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Course Overview Modified Mendelian ratios http://www.erin.utoronto.ca/~w3bio/bio207/index.htm March 1st 2007 Test and Tutorials • You need to do the assigned questions at the end of chapter 6 • As stated on the website you need to present a doctor’s note to the instructor within 48h of missing a test (or a quiz) • If you have a legitimate reason for missing a quiz email IN ADVANCE • NB. Academic conflict is NOT a legitimate reason for missing a quiz • Late penalty is 20% per day for quizzes that are to be handed in to drop boxes. Outline Week 1 2 3 4 5 6 7 8 9 10 11 12 Topic Course objectives and Introduction to genetics Human Pedigrees Patterns of Inheritance: sex-linkage Chromosomal basis of inheritance Changes in chromosome number Gene Mapping Gene to Phenotype Modified Mendelian ratios Model organisms and mutants Genetics of Plant Development (Arabidopsis) Genetics of Animal Development (Drosophila) Behaviour Genetics/Quantitative genetics Chapter Ch. 1 & Ch. 2 Ch. 2 Ch. 2 Ch. 3 Ch. 15 Ch. 4 Ch. 6 Ch. 6 Ch. 6 (Ch. 16) Ch. 18 Ch. 18 Ch. 16 + papers Interaction between alleles • Petal colour in Fouro’clock plants: – P: – F1: Red pink X white X pink • How many loci are involved? 1 Interaction between alleles • Petal colour in Fouro’clock plants: – P: Red pink – F1: gave – ¼ red petals – ½ pink petals – ¼ white petals X white X pink • How many loci are involved? • The monohybrid ratio 3:1 is altered by incomplete dominance to 1:2:1 Incomplete dominance • Petal colour in Fouro’clock plants: let + red and c = different – – – – So F1 X F1: + c X + c ¼++ red petals ½+c pink petals ¼cc white petals • Implies a pigment synthesis pathway: e.g. 2x amount of pigment gives red, 1x amount gives pink and none give white petals. Dihybrid cross • Three of four F2 are round R_ ie 3:1 ratio for seed shape • Three of four F2 are yellow Y_ ie 3:1 ratio for seed colour • F2 progeny in a 9:3:3:1 ratio implies • Two genes independently assorting 2 • • • • Epistasis Epistasis Epistasis: two or more genes interact to influence a trait E.g. Labrador Retriever dogs Black coat colour (B), brown coat colour (b) Another gene (e) is epistatic to these alleles giving a golden coat colour – P: B/B; e/e X b/b ; E/E black Lab X brown Lab – F1: B/b ; E/e Black lab – F1 X F1: B/b ; E/e X B/b; E/e gave….. Genotype B/_;E/_ b/b;E/_ B/_;e/e b/b;e/e Epistasis Genotype Phenotype Ratio B/_;E/_ Black 9 b/b;E/_ Brown 3 B/_;e/e Golden 4 b/b;e/e Golden Phenotype Black Brown Golden Golden Ratio 9 3 4 Coat colour in mice • The two alleles of the pigment gene B or b • And the separate gene E that allows colour deposition in the coat, e/e prevents colour deposition • • • • • The A gene: A= agouti, a= nonagouti The B gene: B= black pigment, b= brown The C gene: C= pigmented, c= albino The D gene: D=nondilution, d= dilution The S gene: S= unspotted, s= pigmented spots on a white background Text p.219 Text p.204 3 Coat colour in mice • • • • • The A gene: A= agouti, a= nonagouti The B gene: B= black pigment, b= brown The C gene: C= pigmented, c= albino The D gene: D=nondilution, d= dilution The S gene: S= unspotted, s= pigmented spots on a white background Agouti • Agouti pattern: a yellow band around each of the hairs, a wildtype pattern • P: A / A X a / a • F1: 3:1 ratio agouti to non agouti: – 1 A / A (agouti) – 2 A / a (agouti) – 1 a / a (nonagouti) Text p.219 Coat colour in mice • • • • • The A gene: