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 P031S090007.

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. 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.

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

12. Sorry!  That is incorrect – be sure your answer is complete.  Try again! Try again! Question 2

13. Congratulations!  You are correct! Question 2

14. 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).

15. 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).

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

17. Sorry!  That is incorrect.  Try again! Try again! Question 3

18. Congratulations!  You are correct! Question 3

19. Now, which of the following pairs of alleles would give white flowers?  PP PP  Pp Pp  pp pp  Both PP and Pp Both PP and Pp Question 4

20. Sorry!  That is incorrect.  Try again! Try again! Question 4

21. Congratulations!  You are correct! Question 4

22. 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

23. 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

24. 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.

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

26. 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

27. 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.

28. 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.

29. 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

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

31. Sorry!  That is incorrect.  Find information on genotype.genotype  Try again! Try again! Question 5

32. Congratulations!  You are correct! Question 5

33. 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

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

35. Sorry!  That is incorrect.  Try again! Try again! Question 6

36. Congratulations!  You are correct! Question 6

37. Which of these Punnett squares is correct? Question 7

38. Sorry!  That is incorrect.  Try again! Try again! Question 7

39. Congratulations!  You are correct! Question 7

40. What are the phenotypes of the offspring?  All normal appearing offspring All normal appearing offspring  All albino offspring All albino offspring  3:1 normal to albino 3:1 normal to albino  2:2 normal to albino 2:2 normal to albino Question 8

41. Sorry!  That is incorrect.  Review Phenotype Review Phenotype  Review Punnett squares Review Punnett squares  Try again! Try again! Question 8

42. Congratulations!  You are correct! Question 8

43. 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.

44. 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).

45. 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 ).

46. FOIL  We can use the FOIL method from math to be sure that we have all of the possible combinations of alleles.  F irst, O uter, I nner, L ast

48. 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

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

51. 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.

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

53. Sorry!  That is incorrect.  Try again! Try again! Question 9

54. Congratulations!  You are correct! Question 9

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

56. Sorry!  That is incorrect.  Try again! Try again! Question 10

57. Congratulations!  You are correct! Question 10

58. 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.

59. 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

60. 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.

61. 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)

62. 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.

63. 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.  ½ ½  ¼ ¼  1/8 1/8  1/16 1/16 Question 11

64. Sorry!  That is incorrect.  Try again! Try again! Question 11

65. Congratulations!  You are correct! Question 11