1. Punnet Squares

2. Important to know before getting started:
Female
Male
XX= Female XY=male

3. Each parent give a set of genes to their offspring.
The different forms of a gene are called alleles.
Heredity: Passing of traits and genes from parents to offspring on chromosomes.

4. Mendel’s Laws (Mendel is the father of Genetics)
Law of Dominance: When the dominant traits overshadows the recessive trait.
Law of Segregation: Alleles for a trait separate when gametes (egg and sperm cells) are formed.
Law of Independent Assortment: Alleles form different traits (eye/hair color) are distributed to sex cells independently of each other.

5. Traits are different forms of a characteristic (hair color, eye color)
Punnett’s Square
Dad’s alleles
Mom’s alleles
Scientists use a Punnett’s square to predict the probability of traits in offspring.
Traits are different forms of a characteristic (hair color, eye color)
Each box represents 25%

6. Punnett Squares:
Each box represents the probability that alleles from each parent could combine in the offspring (babies).

7. Probability The likelihood of a particular event occurring. Chance
Can be expressed as a fraction or a percent.
Example: coin flip.

8. Genotype
The specific combination of alleles (from mom and dad) are called the genotype.
Two forms of the same gene.
Genotypes can be homozygous (aa or AA) or heterozygous (Aa)
They can also be dominant (capital letter) or recessive (lower case letter)
Example: BB Bb bb

9. Genotypes
Homozygous = when an organism possesses two identical alleles. ex.
YY or yy
Heterozygous = when an organism possesses different alleles. ex. Yy

10. PhenoType
The phenotype is the physical appearance/expression of the alleles. Example: BB=brown hair Bb=dirty blonde hair bb= blonde hair

11. Monohybrid crosses show the crossing of one set of alleles from each parent. So we use a Punnett square with four squares/quadrants.

12. Black colored fur is the dominant trait.
We will use the Punnett’s Square to determine the offspring of guinea pigs. The offspring will either be black or white.
Black colored fur is the dominant trait.

13. Generation 1 B B b b
In this case we have a dad with black fur and a mother with white fur. Because black is the dominant gene, we write it with a capital ‘B’.

14. Generation 1 B B b b
White fur is a recessive trait. It is written with a lowercase ‘b’. It does not matter what letter we choose to represent a gene, but capital letter is always dominant and lowercase is always recessive.

15. Generation 1 B B Bb b b
To complete the Punnett’s square we combine the gene from mom with the gene from dad. We always write the dominant gene first.

16. Generation 1 B B Bb b b
We write the dominant gene first because it “masks” the recessive gene. Therefore, the color of the guinea pig with the genes Bb would be black.

17. Generation 1 B B Bb b b
Copy this Punnett’s square into your notebook. Try and fill out the remaining offspring on your own. When you are done, go to the next slide.

18. Generation 1 B B Bb Bb b Bb Bb b
We say an individual is heterozygous when it has two different genes.
What percentage of these offspring are heterozygous?

19. 100% are heterozygous black.
Generation 1 B B Bb Bb b Bb Bb b
100% are heterozygous black.
We will now take one female and one male from this generation to cross for our second generation.

20. What will the gene combinations be for these offspring?
Generation 2 B b B b
What will the gene combinations be for these offspring?
Copy this into your notebook and try to fill out the Punnett’s square. Continue when you are done.

21. Can you figure out what color these offspring will be?
Generation 2 B b BB Bb B bb Bb b
What you have completed is the genotype for all the offspring. Genotype refers to the gene combination that an individual has.
Can you figure out what color these offspring will be?

22. Generation 2 B b BB Bb B bb Bb b black black black white
What you have just determined is called the phenotype. The phenotype is what we see as a result of an individual's genes.

23. Generation 2 B b BB Bb B bb Bb b
We say an individual is homozygous when it has two of the same genes.
Can you find a homozygous black guinea pig?

24. Generation 2 B b BB Bb B bb Bb b This individual is homozygous black.
The entire Punnett’s square represents all possible outcomes. That means each small box represents 25% of the offspring.
What percentage of the offspring are homozygous black?

25. Generation 2 B b BB Bb B bb Bb b
25% of the offspring are homozygous black.
Try the next two on your own:
____% are homozygous white
____% are heterozygous black

26. Generation 2 B b BB Bb B bb Bb b
Notice: This will always add up to 100%.
25% are homozygous black.
25% are homozygous white
50% are heterozygous black

27. Now we will examine some human traits.
Earlobes can be either unattached (A) or attached (B)
Unattached
Attached
The unattached earlobe is the dominant trait.
The attached earlobe is the recessive trait.