A= agouti, a= nonagouti The B gene: B= black pigment, b= brown The C gene: C= pigmented, c= albino The D gene: D=nondilution, d= dilution The S gene: S= unspotted, s= pigmented spots on a white background Black pigment • Black pigmented hair is the wildtype hair colour, and B is dominant to b which confers brown hair colour • P: B / B X b / b • F1: 3:1 ratio black hair colour to brown: – 1 B / B (black) – 2 B / b (black) – 1 b / b (brown) Text p.219 4 Cinnamons • Brown-agouti mice are called cinnamons • They are A/_; b/b • An agouti (black) pure-breeding mouse is crossed with a pure-breeding cinnamon Chocolate • Chocolate mice have a solid rich dark brown colour and are a/a; b/b. A/a ; B/b black agouti X a/a ; B/b chocolate ¼ A/a; B/b wildtype ¼ A/a ; b/b cinnamon ¼ a/a ; B/b black nonagouti ¼ a/a ; b/b chocolate – A/A; b/b X a/a; B/B • The F1 are? – – – – 9 A/ _ ; B/ _ 3 a/a ; B/_ 3 A/_ ; b/b 1 a/a ; b/b Coat colour in mice • • • • • The A gene: A= agouti, a= nonagouti The B gene: B= black pigment, b= brown The C gene: C= pigmented, c= albino The D gene: D=nondilution, d= dilution The S gene: S= unspotted, s= pigmented spots on a white background • A wildtype mouse (A/a; B/b) is testcrossed. • Diagram the cross: Albino • Albino mice are homozygous for the recessive member of the C/c allelic pair. • The c/c constitution is epistatic to the other coat colour genes • An albino mouse is crossed with a truebreeding wildtype (black agouti) mouse. – c/c ; ?/? ; ?/? X C/C, A/A; B/B • The F1 were intercrossed to get the following F2: – Ie. C/c ; A/?; B/? X C/c; A/?; B/? Text p.219 5 ABC of coat colour • P: c/c ; ?/? ; ?/? X C/C, A/A; B/B • F1: C/c ; A/?; B/? X C/c; A/?; B/? • F2: – 87 wildtype – 32 cinnamon – 39 albino • F1 x F1 have cinnamon offspring therefore F1 parents must be B/b • The F1 x F1 do not have chocolate offspring therefore the F1 parents must have been A/A ABC of coat colour • P: c/c ; ?/? ; ?/? X C/C, A/A; B/B • F1: C/c ; A/A; B/b X C/c; A/A; B/b • F2: – 87 wildtype – 32 cinnamon – 39 albino ABC of coat colour • P: c/c ; A/A ; b/b X C/C, A/A; B/B • F1: C/c ; A/A; B/b X C/c; A/A; B/b • F2: – 87 – 32 – 39 C/_ ; A/A; B/_ C/_ ; A/A; b/b c/c ; A/A ; ?/? wildtype cinnamon albino • From the F1 x F1 we would expect ¼ of the offspring to be albino (c/c) • 87+32+39= • 39/158 • Approx 1/4 Coat colour in mice • • • • • The A gene: A= agouti, a= nonagouti The B gene: B= black pigment, b= brown The C gene: C= pigmented, c= albino The D gene: D=nondilution, d= dilution The S gene: S= unspotted, s= pigmented spots on a white background Text p.219 6 D gene • The D gene controls the intensity of pigment specified by other coat colour genes – D/_; A/_; B/_ full expression of coat colour – Dilute agouti dd; A/_; ? – Dilute black: dd; ?; B/_ – Dilute brown dd; ?; b/b – Dilute cinnamon: • dd; a/_ ; b/b • D is a modifier gene, detectable by two grades of mutant phenotypes Coat colour in mice • • • • • The A gene: A= agouti, a= nonagouti The B gene: B= black pigment, b= brown The C gene: C= pigmented, c= albino The D gene: D=nondilution, d= dilution The S gene: S= unspotted, s= pigmented spots on a white background The S gene • The S gene controls the presence or absence of spots on the mouse • No spots: S/_ • Spotted or piebald: s/s • Write the genotype of a dilute spotted cinnamon mouse. • The mouse must be: A/_; b/b; C/_; d/d; s/s • Because to be dilute= dd, to be spotted= ss, to be cinnamon= A/_;b/b and C/_ Text p.219 7 Testing