1. GENETICS PRACTICE Kelly Riedell Brookings Biology
Watch videos BOZEMAN- Beginner’s Guide to Punnett Squares
BOZEMAN- Mendelian genetics
Chromosomal Genetics (Non-Mendelian) BOZEMAN- Advanced genetics

2. GENETICS PRACTICE PROBLEMS
P1 = Parental
F1 = Filial (pffspring)
F2 = Filial (pffspring)

3. Refresh your “Bio Brain” about: GENETICS VOCAB DOMINANT:
RECESSIVE:
gene that hides another represented by capital letter
gene that is hidden by another represented by lower case letter

4. Refresh your “Bio Brain” about: GENETICS VOCAB
HOMOZYGOUS (pure-breeding):
HETEROZYGOUS (hybrid):
Organism with two identical alleles for a gene
TT OR tt
Organism with two different alleles for a gene Tt

5. Refresh your “Bio Brain” about: GENETICS VOCAB GENOTYPE:
PHENOTYPE:
Genetic make up of an organism
“What genes it has”
Appearance of an organism
“Way it looks”

6. Probability is the likelihood that an event will occur
It can be written as a:
Fraction ____
Percent ____
Ratio ____
1/4
25%
1:3

7. COIN FLIP 1/2 50% 1:1 There are 2 possible outcomes: HEADS TAILS
COIN FLIP
There are 2 possible outcomes:
HEADS TAILS
The chance the coin will land on either one is:
____ ____ ____
Alleles segregate randomly just like a coin flip. . . So can use probability to predict outcomes of genetic crosses.
1/2
50%
1:1
NOT 1:2!

8. What is the probability?
AaBBCcDd parent genome What is the probability of producing a gamete with this gene combination? ABCD ______________________________ abcD _______________________________ Abcd _______________________________

9. What is the probability?
AaBBCcDd parent genome What is the probability of producing a gamete with this gene combination? ABCD _____________________________ abcD _______________________________ ABcd _______________________________
X 1
X ½
X ½
= 1/8
X 0
X ½
X ½
= 0
X 1
X ½
X ½
= 1/8

10. What is the probability?
AaBbCcDd parent genome What is the probability of producing a gamete with this gene combination? ABcd _____________________________ AbcD_______________________________
½ X ½ X ½ X ½
= 1/16
½ X ½ X ½ X ½
= 1/16

11. What is the probability?
AaBbCcDd X AaBbCcDd parents What is the probability of producing an offspring with this gene combination? aabbccDd _____________________________ AaBBccDD______________________________ AaBBCCDd______________________________

12. What is the probability?
AaBbCcDd X AaBbCcDd parents What is the probability of producing a offspring with this gene combination? aabbccDd ___________________________ AaBBccDD______________________________ AaBBCCDd______________________________
X ¼
X ¼
X ½
= 1/128
X ¼
X ¼
X ¼
= 1/128
X ¼
= 1/64
X ¼
X ½

13. What is the probability?
AaBbCcDd X AaBbCcDd parents What is the probability of producing a offspring with this gene combination? AABbCcDd _____________________________ AaBbCcDd_______________________________

14. What is the probability?
AaBbCcDd X AaBbCcDd parents What is the probability of producing a offspring with this gene combination? AABbCcDd _____________________________ AaBbCcDd_______________________________
X ½
X ½
X ½
= 1/32
X ½
X ½
X ½
=1/16

15. PEDIGREES
= male; doesn’t show trait = female; doesn’t show trait = shows trait = carrier; doesn’t show trait

16. 1 2 3 4
Circle all males that show the trait BLUE Circle all females that show the trait RED
Circle all carrier females GREEN Circle all carrier males PURPLE
If this is an autosomal recessive trait, what is the genotype of individual #1? If this is an autosomal recessive trait, what is the genotype of individual # 2? If this is an autosomal recessive trait, what is the genotype of individual #3?

17. 1 2 4 3
Circle all males that show the trait BLUE Circle all females that show the trait RED
Circle all carrier females GREEN Circle all carrier males PURPLE
If this is an autosomal recessive trait, what is the genotype of individual #1? If this is an autosomal recessive trait, what is the genotype of individual # 2? If this is an autosomal recessive trait, what is the genotype of individual #3?

18. Write the genotype of each individual next to the symbol.
Is it possible that this pedigree is for an autosomal recessive trait?

19. Write the genotype of each individual next to the symbol.Aa Aa
Aa or AA aa
Is it possible that this pedigree is for an autosomal recessive trait?
YES

20. Write the genotype of each individual next to the symbol.
Is it possible that this pedigree is for an X-linked recessive trait?

