1. Topic 3.4 Inheritance 13.4 Theoretical Genetics

2. Definitions 2 This image shows a pair of homologous chromosomes. Name and annotate the labeled features. 3.4 Theoretical Genetics

3. Definitions 3 This image shows a pair of homologous chromosomes. Name and annotate the labeled features. Centromere Joins chromatids in cell division Gene loci Specific positions of genes on a chromosome Alleles Different versions of a gene Dominant alleles = capital letter Recessive alleles = lower-case letter Homozygous dominant Having two copies of the same dominant allele Homozygous recessive Having two copies of the same recessive allele. Recessive alleles are only expressed when homozygous. Heterozygous Having two different alleles. The dominant allele is expressed. Codominant Pairs of alleles which are both expressed when present. Carrier Heterozygous carrier of a recessive disease-causing allele Genotype The combination of alleles of a gene carried by an organism Phenotype The expression of alleles of a gene carried by an organism 3.4 Theoretical Genetics

4. Explain this 4 Mendel crossed some yellow peas with some yellow peas. Most offspring were yellow but some were green! Mendel from: http://history.nih.gov/exhibits/nirenberg/popup_htm/01_mendel.htm 3.4 Theoretical Genetics

5. Segregation 5 “alleles of each gene separate into different gametes when the individual produces gametes” The yellow parent peas must be heterozygous. The yellow phenotype is expressed. Through meiosis and fertilization, some offspring peas are homozygous recessive – they express a green color. Mendel did not know about DNA, chromosomes or meiosis. Through his experiments he did work out that ‘heritable factors’ (genes) were passed on and that these could have different versions (alleles). Mendel from: http://history.nih.gov/exhibits/nirenberg/popup_htm/01_mendel.htm 3.4 Theoretical Genetics

6. Segregation 6 Mendel from: http://history.nih.gov/exhibits/nirenberg/popup_htm/01_mendel.htm “alleles of each gene separate into different gametes when the individual produces gametes” F0F0 F1F1 Genotype: Y y Gametes: Y or y Alleles segregate during meiosis (anaphase I) and end up in different haploid gametes. Punnett Grid: Genotypes: Phenotypes: Phenotype ratio: 3.4 Theoretical Genetics

7. Monohybrid Cross 7 Mendel from: http://history.nih.gov/exhibits/nirenberg/popup_htm/01_mendel.htm F0F0 F1F1 Genotype: Y y Gametes: Y or y Alleles segregate during meiosis (anaphase I) and end up in different haploid gametes. Fertilization results in diploid zygotes. A punnett square can be used to deduce the potential outcomes of the cross and to calculate the expected ratio of phenotypes in the next generation (F 1 ). Punnett Grid: Genotypes: Phenotypes: Phenotype ratio: Crossing a single trait. 3.4 Theoretical Genetics

8. 8 Mendel from: http://history.nih.gov/exhibits/nirenberg/popup_htm/01_mendel.htm F0F0 F1F1 Genotype: Y y Gametes: Y or y Alleles segregate during meiosis (anaphase I) and end up in different haploid gametes. Fertilization results in diploid zygotes. A punnett square can be used to deduce the potential outcomes of the cross and to calculate the expected ratio of phenotypes in the next generation (F 1 ). Punnett Grid: Genotypes: Phenotypes: Phenotype ratio: Monohybrid Cross Crossing a single trait. 3.4 Theoretical Genetics

9. 9 Mendel from: http://history.nih.gov/exhibits/nirenberg/popup_htm/01_mendel.htm F0F0 F1F1 Genotype: Y y Gametes: Y or y Alleles segregate during meiosis (anaphase I) and end up in different haploid gametes. Fertilization results in diploid zygotes. A punnett square can be used to deduce the potential outcomes of the cross and to calculate the expected ratio of phenotypes in the next generation (F 1 ). Ratios are written in the simplest mathematical form. Punnett Grid: YY Yy yy Genotypes: Phenotypes: Phenotype ratio: 3 : 1 Monohybrid Cross Crossing a single trait. 3.4 Theoretical Genetics

