1. Genetics Tutorial Introduction Punnett Square – 1 Trait
Punnett Square – 2 Traits
Product Rule

2. In this tutorial, you will learn:
Important terms used in genetics.
How to use a Punnett square to determine the outcome of a cross with
1 trait or
2 traits
How to use the product rule to determine the outcome of a cross with any number of traits.
Credits:
Figures and images by N. Wheat unless otherwise noted.
Funded by Title V-STEM grant P031S

3. Introduction
Information that will guide the development of an organism is contained in that organism’s DNA. Every species has a characteristic number of DNA molecules called chromosomes.

4. Introduction
An individual receives one complete set of chromosomes from each parent, resulting in two complete sets. This is the diploid condition (2n).

5. Chromosomes Chromosomes occur in pairs called homologous chromosomes.
One from each parent.

6. Genes are the functional unit of heredity
Chromosomes are made up of genes that code for traits.
A gene is found at a specific location or locus on a chromosome.

7. Genes & Alleles Different versions of genes are called alleles.
Purple flowers vs. white in pea plants
Gene = flower color, allele = white or purple

8. Genes & Alleles
There can be any number of alleles for a given gene, although an individual can have only two alleles(one on each homologous chromosome).
A, B, O blood type in humans

9. Genes & Alleles
Some traits are controlled by just one gene, others are influenced by many genes (polygenic).
Height in humans

10. A, B and O are _______ of the human blood type __________.
Question 1
A, B and O are _______ of the human blood type __________.
Genes, genes
Genes, alleles
Alleles, genes
Alleles, alleles

11. Question 1
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12. Question 1
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13. Homozygous & Heterozygous
Since an individual has two sets of chromosomes, it will have two copies of each gene (one originally coming from each parent). These two copies may be the same allele, or they may be different.
Homozygous – both alleles are the same.
Heterozygous – two different alleles.

14. Which of the following represents the homozygous condition?
Question 2
Which of the following represents the homozygous condition? AA Aa aa
Both AA and aa

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Question 2
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16. Question 2
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17. Dominant & Recessive
A trait is dominant if it is expressed in an individual with one or two copies of the allele:
Purple flower color in peas: P= purple p= white.
The dominant allele is represented by a capitol letter, recessive by the lower case letter.
PP – homozygous dominant – two copies of the dominant (purple) allele.
Pp – heterozygous – one purple allele, one white allele (flowers appear purple).

18. Dominant & Recessive
The trait is said to be recessive if it is necessary for an individual to have two copies of the allele in order to express the trait.
pp – two white flower alleles (homozygous).

19. Question 3
Purple flower color in peas is dominant over white. Which of the following pairs of alleles would give purple flowers? PP Pp pp
Both PP and Pp

20. Question 3
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21. Question 3
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22. Now, which of the following pairs of alleles would give white flowers?
Question 4
Now, which of the following pairs of alleles would give white flowers? PP Pp pp
Both PP and Pp

23. Question 4
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24. Question 4
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25. Genotype Genotype refers to the alleles that are actually present.
PP, Pp, pp in our flower color example.
The purple phenotype may have PP or Pp genotype.
Back to question 5

26. Phenotype
Phenotype refers to the visible or expressed characteristics of the trait.
What does it look like?
Purple or white for our flower color example.
Back to question 8

27. Heredity – Passing on Traits
An individual can pass on genetic information to its offspring. In order to avoid doubling the number of chromosomes in each generation, cells must be created that carry only one set of chromosomes (haploid or 1n).
An individual can pass along either of the two alleles it carries for a trait, but not both.
A Pp individual can pass on either P or p.
These haploid cells (eggs or sperm) are formed during meiosis.

28. Meiosis Meiosis is a type of cell division that occurs in two parts.
Before division starts, each chromosome makes a copy of itself. These identical copies (called sister chromatids) remain connected at the centromere.

29. Meiosis
In the first part of meiosis, homologous chromosomes are separated.
The cells are now haploid.
During the second part of meiosis, the sister chromatids are pulled apart.
This part is like regular cell division (mitosis).
The division of one diploid cell will result in 4 haploid cells.
Each cell is different.

31. Heredity
We can look at how traits are passed from one generation to another individually or two at a time using a Punnett square.

32. Heredity
For our example, we will use the ball python. There are many mutations that breeders want to incorporate into their animals.
Albino – a simple recessive trait
Pinstripe – a dominant pattern mutation

33. Punnett Square – 1 Trait
First let’s focus on the albino trait. It is recessive so:
AA & Aa individuals will have normal coloration.
aa individuals will be albino.

34. Punnett Square – 1 Trait
In a monohybrid cross we will cross two animals that are heterozygous for albino.
Aa x Aa
We want to know, statistically, what kind of offspring to expect.
Each parent can donate only one allele for the albino gene.
A heterozygote (Aa) can donate either an A or an a – not both.
An albino must receive an a from both parents.