28. What percent of offspring will have attached earlobes?
Copy this Punnett’s square into your notebook. Determine both the genotype and phenotype.
What percent of offspring will have attached earlobes?

29. 50% percent of offspring will have attached earlobes.Ee Ee ee ee e
50% percent of offspring will have attached earlobes.

30. The ability to curl your tongue is also genetic.
Tongue curling is the dominant trait.
Non-curling is the recessive trait.

31. Tongue Curling T t T T
Copy into your notebook and complete the Punnett’s Square.
What percentage of offspring will be able to curl their tongue?

32. 100% percent of offspring will be able to curl their tongues.
Tongue Curling T t T TT Tt TT Tt T
100% percent of offspring will be able to curl their tongues.

33. Incomplete dominance A situation in which neither allele is dominant.
When both alleles are present a “new” phenotype appears that is a blend of each allele.
Alleles will be represented by capital letters only.

34. W= white R = red WR = pink Incomplete Dominance
The petal color of pea flowers is an example of incomplete dominance. Both genes for white and red flowers are equally dominant, which results in a new phenotype.
W= white
R = red
WR = pink

35. Pea Flower Petals R R W R
Copy and complete this Punnett’s square in your notebook.
What percentage of offspring will have white flowers?

36. Pea Flower Petals R R WR WR W RR R RR white
0% of the offspring will have flowers.
50% will have red flowers.
50% will have pink flowers.
white

37. Dihybrid Cross a 4x4 representation of crossing TWO traits
monohybrid = 1 trait w/ 2 alleles
= 1 allele gamete each for 2x2 box
dihybrid = 2 traits w/ 2 alleles EACH = 4 alleles
= 2 allele gametes each for 4x4 box
[one allele for each trait in each gamete]

38. Solving a Dihybrid Given to you, or you have to solve for them
Determine the genotypes of each parent
Given to you, or you have to solve for them
Given: Parent #1 AABB
Given: Parent #2 aabb

39. Solving a Dihybrid
Create your dihybrid cross [4x4]

40. Solving a Dihybrid
Determine Gametes:
pair first allele of first trait with first allele of second trait
A A B B
first gamete = AB

41. Solving a Dihybrid
Determine Gametes:
pair first allele of first trait with second allele of second trait
A A B B
second gamete = AB

42. Solving a Dihybrid
Determine Gametes:
pair second allele of first trait with first allele of second trait
A A B B
third gamete = AB

43. Solving a Dihybrid
Determine Gametes:
pair second allele of first trait with second allele of second trait
A A B B
fourth gamete = AB

44. Solving a Dihybrid
Determine Gametes:
pair first allele of first trait with first allele of second trait
a a b b
first gamete = ab

45. Solving a Dihybrid
Determine Gametes:
pair first allele of first trait with second allele of second trait
a a b b
second gamete = ab

46. Solving a Dihybrid
Determine Gametes:
pair second allele of first trait with first allele of second trait
a a b b
third gamete = ab

47. Solving a Dihybrid
Determine Gametes:
pair second allele of first trait with second allele of second trait
a a b b
fourth gamete = ab

48. Solving a Dihybrid
Label cross w/ parent gametes above / to left AB ab

49. Solving a Dihybrid AB ab A B AB ab Fill in boxes:
fill DOWN each column w/ whatever’s there AB ab
A B AB ab

50. Solving a Dihybrid AB ab A B AB ab AaBb Fill in boxes:
fill ACROSS each column with whatever’s there AB ab
A B AB ab
AaBb

51. Solving a Dihybrid record all genotypes you have
Determine Genotype Ratios
record all genotypes you have
count and record number of each genotype
AaBb: 16/16 [or 100%] AB ab
AaBb

52. Solving a Dihybrid count how many of each phenotype you have
Determine Phenotype Ratios
count how many of each phenotype you have
record # of phenotypes represented
purple, round: 16/16 [or 100%] AB ab
AaBb

53. Solving a Dihybrid #2 Given to you, or you have to solve for them
Determine the genotypes of each parent
Given to you, or you have to solve for them
Given – both parents are heterozygous for both traits.
Given: Parent #1 QqRr
Given: Parent #2 QqRr