10. 10 F0F0 F1F1 Genotype: Punnett Grid: Genotypes: Phenotypes: Phenotype ratio: Monohybrid Cross What is the expected ratio of phenotypes in this monohybrid cross? Homozygous recessive Phenotype: Key to alleles: Y = yellow y = green Key to alleles: Y = yellow y = green 3.4 Theoretical Genetics

11. 11 F0F0 F1F1 Genotype: y Punnett Grid: yy Genotypes: Phenotypes: Phenotype ratio: All green Monohybrid Cross What is the expected ratio of phenotypes in this monohybrid cross? Homozygous recessive Phenotype: Key to alleles: Y = yellow y = green Key to alleles: Y = yellow y = green 3.4 Theoretical Genetics

12. 12 F0F0 F1F1 Genotype: Punnett Grid: Genotypes: Phenotypes: Phenotype ratio: Monohybrid Cross What is the expected ratio of phenotypes in this monohybrid cross? Phenotype: Heterozygous Homozygous recessive Key to alleles: Y = yellow y = green Key to alleles: Y = yellow y = green 3.4 Theoretical Genetics

13. 13 F0F0 F1F1 Genotype: y Y y Punnett Grid: Yy yy Genotypes: Phenotypes: Phenotype ratio: 1 : 1 Monohybrid Cross What is the expected ratio of phenotypes in this monohybrid cross? Heterozygous Homozygous recessive Phenotype: Key to alleles: Y = yellow y = green Key to alleles: Y = yellow y = green 3.4 Theoretical Genetics

14. 14 F0F0 F1F1 Genotype: Punnett Grid: Genotypes: Phenotypes: Phenotype ratio: Monohybrid Cross What is the expected ratio of phenotypes in this monohybrid cross? Phenotype: Heterozygous Homozygous dominant Key to alleles: Y = yellow y = green Key to alleles: Y = yellow y = green 3.4 Theoretical Genetics

15. 15 F0F0 F1F1 Genotype: Y Y y Punnett Grid: YY Yy Genotypes: Phenotypes: Phenotype ratio: All yellow Monohybrid Cross What is the expected ratio of phenotypes in this monohybrid cross? Heterozygous Homozygous dominant Phenotype: Key to alleles: Y = yellow y = green Key to alleles: Y = yellow y = green 3.4 Theoretical Genetics

16. 16 F0F0 F1F1 Genotype: R ? r Phenotypes: Test Cross Used to determine the genotype of an unknown individual. The unknown is crossed with a known homozygous recessive. Homozygous recessive unknown Phenotype: Key to alleles: R = Red flower r = white Key to alleles: R = Red flower r = white Unknown parent = RRUnknown parent = Rr Possible outcomes: 3.4 Theoretical Genetics

17. 17 F0F0 F1F1 Genotype: R ? r Phenotypes: All red Test Cross Used to determine the genotype of an unknown individual. The unknown is crossed with a known homozygous recessive. Homozygous recessive unknown Phenotype: Key to alleles: R = Red flower r = white Key to alleles: R = Red flower r = white Some white, some red Unknown parent = RRUnknown parent = Rr Possible outcomes: 3.4 Theoretical Genetics

18. 18 Pedigree Charts Key to alleles: T= Has enzyme t = no enzyme Key to alleles: T= Has enzyme t = no enzyme Pedigree charts can be used to trace family histories and deduce genotypes and risk in the case of inherited gene-related disorders. Here is a pedigree chart for this family history. Looks like 3.4 Theoretical Genetics

19. 19 Pedigree Charts Key to alleles: T= Has enzyme t = no enzyme Key to alleles: T= Has enzyme t = no enzyme Pedigree charts can be used to trace family histories and deduce genotypes and risk in the case of inherited gene-related disorders. Here is a pedigree chart for this family history. Looks like 3.4 Theoretical Genetics

20. 20 Pedigree Charts Key to alleles: T= Has enzyme t = no enzyme Key to alleles: T= Has enzyme t = no enzyme Individuals D and $ are planning to have another child. Calculate the chances of the child having PKU. Looks like $ Genotypes: D = $ = Phenotype ratio Therefore 3.4 Theoretical Genetics