21. Write the genotype of each individual next to the symbol.
XB Y
XB Xb
XB Y
Xb Xb
 Is it possible that this pedigree is for an X-linked recessive trait? NO

22. Could this trait be inherited as AUTOSOMAL RECESSIVE?

23. Fill in genotypes of affected individuals
Is this trait inherited as AUTOSOMAL RECESSIVE?  
Fill in genotypes of affected individuals aa aa aa aa aa aa aa aa aa

24. Yes Possible if . . . Is this trait inherited as AUTOSOMAL RECESSIVE? aa Aa
A ? aa aa aa aa aa aa aa aa

25. Is this trait inherited as AUTOSOMAL DOMINANT?

26. Fill in genotypes of affected individuals
Is this trait inherited as AUTOSOMAL DOMINANT?  
Fill in genotypes of affected individuals
A ?
A ?
A ?
A ?
A ?
A ?
A ?
A ?

27. NO If individual shows trait one of parents must also have trait.
Is this trait inherited as AUTOSOMAL DOMINANT?   NO
If individual shows trait one of parents must also have trait.
A ?
A ?
A ?
A ?
A ?
A ?
A ?
A ?

28. Is this trait inherited as X-LINKED RECESSIVE?

29. Fill in genotypes of affected individuals
Is this trait inherited as X-LINKED RECESSIVE?  
Fill in genotypes of affected individuals
XaXa
XaY
Xa Y
XaXa
XaY
XaY
XaXa
XaY
XaXa

30. NO If female has trait dad must show it too
Is this trait inherited as X-LINKED RECESSIVE?   NO
If female has trait dad must show it too
XAXa
XA Y
XaXa
XaY
XaY
Xa Y
XaXa
Daughter has trait, but dad doesn’t have allele or would be filled in
XaY
XaXa
XaY
XaXa

31. Is this trait inherited as X-LINKED DOMINANT?

32. Fill in genotypes of affected individuals
Is this trait inherited as X-LINKED DOMINANT?  
Fill in genotypes of affected individuals
XAX?
XAY
XA Y
XAX?
XAY
XAY
XAX?
XAY
XAX?

33. NO If male has it came from mother If dad has it, so should daughters
Is this trait inherited as X-LINKED DOMINANT?   NO
If male has it came from mother If dad has it, so should daughters
XAX?
XAY
Dad has it but not daughters
XAY
XA Y
XAX?
XAY
XAX?
XAY
Son has it, but mom doesn’t
XAX?

34. If dad passed to son it must be AUTOSOMAL
The inheritance of the disorder in II-3 from his father rules out what form of inheritance?
Xb Y
Can’t be X-linked recessive Males get their X-linked allele from their mother
If dad passed to son it must be AUTOSOMAL
Xb Y

35. If CLOSE TOGETHER on homologous chromosomes, STAY TOGETHER and end up together 100% of time Act like one gene

36. http://image. slidesharecdn
CROSSING OVER If far apart on homologous chromosomes, end up together 50% of time
RECOMBINANTS- Put different maternal/paternal alleles together on different chromosomes

37. INDEPENDENT ASSORTMENT
INDEPENDENT ASSORTMENT
If on different chromosomes, END UP TOGETHER 50% of time.

38. A Wild type fruit fly (heterozygous for gray body and normal wings) is mated with a black fly with vestigial wings.
OFFSPRING: wild type 785- black-vestigial 158- black- normal wings 162- gray body-vestigial wings
Is it 9:3:3:1? (2 genes on 2 different chromosomes)
Is it 3 Wild type : 1 black vestigial? (linked on homologous chromosomes)

39. A Wild type fruit fly (heterozygous for gray body and normal wings) is mated with a black fly with vestigial wings.
OFFSPRING: wild type 785- black-vestigial 158- black- normal wings 162- gray body-vestigial wings
What is the recombination frequency between these genes?

40. A Wild type fruit fly (heterozygous for gray body and normal wings) is mated with a black fly with vestigial wings.
OFFSPRING: wild type 785- black-vestigial 158- black- normal wings 162- gray body-vestigial wings
Recombinants = Total 320 = 17% 1883

41. When F1’s make gametes PROPHASE I Crossing over makes recombinants
Wild type wings
Wild type body (yellow-brown)
Vestigial wings
Black body
Parents (P)
F1 offspring
When F1’s make gametes PROPHASE I Crossing over makes recombinants
CROSS WITH HOMOZYGOUS RECESSIVE

42. A Wild type fruit fly (heterozygous for gray body and normal wings) is mated with a black fly with vestigial wings.
OFFSPRING: wild type 785- black-vestigial 158- black- normal wings 162- gray body-vestigial wings
What is the recombination frequency between these genes?

