allele
Achondroplasia
Autosomal dominant “A” mutant allele
“a” normal allele
Ron
Peggy
Gordon
Pat
Matt
Amy
What is Matt’s genotype (Matt has diastrophic dysplasia)?
A. ddaa
B. ddAa
C. Ddaa
D. DdAa
E. ddAA
Fall ’07 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Michelle Smith
92% correct 11

12. DDAa (note AA embryos are not viable)
Diastrophic dysplasia
Autosomal recessive SLC26A2 gene
“D” normal allele
“d” mutant allele
Achondroplasia
Autosomal dominant FGFR3 gene
“A” mutant allele
“a” normal allele
Ddaa
Ron
Peggy
Gordon
Pat
Matt
Amy
ddaa
DDAa (note AA embryos are not viable)
Fall ’07 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Michelle Smith
98% correct
What is Ron’s genotype?
A. ddaa
B. Ddaa
C. DDaa 12

13. DDAa (note AA embryos are not viable)
Diastrophic dysplasia
Autosomal recessive SLC26A2 gene
“D” normal allele
“d” mutant allele
Achondroplasia
Autosomal dominant FGFR3 gene
“A” mutant allele
“a” normal allele
Ddaa
DDaa
Ron
Peggy
Ddaa
Gordon
Pat
Matt
Amy
ddaa
DDAa (note AA embryos are not viable)
Fall ’07 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Michelle Smith
62% correct
What is Pat’s genotype?
A. DDAa
B. DDAA
C. Ddaa
D. None of the above 13

14. DDAa (note AA embryos are not viable)
Diastrophic dysplasia
Autosomal recessive SLC26A2 gene
“D” normal allele
“d” mutant allele
Achondroplasia
Autosomal dominant FGFR3 gene
“A” mutant allele
“a” normal allele
Matt
Amy
ddaa
DDAa (note AA embryos are not viable)
What is Zach’s genotype?
A. Ddaa
B. DdAa
C. DdAA
Fall ’07 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Michelle Smith
95% correct
Jeremy
Zach
Molly
Jacob 14

15. 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. ? ?
Spring ’08 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Christy Fillman.
What is the phenotype of the twins’ father?
A) RR
B) Rr
C) rr
D) red

16. What is the genotype of the twins’ father? A) RR B) Rr C) 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. ? ?
Spring ’08 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Christy Fillman.
What is the genotype of the twins’ father?
A) RR
B) Rr
C) rr
D) 1/2 Rr, 1/2 RR

17. What is the genotype of the twins' mother? A) RR B) Rr C) ½ Rr, ¼ 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. ? ?
Spring ’08 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Christy Fillman.
What is the genotype of the twins' mother?
A) RR
B) Rr
C) ½ Rr, ¼ RR
D) 2/3 Rr, 1/3 RR

18. What is the probability that the first twin born
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. ? ?
Spring ’08 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Christy Fillman.
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

19. The next few questions are not about pedigrees, but follow the cootie example

20. What is the phenotype of the F1 generation? A) All red with antennae
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. X P
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
Spring ’08 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Christy Fillman.

21. What is the probability that an F2 cootie will be red? A) 1/4 B) 1/2
You allow the F1 generation to mate and produce offspring (F2 generation). X P F1
RrAa
What is the probability that an F2 cootie will be
red?
A) 1/4
B) 1/2
C) 3/4
D) 1
Spring ’08 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Christy Fillman.

22. 1767 543 598 221 What is expected number of red cooties with antennae?
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
Spring ’08 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Christy Fillman.

23. Here is the F2 generation (RrAa X RrAa)
observed
expected
(O-E)2/E
1767
---
543
586.7
598
221
196
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
Spring ’08 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Christy Fillman.

24. Here is the F2 generation (RrAa X RrAa) 3138 total
observed
expected
(O-E)2/E
1767
---
0.02
543
586.7
3.3
598
0.21
221
196
Spring ’08 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Christy Fillman.
Do the red and antenna gene follow rules of independent assortment?
Yes, accept hypothesis – differences are likely due to chance
Yes, accept hypothesis – differences are not likely due to chance
No, reject hypothesis – differences are likely due to chance
No, reject hypothesis – differences are not likely due to chance

25. Calculating probability of inheritance (monhybrid, dihybrid crosses)

26. 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
Spring ’08 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Christy Fillman.
91% correct in Tin Tin Su’s course 26

27. 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
Spring ’08 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Christy Fillman. 27