21. 21 Pedigree Charts Key to alleles: T= Has enzyme t = no enzyme Key to alleles: T= Has enzyme t = no enzyme Individuals D and $ are planning to have another child. Calculate the chances of the child having PKU. Looks like $ Genotypes: D = Tt (carrier) $ = tt (affected) Phenotype ratio 1 : 1 Normal : PKU Therefore 50% chance of a child with PKU 3.4 Theoretical Genetics

22. 22 Codominance Some genes have more than two alleles. Where alleles are codominant, they are both expressed. Human ABO blood typing is an example of multiple alleles and codominance. The gene is for cell-surface antigens (immunoglobulin receptors). These are either absent (type O) or present. If they are present, they are either type A, B or both. Where the genotype is heterozygous for I A and I B, both are expressed. This is codominance. Key to alleles: i = no antigens present I A = type A antigens present I B = type B antigens present Key to alleles: i = no antigens present I A = type A antigens present I B = type B antigens present 3.4 Theoretical Genetics

23. 23 More about blood typing A Nobel breakthrough in medicine. Images and more information from: http://learn.genetics.utah.edu/content/begin/traits/blood/ Antibodies (immunoglobulins) are specific to antigens. The immune system recognizes 'foreign' antigens and produces antibodies in response - so if you are given the wrong blood type your body might react fatally as the antibodies cause the blood to clot. Blood type O is known as the universal donor, as it has not antigens against which the recipient immune system can react. Type AB is the universal recipient, as it has no antibodies which will react to AB antigens. Blood typing game from Nobel.org: http://www.nobelprize.org/educational/medicine/bloodtypinggame/gam ev2/index.html 3.4 Theoretical Genetics

24. 24 Sickle Cell Another example of codominance. Remember the notation used: superscripts represent codominant alleles. In codominance, heterozygous individuals have a mixed phenotype. The mixed phenotype gives protection against malaria, but does not exhibit full-blown sickle cell anemia. Complete the table for these individuals: 3.4 Theoretical Genetics

25. 25 Sickle Cell Another example of codominance. Remember the notation used: superscripts represent codominant alleles. In codominance, heterozygous individuals have a mixed phenotype. The mixed phenotype gives protection against malaria, but does not exhibit full-blown sickle cell anemia. Complete the table for these individuals: 3.4 Theoretical Genetics

26. 26 Sickle Cell Another example of codominance. Predict the phenotype ratio in this cross: Key to alleles: Hb A = Normal Hb Hb S = Sickle cell Key to alleles: Hb A = Normal Hb Hb S = Sickle cell Therefore 50% chance of a child with sickle cell disease. F1F1 Punnett Grid: Genotypes: Phenotypes: F0F0 Genotype: Phenotype: carrier affected Phenotype ratio: : 3.4 Theoretical Genetics

27. 27 Sickle Cell Another example of codominance. Predict the phenotype ratio in this cross: Key to alleles: Hb A = Normal Hb Hb S = Sickle cell Key to alleles: Hb A = Normal Hb Hb S = Sickle cell Therefore 50% chance of a child with sickle cell disease. F1F1 Punnett Grid: Genotypes: Phenotypes: F0F0 Genotype: Hb A Hb s Hb S Hb s Phenotype: carrier affected Phenotype ratio: Carrier & Sickle cell 1 : 1 Hb A Hb S & Hb S Hb S 3.4 Theoretical Genetics

28. 28 Sickle Cell Another example of codominance. Predict the phenotype ratio in this cross: Key to alleles: Hb A = Normal Hb Hb S = Sickle cell Key to alleles: Hb A = Normal Hb Hb S = Sickle cell F1F1 Punnett Grid: Genotypes: Phenotypes: F0F0 Genotype: Phenotype: carrier Phenotype ratio: 3.4 Theoretical Genetics

29. 29 Sickle Cell Another example of codominance. Predict the phenotype ratio in this cross: Key to alleles: Hb A = Normal Hb Hb S = Sickle cell Key to alleles: Hb A = Normal Hb Hb S = Sickle cell Therefore 25% chance of a child with sickle cell disease. F1F1 Punnett Grid: Genotypes: Phenotypes: F0F0 Genotype: Hb A Hb S Phenotype: carrier Phenotype ratio: Unaffected & Carrier & Sickle cell 1: 2 : 1 Hb A Hb & 2 Hb A Hb S & Hb S Hb S 3.4 Theoretical Genetics