43. A Wild type fruit fly (heterozygous for gray body and normal wings) is mated with a black fly with vestigial wings.
OFFSPRING: wild type 785- black-vestigial 158- black- normal wings 162- gray body-vestigial wings
What is the recombination frequency between these genes?
Recombinants = Total 314 = 16.7% 1877

44. A Wild type fruit fly (heterozygous for gray body and red eyes) is mated with a black fly with purple eyes.
OFFSPRING: gray body/red eyes black body/purple eyes 49- gray body/purple eyes 45- black body/red-eyes
What is the recombination frequency between these genes?

45. A Wild type fruit fly (heterozygous for gray body and red eyes) is mated with a black fly with purple eyes.
OFFSPRING: gray body/red eyes black body/purple eyes 49- gray body/purple eyes 45- black body/red-eyes
What is the recombination frequency between these genes?
Recombinants = Total = 6 % 1566

46. A Wild type fruit fly (heterozygous for normal bristles and red eyes) is mated with a spineless bristle fly with sepia eyes.
OFFSPRING: normal bristles/red eyes spineless bristles/sepia eyes 72- normal bristles/sepia eyes 83- spineless bristles/red-eyes
What is the recombination frequency between these genes?

47. A Wild type fruit fly (heterozygous for normal bristles and red eyes) is mated with a spineless bristle fly with sepia eyes.
OFFSPRING: normal bristles/red eyes spineless bristles/sepia eyes 72- normal bristles/sepia eyes 83- spineless bristles/red-eyes
What is the recombination frequency between these genes?
Recombinants = Total 155 = 10.4% 1484

48. Determine the sequence of genes along a chromosome based on the following recombination frequencies
A-D 10%
B-C 15%
B-D 5%

49. Determine the sequence of genes along a chromosome based on the following recombination frequencies
A-D 10%
B-C 15%
B-D 5%

50. Determine the sequence of genes along a chromosome based on the following recombination frequencies
A-D 30%
B-C 24%
B-D 16%

51. Determine the sequence of genes along a chromosome based on the following recombination frequencies
A-D 30%
B-C 24%
B-D 16%

52. Determine the sequence of genes along a chromosome based on the following recombination frequencies
A-D 30%
B-C 24%
B-D 16%

53. Determine the sequence of genes along a chromosome based on the following recombination frequencies
A-D 30%
B-C 24%
B-D 16%
CABD

54. Determine the sequence of genes along a chromosome based on the following recombination frequencies
A-B 8%
A-C 28%
A-D 25%
B-C 20%
B-D 33%

55. Determine the sequence of genes along a chromosome based on the following recombination frequencies
A-B 8%
A-C 28%
A-D 25%
B-C 20%
B-D 33%

56. Determine the sequence of genes along a chromosome based on the following recombination frequencies
A-B 8%
A-C 28%
A-D 25%
B-C 20%
B-D 33%

57. Determine the sequence of genes along a chromosome based on the following recombination frequencies
A-B 8%
A-C 28%
A-D 25%
B-C 20%
B-D 33%

58. Determine the sequence of genes along a chromosome based on the following recombination frequencies
A-B 8%
A-C 28%
A-D 25%
B-C 20%
B-D 33%

59. A large ear of corn has a total of 500 kernals, including 312 purple & starchy, 73 purple & sweet,
76 yellow & starchy, and 39 yellow & sweet.
HYPOTHESIS: This ear of corn was produced by a dihybrid cross (PpSs X PpSs) involving two pairs of heterozygous genes resulting in a theoretical (expected) ratio of 9:3:3:1.
Test your hypothesis using Chi-square and probability values. SHOW YOUR WORK!

60. TOTAL = 500 offspring IF 9:3:3:1 then expect:
H0- There is no difference between the frequencies observed and the frequencies expected for a HETEROZYGOUS DIHYBRID (9:3:3:1) cross.
TOTAL = 500 offspring IF 9:3:3:1 then expect:
Purple & Starchy = 500 X 9/16 = Purple & Sweet = 500 X 3/16 = Yellow & Starchy = 500 X 3/16 = 93.75
Yellow & Sweet = 500 X 1/16 =
312 73 76 39
281.25
93.75
31.25
30.75
-20.75
-17.75
7.75
945.56
430.56
315.06
60.06
3.36
4.59
1.92
13.23
4-1= 3
is larger than REJECT THE NULL There is a difference between observed and expected for 9:3:3:1 = NOT A 9:3:3:1 cross