interactions The S gene? • A history student ask you what might be the genetic basis of a story in an ancient text: – Two people Jacob and Laban divided their flock of animals by coat colour Jacob took the unspotted goats and Laban the spotted goats but only Jacob’s flock increased in number giving both spotted and unspotted goats. • From the coat colour genetics in mice: S_ gives no spots and ss gives spots. So perhaps in goats Ss x Ss goats that would produce both spotted and unspotted progeny while the spotted goats (ss) would only give spotted progeny • P: c/c ; A/A ; b/b X C/C, A/A; B/B • F1: C/c ; A/A; B/b X C/c; A/A; B/b • F2: – 87 – 32 – 39 Phenotype Wildtype Cinnamon Albino total Chi-Square (Χ2 ) Test • The chi-square test is useful for comparing observed results against those predicted by a hypothesis • Various deviations from the expected occur by chance even if the hypothesis is true • How likely is it that the deviations from the expected occur by chance alone? • The chi-square test is a way of quantifying the various deviations expected by chance if the hypothesis is true. C/_ ; A/A; B/_ C/_ ; A/A; b/b c/c ; A/A ; ?/? Observed 87 32 39 158 wildtype cinnamon Albino albino x wildtype wildtype F1 x F1 = C/_ ; B/_ = C/_ ; b/b = c/c ; b/b Expected 9/16*158 3/16*158 4/16 *158 Chi-Square (Χ2 ) Test • Procedure: 1. State the simple hypothesis that gives the precise expectation: the null hypothesis 2. Calculate the chi-square, (using observed values not percentages): Χ2 = Σ (O-E)2 /E 3. Estimate the probability of obtaining a deviation from the expected at least this large by chance alone 4. Report the chi-square value, the degrees of freedom, and the probability and the rejection level. 8 Testing interactions Chi-Square (Χ2 ) Test F2: – 87 – 32 – 39 C/_ ; A/A; B/_ C/_ ; A/A; b/b c/c ; A/A ; ?/? wildtype = C/_ ; B/_ cinnamon = C/_ ; b/b Albino = c/c ; b/b • H0 : The observed coat colour ratio is not different from the 9:3:4 ratio expected for recessive epistasis. Phenotype Observed Expected (O-E)2 (O-E)2 /E Wildtype 87 88.9 3.61 0.041 Cinnamon 32 29.6 5.76 0.194 Albino 39 39.5 0.25 0.063 total 158 0.298 4. Report the chi-square value, the degrees of freedom, and the probability and the rejection level. • Χ2 =0.30 • Degrees of freedom (df) = 3-1 = 2 • Probability of observing a deviation from the expected results at least this large on the basis of chance alone, in our example 0.9>p>0.5 • Rejection level is p=0.05 • Conclusion: At the 5% rejection level we fail to reject the null hypothesis that the observed progeny is 9 C/_;B_ : 3 C/_; b/b :4 c/c;b/b, the recessive epistasis ratio of wildtype to cinnamon to albino mice. We conclude that observed data support the hypothesis of recessive epistasis. 3. Estimate the probability of obtaining a deviation from the expected at least this large by chance alone • • • Χ2 = 0.30 Degrees of freedom (df) = 3-1=2 0.9>p>0.5 Testing interactions F2: – 32 – 87 – 39 cinnamon = C&/C& wildtype = C&/C$ Albino = C$/C$ 1 2 1 • H0 : The observed coat colour ratio is not different from the 1:2:1 ratio expected for incomplete dominance. Phenotype O Cinnamon Wildtype Albino total Expected (O-E)2 32 ⅓ * 158 87 ⅔ * 158 39 ⅓ * 158 158 (32-52.6)2 (87-105.3)2 (39-52.6)2 (O-E)2 /E 8.07 3.18 3.52 14.77 9 4. Report the chi-square value, the degrees of freedom, and the probability and the rejection level. Chi-Square (Χ2 ) Test 3. Estimate the probability of obtaining a deviation from the expected