30. 30 Sickle Cell Another example of codominance. Predict the phenotype ratio in this cross: Key to alleles: Hb A = Normal Hb Hb S = Sickle cell Key to alleles: Hb A = Normal Hb Hb S = Sickle cell F1F1 Punnett Grid: Genotypes: Phenotypes: F0F0 Genotype: Hb A Hb S Phenotype: carrier unknown Phenotype ratio: 3.4 Theoretical Genetics

31. 31 Sickle Cell Another example of codominance. Predict the phenotype ratio in this cross: Key to alleles: Hb A = Normal Hb Hb S = Sickle cell Key to alleles: Hb A = Normal Hb Hb S = Sickle cell F1F1 Punnett Grid: Genotypes: Phenotypes: F0F0 Genotype: Hb A Hb S Hb A Hb A or Hb A Hb S Phenotype: carrier unknown Phenotype ratio: 3.4 Theoretical Genetics

32. 32 Sickle Cell Another example of codominance. Predict the phenotype ratio in this cross: Key to alleles: Hb A = Normal Hb Hb S = Sickle cell Key to alleles: Hb A = Normal Hb Hb S = Sickle cell Therefore 12.5% chance of a child with sickle cell disease. F1F1 Punnett Grid: Genotypes: Phenotypes: F0F0 Genotype: Hb A Hb S Hb A Hb A or Hb A Hb S Phenotype: carrier unknown Phenotype ratio: 3 Unaffected & 4 Carrier & 1 Sickle cell 3 : 4 : 1 3 Hb A Hb A & 4 Hb A Hb S & 1 Hb S Hb S 3.4 Theoretical Genetics

33. 33 Sex Determination It’s all about X and Y… Karyotype of a human male, showing X and Y chromosomes: http://en.wikipedia.org/wiki/Karyotype Humans have 23 pairs of chromosomes in diploid somatic cells (n=2). 22 pairs of these are autosomes, which are homologous pairs. One pair is the sex chromosomes. XX gives the female gender, XY gives male. The X chromosome is much larger than the Y. X carries many genes in the non-homologous region which are not present on Y. The presence and expression of the SRY gene on Y leads to male development. SRY Chromosome images from Wikipedia: http://en.wikipedia.org/wiki/Y_chromosome 3.4 Theoretical Genetics

34. 34 Sex Determination It’s all about X and Y… Segregation of the sex chromosomes in meiosis. Chromosome pairs segregate in meiosis. Females (XX) produce only eggs containing the X chromosome. Males (XY) produce sperm which can contain either X or Y chromosomes. Therefore there is an even chance* of the offspring being male or female. SRY gene determines maleness. Find out more about its role and just why do men have nipples? http://www.hhmi.org/biointeractive/gender/lectures.html Chromosome images from Wikipedia: http://en.wikipedia.org/wiki/Y_chromosome 3.4 Theoretical Genetics

35. 35 Sex Determination Non-disjunction can lead to gender disorders. XYY Syndrome: Fertile males, with increased risk of learning difficulties. Some weak connections made to violent tendencies. XO: Turner Syndrome Monosomy of X, leads to short stature, female children. XXX Syndrome: Fertile females. Some X-carrying gametes can be produced. XXY: Klinefelter Syndrome: Males with enhanced female characteristics Image from NCBI: http://www.ncbi.nlm.nih.gov/bookshelf/br.fcgi?book=mga&part=A1179 Interactive from HHMI Biointeractive: http://www.hhmi.org/biointeractive/gender/click.html 3.4 Theoretical Genetics

36. 36 Colour Blind cartoon from: http://www.almeidacartoons.com/Med_toons1.html 3.4 Theoretical Genetics