35. Punnett Square – 1 Trait
Place the alleles that may be donated by each parent across the top and along the sides.
Fill in the boxes:
1AA - normal
2Aa – normal, heterozygous for albino
1aa - albino
Back to question 8

36. Which is the genotype? Normal or albino AA, Aa, or aa
Question 5
Which is the genotype?
Normal or albino
AA, Aa, or aa
Both are considered to be the genotype.
Neither of these is the genotype.

37. Sorry! That is incorrect. Find information on genotype. Try again!
Question 5
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Find information on genotype.
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38. Question 5
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39. Let’s try another example!
If we have a male that is heterozygous for albino and an albino female, what kind of offspring do we get? X

40. What is the genotype of this pairing?
Question 6
What is the genotype of this pairing?
Aa x Aa
AA x aa
Aa x aa
aa x aa

41. Question 6
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42. Question 6
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43. Which of these Punnett squares is correct?
Question 7
Which of these Punnett squares is correct?

44. Question 7
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45. Question 7
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46. What are the phenotypes of the offspring?
Question 8
What are the phenotypes of the offspring?
All normal appearing offspring
All albino offspring
3:1 normal to albino
2:2 normal to albino

47. Sorry! That is incorrect. Review Phenotype Review Punnett squares
Question 8
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Review Phenotype
Review Punnett squares
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48. Question 8
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49. Punnett Square – 2 Traits
We can also use the Punnett square to track two traits at once.
Remember each gamete (egg or sperm) will contain one allele for each trait.
So, the possible combinations of alleles that we will place on our Punnett squares will always have one letter for each trait.

50. Punnett Square – 2 Traits
In a dihybrid cross, both animals are heterozygous for two traits – here, albino (recessive) and pinstripe (dominant).
AaPp x AaPp
The parents will have normal coloration (Aa) and they will be Pinstripes (Pp).

51. Punnett Square – 2 Traits
Again, we want to know, statistically, what kind of offspring to expect.
Each parent (AaPp) will donate either an A or an a allele for the albino gene and either a P or a p allele for the pinstripe gene.
So every gamete will always contain ONE A(or a) and one P(or p).

52. FOIL
We can use the FOIL method from math to be sure that we have all of the possible combinations of alleles.
First, Outer, Inner, Last

54. Punnett Square – 2 Traits
Next, we fill in each square. By convention, we put the alleles for one gene together followed by the second:
Aapp not Apap
Also, any dominant alleles are placed before recessives.
AaPp not aApP

56. Genotypes from Dihybrid Cross
The Punnett square gives us the genotypes that result from the cross.

57. Phenotypes from Dihybrid Cross
The phenotypes would be:
9 Pinstripe (A_P_)
3 Normal (A_pp)
3 Albino pinstripe (aaP_)
1 Albino (aapp)
Albino is a recessive trait, while pinstripe is a dominant trait.
The 9:3:3:1 phenotypic ratio is characteristic of a dihybrid cross.

58. Question 9
Which of the following Punnett squares is correct for this cross: AaPp x aapp

59. Question 9
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60. Question 9
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61. Question 10
What is the ratio of phenotypes that would result from the cross? AaPp x aapp
9 normal: 3 albino: 3 pinstripe: 1albino pinstripe
4 pinstripe: 4 normal: 4 albino pinstripe: 4 albino
All albino pinstripe
AaPp: 4 Aapp: 4 aaPp: 4 aapp

62. Question 10
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63. Question 10
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64. The Product Rule
Punnett squares are very useful for tracking one or two traits, but they can become unwieldy when looking at more than two traits.
The product rule is a simple way to determine the likelihood of getting a particular result from any cross, regardless of the number of traits involved.

65. The Product Rule
To use the product rule, we determine the likelihood of getting each trait individually, then multiply those probabilities together.
We’ll use our dihybrid cross example to start with.
AaPp x AaPp

66. The Product Rule We need to look at the traits separately: Aa x Aa
There would be a 1 in 4 chance of hatching an albino from this cross.
Pp x Pp
Pinstripe is dominant so ¾ of the offspring will be pinstripes.
What is the chance of getting an albino pinstripe from this cross?
¼ x ¾ = 3/16
This is the same result that we got using the Punnett square.

67. The Product Rule
We can look at as many traits as we want using the product rule.
Say we are interested in combining these 4 traits:
Pinstripe (dominant) – (PP, Pp, pp)
Albino (recessive) – (AA, Aa, aa)
Piebald (recessive) – (BB, Bb, bb)
Hypo (recessive) – (HH, Hh, hh)

68. The Product Rule The parents have the following genotypes:
AaPpBbhh x aappBbHh
Calculate probability of getting individual traits:
Albino – Aa x aa = ½
Pinstripe – Pp x pp = ½
Piebald – Bb x Bb = ¼
Hypo – hh x Hh = ½
½ x ½ x ¼ x ½ = 1/32 = chance of getting an animal that shows all 4 traits from this pairing.

69. Question 11
What is the probability of getting a hypo albino piebald animal from these parents: AaBbhh x Aabbhh
Remember, albino, piebald, and hypo are all recessive traits.
1/8
1/16

70. Question 11
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71. Question 11
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