at least this large by chance alone • • • Χ2 = 14.77 Degrees of freedom (df) = 3-1=2 0.005>p i.e. p< 0.05 • Χ2 =14.77 • Degrees of freedom (df) = 3-1 = 2 • Probability of observing a deviation from the expected results at least this large on the basis of chance alone, in our example p<< 0.05 • Rejection level is p=0.05 • Conclusion: At the 5% rejection level we reject the null hypothesis that the observed progeny is 1 C&/C& : 2 C&/C$ : 1 C$/C$, the incomplete dominance ratio of cinnamon to wildtype to albino mice. We conclude that observed data does NOT support the hypothesis of incomplete dominance. Testing interactions • Pure strains of pineapples were crossed and the following types of leaves were noted: – S= spiny, ST= spiny tip P= non-spiny; piping Cross 1 2 3 Parental ST x S P x ST PxS Phenotypes F1 F2 ST 99 ST: 34 S P 120 P: 39 ST P 95 P: 25 ST: 8S Testing interactions Cross 1 2 3 • • • • Parental ST x S P x ST PxS Phenotypes F1 F2 ST 99 ST: 34 S P 120 P: 39 ST P 95 P: 25 ST: 8 S Cross 1 indicates a single gene, note 3:1 ratio Cross 2 indicates a single gene, note 3:1 ratio Cross 3 indicates the two genes are interacting What is the modified ratio? 10 Testing interactions Cross 1 2 3 Parental ST x S P x ST PxS Testing interactions Phenotypes F1 F2 ST 99 ST: 34 S P 120 P: 39 ST P 95 P: 25 ST: 8 S • Using the assigned gene symbols P for piping and T for tip • Cross 3: ? • Hypothesis: independent assortment – – – – • Assign gene symbols: – P for piping – T for tip • So for cross 1: ST x S could be written – P: T/T x t/t – F1: T/t – F2: 99 T/_ : 34 t/t – – – – 9: P/_; T/_ 3 P/_ ; t/t 3 p/p ; T/_ 1 p/p ; t/t – 95 P/_ ; T/_ – 25 p/p; T/_ – 8 p/p; t/t =P =ST =S • So it seems that only whenever P is dominant we can see” the piping phenotype Text p.213 Dominant epistasis Testing interactions • Using the assigned gene symbols P for piping and T for tip • Cross 3: ? • Hypothesis: independent assortment 9: P/_; T/_ 3 P/_ ; t/t 3 p/p ; T/_ 1 p/p ; t/t • The dihybrid ratio is modified • The dihybrid ratio is modified – 95 P/_ ; T/_ – 25 p/p; T/_ – 8 p/p; t/t =P =ST =S • So it seems that only whenever P is dominant we can “see” the piping phenotype • Dominant epistasis: – – – – 9: P/_; T/_ 3 P/_ ; t/t 3 p/p ; T/_ 1 p/p ; t/t Phenotype O P ST S 12 3 1 Expected (O-E)2 95 12/16 * 128 (95-96)2 25 3/16 * 128 (25-24)2 8 1/16 * 128 (8-8)2 total 128 (O-E)2 /E 0.01 0.042 0 0.052 11 Chi-Square (Χ2 ) Test 3. Estimate the probability of obtaining a deviation from the expected at least this large by chance alone • • • Χ2 = 0.052 Degrees of freedom (df) = 3-1=2 0.975>p< 0.9 4. Report the chi-square value, the degrees of freedom, and the probability and the rejection level. • Χ2 =0.052 • Degrees of freedom (df) = 3-1 = 2 • Probability of observing a deviation from the expected results at least this large on the basis of chance alone, in our example 0.975>p< 0.9 i.e. p >> 0.05 • Rejection level is p=0.05 • Conclusion: – At the 5% rejection level we fail to reject the null hypothesis that the observed progeny is 12 P: 3 ST: 1S, the dominant epistasis ratio of piping to spiny tip to spiny pineapple leaf types. We conclude that observed data supports the hypothesis of dominant epistasis. Testing interactions • Hypothesis: independent assortment – – – – 9: P/_; T/_ 3 P/_ ; t/t 3 p/p ; T/_ 1 p/p ; t/t • The dihybrid ratio is modified – 95 P/_ ; T/_ – 25 p/p; T/_ – 8 p/p; t/t =12 P =3 ST =1 S • So it seems that only whenever P is dominant we can “see” the piping phenotype • The data support a 12:3:1 ratio typical of dominant epistasis – Piping P/_ (T or t irrelevant) – Spiny tip p/p ; T/_ – Spiny p/p ; t/t Two genes, duplicate action • A graduate student crosses wheat plants that are homozygous for spring growth with plants that are homozygous for winter growth • The F1 all have the spring growth phenotype • How would you write the cross? 