37. 37 Sex Linkage X and Y chromosomes are non-homologous. Non-homologous region The sex chromosomes are non-homologous. There are many genes on the X-chromosome which are not present on the Y-chromosome. Sex-linked traits are those which are carried on the X-chromosome in the non-homologous region. They are more common in males. Examples of sex-linked genetic disorders: - hemophilia - color blindness X and Y SEM from http://www.angleseybonesetters.co.uk/bones_DNA.html Chromosome images from Wikipedia: http://en.wikipedia.org/wiki/Y_chromosome 3.4 Theoretical Genetics

38. 38 Sex Linkage X and Y chromosomes are non-homologous. What number do you see? Chromosome images from Wikipedia: http://en.wikipedia.org/wiki/Y_chromosome 3.4 Theoretical Genetics

39. 39 Sex Linkage X and Y chromosomes are non-homologous. What number do you see? 5 = normal vision 2 = red/green color blindness Chromosome images from Wikipedia: http://en.wikipedia.org/wiki/Y_chromosome 3.4 Theoretical Genetics

40. 40 Sex Linkage X and Y chromosomes are non-homologous. How is color-blindness inherited? The red-green gene is carried at locus Xq28. This locus is in the non-homologous region, so there is no corresponding gene (or allele) on the Y chromosome. Normal vision is dominant over color-blindness. Xq28 Key to alleles: N = normal vision n = red/green color blindness Key to alleles: N = normal vision n = red/green color blindness X N X n X N X n X N Y X n Y no allele carried, none written Normal female Normal male Affected femaleAffected male Carrier female Human females can be homozygous or heterozygous with respect to sex-linked genes. Heterozygous females are carriers. Chromosome images from Wikipedia: http://en.wikipedia.org/wiki/Y_chromosome 3.4 Theoretical Genetics

41. 41 Sex Linkage X and Y chromosomes are non-homologous. What chance of a color-blind child in the cross between a normal male and a carrier mother? Key to alleles: N = normal vision n = red/green color blindness Key to alleles: N = normal vision n = red/green color blindness X N X n X N Y Normal male Carrier female X F1F1 Punnett Grid: F0F0 Genotype: Phenotype: Chromosome images from Wikipedia: http://en.wikipedia.org/wiki/Y_chromosome 3.4 Theoretical Genetics

42. 42 Sex Linkage X and Y chromosomes are non-homologous. What chance of a color-blind child in the cross between a normal male and a carrier mother? Key to alleles: N = normal vision n = red/green color blindness Key to alleles: N = normal vision n = red/green color blindness X N X n X N Y Normal male Carrier female X XNXN XnXn XNXN Y XN XNXN XN XN XnXN Xn X N Y X n Y F1F1 Punnett Grid: F0F0 Genotype: Phenotype: Chromosome images from Wikipedia: http://en.wikipedia.org/wiki/Y_chromosome 3.4 Theoretical Genetics

43. 43 Sex Linkage X and Y chromosomes are non-homologous. What chance of a color-blind child in the cross between a normal male and a carrier mother? Key to alleles: N = normal vision n = red/green color blindness Key to alleles: N = normal vision n = red/green color blindness X N X n X N Y Normal male Carrier female X XNXN XnXn XNXN Y XN XNXN XN XN XnXN Xn X N Y X n Y F1F1 Punnett Grid: F0F0 Genotype: Phenotype: There is a 1 in 4 (25%) chance of an affected child. Carrier female Normal female Normal male Affected male What ratios would we expect in a cross between: a. a color-blind male and a homozygous normal female? b. a normal male and a color-blind female? Chromosome images from Wikipedia: http://en.wikipedia.org/wiki/Y_chromosome 3.4 Theoretical Genetics

44. 44 Red-Green Color Blindness How does it work? Xq28 The OPN1MW and OPN1LW genes are found at locus Xq28. They are responsible for producing photoreceptive pigments in the cone cells in the eye. If one of these genes is a mutant, the pigments are not produced properly and the eye cannot distinguish between green (medium) wavelengths and red (long) wavelengths in the visible spectrum. Because the Xq28 gene is in a non-homologous region when compared to the Y chromosome, red-green color blindness is known as a sex- linked disorder. The male has no allele on the Y chromosome to combat a recessive faulty allele on the X chromosome. Chromosome images from Wikipedia: http://en.wikipedia.org/wiki/Y_chromosome 3.4 Theoretical Genetics