54. Solving a Dihybrid
Create your dihybrid cross [4x4]

55. Solving a Dihybrid
Determine Gametes:
pair first allele of first trait with first allele of second trait
Q q R r
first gamete = QR

56. Solving a Dihybrid
Determine Gametes:
pair first allele of first trait with second allele of second trait
Q q R r
second gamete = Qr

57. Solving a Dihybrid
Determine Gametes:
pair second allele of first trait with first allele of second trait
Q q R r
third gamete = qR

58. Solving a Dihybrid
Determine Gametes:
pair second allele of first trait with second allele of second trait
Q q R r
fourth gamete = qr

59. Solving a Dihybrid QR Qr qR qr
Label cross w/ parent gametes above / to left QR Qr qR qr

60. Solving a Dihybrid Q R Q r q r QR Qr qR qr Fill in boxes:
fill DOWN each column w/ whatever’s there QR Qr qR qr
Q R
Q r
q r

61. Solving a Dihybrid QQRR QQRr QqRR QqRr QQrr Qqrr qqRR qqRr qqrr Q R
Fill in boxes:
fill ACROSS each column with whatever’s there QR Qr qR qr
QQRR
QQRr
QqRR
QqRr
QQrr
Qqrr
qqRR
qqRr
qqrr QR Qr qR qr
Q R
Q r
q r

62. Solving a Dihybrid record all genotypes you have
Determine Genotype Ratios
record all genotypes you have
count and record number of each genotype
Write out each
Genotype
EX: QQRR
QQRr QR Qr qR qr
QQRR
QQRr
QqRR
QqRr
QQrr
Qqrr
qqRR
qqRr
qqrr

63. Solving a Dihybrid count how many of each phenotype you have
Determine Phenotype Ratios
count how many of each phenotype you have
record # of phenotypes represented
9/16 - purple, round
3/16 - purple, wrinkled
3/16 - white, round
1/16 - white, wrinkled QR Qr qR qr
QQRR
QQRr
QqRR
QqRr
QQrr
Qqrr
qqRR
qqRr
qqrr

64. Pedigree Charts
These charts are used to show how family members are related to each other and it includes atleast two generations.

65. Pedigree What do you think of when you read the above word? Dog food
Family
Lineage
inheritance

66. Pedigree
Chart that shows how a trait and the genes that control it are inherited in a family.

67. Pedigree -Females are circles -Males are squares
-Shading is an individual with the trait
-Individual carrying the trait

68. Pedigree
Marriage / mating
Offspring
(in order of birth)

69. Finding a Gene on the Chromosome Map
We can examine inheritance in a family by constructing a pedigree.
In this family certain members (in black) have Whirling disorder.

70. Finding a Gene on the Chromosome Map
How do scientists find the genes responsible for a hereditary disease in a family?
Think of all the genes in a person’s genome as puzzle pieces in a jigsaw puzzle.
Each piece represents a different gene.
Because all humans have the same set of genes in the same order, every family member has the same basic jigsaw puzzle arrangement.
Shown here is a generic human jigsaw puzzle.

71. Make a pedigree based on the following passage about freckles.
Although Jane and Joe Smith have dimples, their daughter, Clarissa, does not. Joe’s dad has dimples, but his mother and his sister, Grace, do not. Jane’s dad, Mr. Renaldo, her brother, Jorge, and her sister, Emily, do not have dimples, but her mother does.

72. Mrs. Renaldo Mr. Renaldo Mrs. Smith Mr. Smith
Emily Jorge Jane
Joe Grace
Clarisa

73. Andy, Penny, and Delbert have freckles, but their mother, Mrs
Andy, Penny, and Delbert have freckles, but their mother, Mrs. Cummins, does not. Mrs. Biodano, Mrs. Cummins’s sister, has freckles, but her parents, Mr. And Mrs. Lutz, do not. Deidra and Darlene Giordano are freckled, but their sister, Dixie, like her father, is not freckled.

75. Your Pedigree
Make a Pedigree based on your family. Include the following:
3 generations
One trait (hair, dimples, eye color, etc)
Names
Key

76. A pedigree is a chart for tracing genes in a family.
Phenotypes are used to infer genotypes on a pedigree.
Autosomal genes show different patterns on a pedigree than sex-linked genes.

77. If the phenotype is more common in males, the gene is likely sex-linked.