12 Two genes, duplicate action • The graduate student then crossed two F1 plants and got the following progeny ratio: – 15 : 1 – Spring growth : Winter growth • How would she explain what happened? – Let S and H represent the dominant alleles Cross: ¼ SH ¼ Sh ¼ sH ¼ sh ¼ SH SS; HH SS; Hh sS; HH sS; Hh S/s; H/h x S/s; H/h ¼ Sh ¼ sH SS; Hh sS; HH SS; hh sS; Hh sS; Hh Ss; Hh sS; hh ss; Hh ¼ sh Ss ; Hh Ss; hh Ss; Hh ss; hh Modified ratios from the cross AaBb x AaBb A_ ; B _ A_ ; bb aa ; B_ aa ; bb Independent assortment 9 3 3 1 Complementary gene action 9 7 Duplicate gene action Recessive epistasis Dominant epistasis 15 9 1 3 12 4 3 1 Duplicate gene action • Spring growth is the result of duplicate gene action between two independent genes S and H • Spring growth: S_; H_ , S_; hh , ss; H_ • Winter growth: ss;hh • In duplicate gene action the dominant phenotype is evident when at least one dominant allele is present at each locus, the recessive phenotype is only visible when both genes are homozygous recessive Suppression ratio • Suppressors cancel the expression of the mutant allele of another gene, resulting in normal wildtype phenotype • Ratio typically of a suppressor is 13:3 • Consider the recessive allele that confers purple eye colour (pd) in Drosophila • Suppressed by a recessive allele su. – pd /pd ; su+ / su+ X pd+/ pd+ ; su / su • Gives all red eyed (wildtype) F1: – pd+ /pd; su+ / su 13 recessive suppressor • F1 X F1 red eyed (wildtype) : • F1 X F1 red eyed (wildtype) : – pd+ /pd; su+ / su X pd+ /pd; su+ / su • • • • 9 3 1 3 pd+ /_; su+ / _ pd+ /_; su / su pd /pd; su / su pd /pd; su +/ _ Suppressors 13 3 – i.e. 9+3+1= 13 red eyed to 1 purple eyed – pd+ /pd; su+ / su X pd+ /pd; su+ / su • • • • 9 3 1 3 pd+ /_; su+ / _ pd /pd; su +/ _ pd /pd; su / su pd+ /_; su / su The purple eye colour can only be expressed when the suppressor is inactive (su+/_) AND the recessive mutation purple (pd) homozygous Text p.206 Introduction to genetic analysis Griffiths, A., Wessler, S.R., Lewontin,R.C., Gelbart, W.M.,Suzuki, D.T. and Miller, J.H. Eighth Edition, W.H. Freeman and Company NY • Part I Transmission genetic analysis – – – – Chapter 1: all questions p. 24-26 Chapter 2: all the questions p. 62-72 Chapter 3: questions #1-12,18,19, 22, 25-27, 29, 30, 32, 40-42. Chapter 4: sections 4.1- 4.4 and 4.6, questions # 1-4, 6-13,1522,24-43. • Part II The relationship of DNA and phenotype – Chapter 6: questions# 1,3-5,7,9,11,14,18,19,21,23,24, 26,29,32,33,35,39,41,44,45,47,49,51,53,55,63-66. • Part IV The nature of heritable change – Chapter 15: sections 15.1 and 15.3; questions #1-3,1113,19,21,22, 32, 38, 52. Test and Tutorials • You need to do the assigned questions at the end of chapter 6 • As stated on the website you need to present a doctor’s note to the instructor within 48h of missing a test (or a quiz) • If you have a legitimate reason for missing a quiz email IN ADVANCE • NB. Academic conflict is NOT a legitimate reason for missing a quiz • Late penalty is 20% per day for quizzes that are to be handed in to drop boxes. . 14