45. 45 Hemophilia Another sex-linked disorder. Blood clotting is an example of a metabolic pathway – a series of enzyme-controlled biochemical reactions. It requires globular proteins called clotting factors. A recessive X-linked mutation in hemophiliacs results in one of these factors not being produced. Therefore, the clotting response to injury does not work and the patient can bleed to death. X H X h X H X h X H Y X h Y no allele carried, none written Normal female Normal male Affected femaleAffected male Carrier female Human females can be homozygous or heterozygous with respect to sex-linked genes. Heterozygous females are carriers. Key to alleles: X H = healthy clotting factors X h = no clotting factor Key to alleles: X H = healthy clotting factors X h = no clotting factor Chromosome images from Wikipedia: http://en.wikipedia.org/wiki/Y_chromosome 3.4 Theoretical Genetics

46. 46 Hemophilia results from a lack of clotting factors. These are globular proteins, which act as enzymes in the clotting pathway. Read/ research/ review: How can gene transfer be used to treat hemophiliacs? What is the relevance of “the genetic code is universal” in this process? Chromosome images from Wikipedia: http://en.wikipedia.org/wiki/Y_chromosome

47. 47 Hemophilia results from a lack of clotting factors. These are globular proteins, which act as enzymes in the clotting pathway. Chromosome images from Wikipedia: http://en.wikipedia.org/wiki/Y_chromosome

48. Hemophilia This pedigree chart of the English Royal Family gives us a picture of the inheritance of this X-linked disorder. Royal Family Pedigree Chart from: http://www.sciencecases.org/hemo/hemo.asp 3.4 Theoretical Genetics

49. 49 Hemophilia Pedigree chart practice Royal Family Pedigree Chart from: http://www.sciencecases.org/hemo/hemo.asp Key to alleles: H = healthy clotting factors h = no clotting factor Key to alleles: H = healthy clotting factors h = no clotting factor State the genotypes of the following family members: 1.Leopold 2.Alice 3.Bob was killed in a tragic croquet accident before his phenotype was determined. 4.Britney 3.4 Theoretical Genetics

50. 50 Hemophilia Pedigree chart practice Royal Family Pedigree Chart from: http://www.sciencecases.org/hemo/hemo.asp Key to alleles: H = healthy clotting factors h = no clotting factor Key to alleles: H = healthy clotting factors h = no clotting factor State the genotypes of the following family members: 1.Leopold 2.Alice 3.Bob was killed in a tragic croquet accident before his phenotype was determined. 4.Britney X h Y 3.4 Theoretical Genetics

51. 51 Hemophilia Pedigree chart practice Royal Family Pedigree Chart from: http://www.sciencecases.org/hemo/hemo.asp Key to alleles: H = healthy clotting factors h = no clotting factor Key to alleles: H = healthy clotting factors h = no clotting factor State the genotypes of the following family members: 1.Leopold 2.Alice 3.Bob was killed in a tragic croquet accident before his phenotype was determined. 4.Britney X h Y X H X h 3.4 Theoretical Genetics

52. 52 Hemophilia Pedigree chart practice Royal Family Pedigree Chart from: http://www.sciencecases.org/hemo/hemo.asp Key to alleles: H = healthy clotting factors h = no clotting factor Key to alleles: H = healthy clotting factors h = no clotting factor State the genotypes of the following family members: 1.Leopold 2.Alice 3.Bob was killed in a tragic croquet accident before his phenotype was determined. 4.Britney X h Y X H X h X H Y or X h Y 3.4 Theoretical Genetics

53. 53 Hemophilia Pedigree chart practice Royal Family Pedigree Chart from: http://www.sciencecases.org/hemo/hemo.asp Key to alleles: H = healthy clotting factors h = no clotting factor Key to alleles: H = healthy clotting factors h = no clotting factor State the genotypes of the following family members: 1.Leopold 2.Alice 3.Bob was killed in a tragic croquet accident before his phenotype was determined. 4.Britney X h Y X H X h X H Y or X h Y X H X H or X H X h 3.4 Theoretical Genetics