28. Plant 1: Yellow, round peas Plant 2: Green, wrinkled peasX P:
F1:
1/2 Yellow, round peas
1/2 Yellow, wrinkled peas
Y - Yellow
y - Green
R - Round
r - wrinkled
What is the genotype of the yellow, round parent?
Spring ’08 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Christy Fillman.
A: YYRR
B: YyRR
C: YYRr
D: YyRr
E: Cannot be determined 28

29. Use Mendel’s Dihybrid cross results:P X F1 F2
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
Spring ’08 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Tin Tin Su and Dr. Michelle Smith
98% correct 29

30. 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 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
Spring ’08 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Christy Fillman.
87% correct in Tin Tin Su’s course 30

31. Genotype and phenotype
Phenotypes
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? YY Yy yy
Need more information
Fall ’07 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Kenneth Krauter 31

32. Mating between individuals that differ in two traits
Dihybrid cross
Mating between individuals that differ in two traits
Round, Yellow
Wrinkled, Green P X
RRYY
rryy F1
100% Round, Yellow
RrYy
Spring ’08 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Christy Fillman.
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

33. Dihybrid cross X F1 RrYy Question 6: What fraction of the
F2 generation is green?
A) 1/16
B) 1/2
C) 1/9
D) 1/4
RY Ry rY ry RY
F2 Generation
RRYY
RYRy
RrYY
RrYy
RRYy
RRyy
Rryy
rrYY
rrYy
rryy Ry rY
RrYy
Spring ’08 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Christy Fillman. ry

34. 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 9 3 1 3 1
Round, Yellow
Wrinkled, Yellow
Round, Green
Wrinkled, Green 1 1 3 9
Spring ’08 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Christy Fillman.

35. 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
Fall ’07 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Kenneth Krauter 35

36. Monohybrid cross probability
Consider Yy X Yy cross
What is the probability of getting a Y from parent 1?
1/4
1/2 1
1/16
Fall ’07 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Kenneth Krauter 36

37. 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)?
1/4
1/2 1
1/16
Fall ’07 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Kenneth Krauter 37

38. Monohybrid cross probability
Consider Yy X Yy cross
What is the probability of being Yellow (i.e. YY OR Yy)?
1/4
1/2
3/4 1
Fall ’07 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Kenneth Krauter 38

39. Consider the following cross: AaBBCcddEe X aabbCCDdEe
What is the probability their first offspring will be
aaBbCCDdee?
1/8
1/16
1/32
1/64
Cannot be determined
Spring ’08 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Christy Fillman.

40. What is the probability of rolling a two OR a three with one role of a six-sided die?
C)1/6
D)1/36
E)1/64
Spring ’08 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Christy Fillman.

41. 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
Spring ’08 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Christy Fillman.

42. 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
Spring ’08 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Christy Fillman.

43. Statistical Analysis of Crosses

44. Use Mendel’s Dihybrid cross results:P X
Total seeds observed = 556 F1 F2
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
Spring ’08 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Tin Tin Su and Dr. Michelle Smith
82% correct 44

45. Calculate Expected (e) numbers for each class if hypothesis correct
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
( )2/312=0.029
X2 =
0.527
Spring ’08 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Tin Tin Su and Dr. Michelle Smith
96% correct
5. Calculate Degree of Freedom
A: 1
B: 4
C: 3 45

46. 6. Look up probability (p) for X2 at a given df in the table
Spring ’08 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Tin Tin Su and Dr. Michelle Smith
93% correct
A: Accept the hypothesis B: Reject the hypothesis 46

47. How many degrees of freedom are there in the F2 generation
of the following cross? X P F1 F2
Spring ’08 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Christy Fillman.
A) 1
B) 2
C) 3
D) 4
E) 5

48. What if his results had been 5120 yellow and 2903 green?
Could Mendel still accept his hypothesis?
X2 = 535
Spring ’08 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Christy Fillman.
A) Accept the hypothesis
B) Reject the hypothesis

49. What does P = 0.005mean for the 28:20 ratio?
28:20 is likely to be 3:1
28:20 is NOT likely to be 3:1
28:20 is not statistically “significant” and so cannot be used to assess 3:1 ratio
This experiment is totally flawed and cannot be interpretted
Fall ’07 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Kenneth Krauter 49

50. Exceptions to Mendel’s Laws (maternal, cytoplasmic/mitochondrial, sex-limited, co-dominance, incomplete dominance, lethal, epistatsis, heterozygous advantage, imprinting)

51. Maternal

52. 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
Spring ’08 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Christy Fillman.

53. 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
Spring ’08 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Christy Fillman.

54. 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%
Spring ’08 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Tin Tin Su and Dr. Michelle Smith
74% correct 54

55. 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%
Spring ’08 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Tin Tin Su and Dr. Michelle Smith
88% correct 55

56. 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?
100%
50%
25% 0%
Spring ’08 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Tin Tin Su and Dr. Michelle Smith
20% correct 56

57. 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?
100%
50%
25% 0%
Spring ’08 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Tin Tin Su and Dr. Michelle Smith
70% correct 57

58. Cytoplasmic/mitochondrial

59. Sex Limited

60. Co-dominance/incomplete dominance Variable Expression Conditional

61. 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? 1 2 4
Not enough information
Spring ’08 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Tin Tin Su and Dr. Michelle Smith
66% correct 61

62. The alleles for buffalo color show: A) complete dominance
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
Spring ’08 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Christy Fillman.

63. B. Incomplete dominance C. Complete dominance D. Recessive epistasis
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. Codominance
B. Incomplete dominance
C. Complete dominance
D. Recessive epistasis X
Long haired Ralphie buffalo
Short haired Ralphie buffalo
Ralphie buffalo with a mix of short and long hairs
Spring ’08 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Tin Tin Su and Dr. Michelle Smith
54% correct 63

64. 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
Spring ’08 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Christy Fillman.

65. A: O only B: AB and O C: A, B and O D: A, B, AB and O
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
Spring ’08 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Tin Tin Su and Dr. Michelle Smith
89% correct 65

66. 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
Fall ’07 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Kenneth Krauter 66

67. Mr. Chaplin’s case 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
Fall ’07 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Kenneth Krauter 67

68. A: shows incomplete penetrance B: shows variable expressivity C: both
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
Spring ’08 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Tin Tin Su and Dr. Michelle Smith
97% correct 68

69. Lethal

70. 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?
Spring ’08 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Christy Fillman.
YYss X yySS
YY, Yy = yellow
yy = black
SS, Ss = star
ss = no star
A) 1
B) 1/2
C) 1/4
D) 3/16

71. Lethal Alleles YS Ys yS ys YYSS YYSs YySs YYss Yyss YySS yySS yySs
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 YS Ys yS ys
YYSS
YYSs
YySs
YYss
Yyss
YySS
yySS
yySs
yyss
Spring ’08 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Christy Fillman.
A) 1
B) 1/4
C) 1/5
D) 3/16

72. DdAa DA Da dA da DA DDAA DdAA DdAa DDAa DDaa Ddaa ddAA ddAa ddaa DDAa
Sue
DdAa
SLC26A gene
FGFR3 gene
ttp://images.icnetwork.co.uk/upl/icnewcastle/aug2005/4/8/ E719-12F1-88F40C01AC1BF814.jpg DA Da dA da
Zach DA
DDAA
DdAA
DdAa
DDAa
DDaa
Ddaa
ddAA
ddAa
ddaa
DDAa Da dA
DdAa
SLC26A gene
FGFR3 gene da
What is the chance that Zach and Sue will have a child with diastrophic dysplasia and achondroplasia?
1/6
1/8
3/16
9/16
None of the above
Fall ’07 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Michelle Smith
66% correct
Note: AA is lethal 72

73. DdAa DA Da dA da DA DdAA DdAa DDAa DDaa Ddaa ddAA ddAa ddaa DDAA DDAa
Sue
DdAa
SLC26A gene
FGFR3 gene
ttp://images.icnetwork.co.uk/upl/icnewcastle/aug2005/4/8/ E719-12F1-88F40C01AC1BF814.jpg DA Da dA da
Zach DA
DdAA
DdAa
DDAa
DDaa
Ddaa
ddAA
ddAa
ddaa
DDAA
DDAa Da dA
DdAa
SLC26A gene
FGFR3 gene da
Fall ’07 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Michelle Smith
91% correct
Note: AA is lethal
What is the chance that Zach and Sue will have a child with only diastrophic dysplasia?
1/3
1/6
1/12
1/16 73

74. Epistasis

75. Partial Dominance Model 1: BB -> Brown 2: Bb -> Black
P Brown X Yellow
F 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
Spring ’08 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Tin Tin Su and Dr. Michelle Smith
68% correct
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 75

76. 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)
Fall ’07 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Michelle Smith
98% correct
What is the phenotype of a BbEe lab?
Black
Brown
Yellow 76

77. 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?
Black
Brown
Yellow
Fall ’07 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Michelle Smith
100% correct 77

78. 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?
Black
Brown
Yellow
Fall ’07 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Michelle Smith
100% correct 78

79. 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?
Black
Brown
Yellow
Fall ’07 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Michelle Smith
97% correct 79