54. 54 Pedigree Chart Practice Dominant or Recessive?Autosomal or Sex-linked? 3.4 Theoretical Genetics

55. 55 Pedigree Chart Practice Dominant or Recessive? Dominant. A and B are both affected but have produced unaffected (D & F). Therefore A and B must have been carrying recessive healthy alleles. If it were recessive, it would need to be homozygous to be expressed in A & B – and then all offspring would be homozygous recessive. Autosomal or Sex-linked? 3.4 Theoretical Genetics

56. 56 Pedigree Chart Practice Dominant or Recessive? Dominant. A and B are both affected but have produced unaffected (D & F). Therefore A and B must have been carrying recessive healthy alleles. If it were recessive, it would need to be homozygous to be expressed in A & B – and then all offspring would be homozygous recessive. Autosomal or Sex-linked? Autosomal. Male C can only pass on one X chromosome. If it were carried on X, daughter H would be affected by the dominant allele. Tip: Don’t get hung up on the number of individuals with each phenotype – each reproductive event is a matter of chance. Instead focus on possible and impossible genotypes. Draw out the punnett squares if needed. 3.4 Theoretical Genetics

57. 57 Super Evil Past Paper Question In this pedigree chart for hemophilia, what is the chance that offspring ? will be affected? A. 0% B. 12.5% C. 25% D. 50% 3.4 Theoretical Genetics

58. 58 Super Evil Past Paper Question In this pedigree chart for hemophilia, what is the chance that offspring ? will be affected? A. 0% B. 12.5% C. 25% D. 50% Key to alleles: X H = healthy clotting factors X h = no clotting factor Key to alleles: X H = healthy clotting factors X h = no clotting factor 3.4 Theoretical Genetics

59. 59 Super Evil Past Paper Question In this pedigree chart for hemophilia, what is the chance that offspring ? will be affected? A. 0% B. 12.5% C. 25% D. 50% Key to alleles: X H = healthy clotting factors X h = no clotting factor Key to alleles: X H = healthy clotting factors X h = no clotting factor What do we know? A = X H Y B = X H X h (because G = X h Y) E = X H Y 3.4 Theoretical Genetics

60. 60 Super Evil Past Paper Question In this pedigree chart for hemophilia, what is the chance that offspring ? will be affected? A. 0% B. 12.5% C. 25% D. 50% Key to alleles: X H = healthy clotting factors X h = no clotting factor Key to alleles: X H = healthy clotting factors X h = no clotting factor What do we know? A = X H Y B = X H X h (because G = X h Y) E = X H Y There is an equal chance of F being X H X H or X H X h So: 3.4 Theoretical Genetics

61. 61 Super Evil Past Paper Question In this pedigree chart for hemophilia, what is the chance that offspring ? will be affected? A. 0% B. 12.5% C. 25% D. 50% Key to alleles: X H = healthy clotting factors X h = no clotting factor Key to alleles: X H = healthy clotting factors X h = no clotting factor What do we know? A = X H Y B = X H X h (because G = X h Y) E = X H Y There is an equal chance of F being X H X H or X H X h So: 3.4 Theoretical Genetics

62. 62 Super Evil Past Paper Question In this pedigree chart for hemophilia, what is the chance that offspring ? will be affected? A. 0% B. 12.5% C. 25% D. 50% Key to alleles: X H = healthy clotting factors X h = no clotting factor Key to alleles: X H = healthy clotting factors X h = no clotting factor What do we know? A = X H Y B = X H X h (because G = X h Y) E = X H Y There is an equal chance of F being X H X H or X H X h So: So there is a 1 in 8 (12.5%) chance of the offspring being affected! 3.4 Theoretical Genetics

63. Cystic Fibrosis is a recessive, autosomal disorder. If you have two unaffected parents, and a child with cystic fibrosis, what must their genotype be for the trait? Cystic Fibrosis affects these protein channels all over your body! Lungs, Pancreas, Sweat Glands This reduces their capacity to deal with the build up of mucus. Check out all the mucus build up! This mucus eventually builds up and creates a great environment for bacteria to grow. Most people die of infections of Pseudomonas aeruginosa (Option D)