80. What is the phenotype of the aabb mice in the F2 generation?
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
Spring ’08 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Tin Tin Su and Dr. Michelle Smith
36% correct 80

81. hairless mouse aaBB X curly hair AAbb F1s have straight hair AaBb F2
What is the order of function?
A, then B
B. B, then A
C. A and B act simultaneously
D. Not enough data
18 straight hair mice A-B-
8 hairless mice aaB- & aabb
6 curly hair mice A-bb
Spring ’08 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Tin Tin Su and Dr. Michelle Smith
75% correct 81

82. Recessive epistasis Model
9 B_E_ -> Black
3 bbE_ -> Brown
4 _ _ee -> Yellow
What is the order of function?
B, then E
B. E, then B
C. E and B act simultaneously
D. Not enough data
Spring ’08 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Tin Tin Su and Dr. Michelle Smith
83% correct 82

83. 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?
intermediate 1
intermediate 2
intermediate 3
Spring ’08 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Christy Fillman.

84. Epistasis X What are the possible gametes produced by each chuzzel?
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
Spring ’08 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Christy Fillman.

85. 2. You find out that all of the F1 buffalos are gold
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)
Fall ’07 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Michelle Smith
76% correct
What is the genotype of the tan buffalo in the P generation?
AAtt
Aatt
Either AAtt or Aatt 85

86. 2. You find out that all of the F1 buffalos are gold
1.You cross a true-breeding white buffalo to a true-breeding tan buffalo in the P generation
aaTT
AaTt
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: 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)
Fall ’07 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Michelle Smith
90% correct
What is the genotype of the white buffalo in the P generation?
aatt
aaTT
aaTt 86

87. 2. You find out that all of the F1 buffalos are goldX
AaTt
AaTt
3. You cross two of the F1 gold buffalo together and in the F2 generation get: 91 47 33
How many of the genotypes in the Punnett square will result in a gold buffalo? 9 4 3 2 1
Punnett square for cross of two gold buffalo AaTt x AaTt: AT At aT at
AATT
AATt
AaTT
AaTt
AAtt
Aatt
aaTT
aaTt
aatt
Fall ’07 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Michelle Smith
80% correct 87

88. 2. You find out that all of the F1 buffalos are goldX
AaTt
AaTt
3. You cross two of the F1 gold buffalo together and in the F2 generation get: 91 47 33
How many of the genotypes in the Punnett square will result in a white buffalo? 9 4 3 2 1 AT At aT at
AATT
AATt
AaTT
AaTt
AAtt
Aatt
aaTT
aaTt
aatt
Fall ’07 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Michelle Smith
95% correct 88

89. 2. You find out that all of the F1 buffalos are goldX
AaTt
AaTt
3. You cross two of the F1 gold buffalo together and in the F2 generation get: 91 47 33
How many of the genotypes in the Punnett square will result in a tan buffalo? 9 4 3 2 1 AT At aT at
AATT
AATt
AaTT
AaTt
AAtt
Aatt
aaTT
aaTt
aatt
Fall ’07 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Michelle Smith
92% correct 89

90. 3. You cross two of the F1 gold buffalo together and in the F2 generation get:91 47 33
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 AT At aT at
AATT
AATt
AaTT
AaTt
AAtt
Aatt
aaTT
aaTt
aatt
Fall ’07 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Michelle Smith
51% correct 90

91. Heterozygous Advantage

92. 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
Spring ’08 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Tin Tin Su and Dr. Michelle Smith 92

93. Imprinting

94. Which of the offspring will be affected? A) 1 and 3 B) 2 and 3 4
A’a
Aa’
aa’
AA’
Spring ’08 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Christy Fillman.
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

95. 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’ 1 2 3
Spring ’08 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Christy Fillman.
A’a
AA’
aa’
Which disorder does the mother have?
A) None
B) PWS
C) AS
D) need more information

96. 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’ 1 2 3
Spring ’08 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Christy Fillman.
A’a
AA’
aa’
Which disorder does the offspring 1 have?
A) None
B) PWS
C) AS
D) need more information

97. 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’ 1 2 3
Spring ’08 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Christy Fillman.
A’a
AA’
aa’
Which disorder does the offspring 3 have?
A) both syndromes
B) PWS
C) AS
D) need more information

98. 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
Spring ’08 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Christy Fillman.

99. Dominance vs. recessive

100. 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.
Spring ’08 Clicker Question used in MCDB 2150, Principles of Genetics, written by Dr. Christy Fillman.
Which trait is dominant? A) red
B) green