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Probability Probability = the # of times an event happens ⁄ the # of opportunities for an event to happen Pedigree analysis Probability: Product rule • The probability of two independent events occurring simultaneously is the product of each of their respective probabilities – Ie. RrYy x RrYy Jan 11 • What is the probability of getting rryy genotype ? • ¼ * ¼ = 1/16 • The probability of rolling a 3 on a die in one trial is: – P (of a 3) = 1/6 – P (of two 3’s) = (1/6) x (1/6) = 1/36 • The second example show the product rule: – The probability of two independent events occurring simultaneously is the product of each of their respective probabilities Text p.39 Probability: Product rule • The probability of two independent events occurring simultaneously is the product of each of their respective probabilities – Ie. RrYy x RrYy • • • • Probability: sum rule • The probability of either of two mutually exclusive events occurring is the sum of their individual probabilities – Probability of a 3 or a 5 when rolling a die is? • The probability of either of two mutually exclusive events occurring is the sum of their individual probabilities – Probability of a 3 or a 5 when rolling a die is What is the probability of getting rryy genotype ? ¼ * ¼ = 1/16 What is the probability of getting RRYY genotype? ¼ * ¼ = 1/16 • P (of a 3)= 1/6 and p (of a 5) = 1/6 so • P (of a 3 or a 5) = 1/6 + 1/6 =2/6= ⅓ . Text p.40 Text p.39 Probability: sum rule • The probability of either of two mutually exclusive events occurring is the sum of their individual probabilities – Probability of a 3 or a 5 when rolling a die is • P (of a 3)= 1/6 and p (of a 5) is 1/6 so • P (of a 3 or a 5) = 1/6 + 1/6 =2/6= ⅓ . – In the cross RrYy x RrYy what is the probability of getting either the rryy or the RRYY genotype? Text p.40 Probability: sum rule Text p.40 Probability: sum rule • The probability of either of two mutually exclusive events occurring is the sum of their individual probabilities – Probability of a 3 or a 5 when rolling a die is • P (of a 3)= 1/6 and p (of a 5) is 1/6 so • P (of a 3 or a 5) = 1/6 + 1/6 =2/6= ⅓ . – In the cross RrYy x RrYy what is the probability of getting either the rryy or the RRYY genotype? • 1/16 + 1/16 = 2/16 = ⅛ Probability and pedigrees eg: albinism • IV-3 came to the genetic consult visit he is planning to have a child with a normally pigmented woman who also has an albino brother. What is the probability of their first son being albino? Text p.40 1 Probability and pedigrees eg: albinism • IV-3 came to the genetic consult visit he is planning to have a child with a normally pigmented woman who also has an albino brother. What is the probability of their first son being albino? • P (both parents being heterozygous) = P (Aa) * P (Aa) = (2/3)*(2/3) • And P (having an albino child) = P (aa)= ¼ • And P (having a son)= ½ • Thus P (of first son being aa)= (2/3)*(2/3)*(1/4)*(1/2) =4/72=1/8 Probability and pedigrees eg: albinism • IV-3 came to the genetic consult visit he is planning to have a child with a normally pigmented woman who also has an albino brother. What is the probability of their first son being albino? • P (both parents being heterozygous) = P (Aa) * P (Aa) = (2/3)*(2/3) • And P (having an albino child) = P (aa)= ¼ • And P (having a son)= ½ • Thus P (of first son being aa)= (2/3)*(2/3)*(1/4)*(1/2) =4/72=1/8 Albinism in mice • In a test cross what is the probability of getting a male albino mouse? • C/c X c/c • P (of being male)= ½ * p (of being cc)= ½ =¼ Probability and pedigrees eg: albinism • IV-3 came to the genetic consult visit he is planning to have a child with a normally pigmented woman who also has an albino brother. What is the probability of their first son being albino? • P (both parents being heterozygous) = P (Aa) * P (Aa) = (2/3)*(2/3) • And P (having an albino child) = P (aa)= ¼ • And P (having a son)= ½ • Thus P (of first son being aa)= (2/3)*(2/3)*(1/4)*(1/2) =4/72=1/8 Albinism in mice • The allele c causes albinism in mice (C causes mice to be black). In a test cross what is the probability of getting a male albino mouse? Do you recognize the pattern? • The seed colour Yellow is dominant to green • The F1 heterozygotes crossed to plants homozygous recessive for seed colour • From a monohybrid test cross a 1:1 progeny ratio is expected Probability and pedigrees eg: albinism • IV-3 came to the genetic consult visit he is planning to have a child with a normally pigmented woman who also has an albino brother. What is the probability of their first son being albino? • P (both parents being heterozygous) = P (Aa) * P (Aa) = (2/3)*(2/3) • And P (having an albino child) = P (aa)= ¼ • And P (having a son)= ½ • Thus P (of first son being aa)= (2/3)*(2/3)*(1/4)*(1/2) =4/72=1/8 Albinism in mice • In a test cross what is the probability of getting a male albino mouse? • C/c X c/c Albino mice • The allele c causes albinism in mice (C causes mice to be black). A student tested the hypothesis that a black mouse is a heterozygote. 2 Albino mice Chi-Square (Χ2 ) Test Chi-Square (Χ2 ) Test • The allele c causes albinism in mice (C causes mice to be black). A student tested the hypothesis that a black mouse is a heterozygote. • C/c X c/c • What is the expected ratio of black to albino progeny • The allele c causes albinism in mice (C causes mice to be black). A student tested the hypothesis that a black mouse is a heterozygote. • The allele c causes albinism in mice (C causes mice to be black). A student tested the hypothesis that a black mouse is a heterozygote. – H0: based on Mendel’s first law of equal segregation the observed progeny is 50% C/c and 50% c/c. – H0: based on Mendel’s first law of equal segregation the observed progeny is 50% C/c and 50% c/c. Class O E Class O E Albino 55 60 Albino 55 60 Black 65 60 Black 65 60 120 Chi-Square (Χ2 ) Test Chi-Square (Χ2 ) Test • The chi-square test is useful for comparing observed results against those predicted by a hypothesis • 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 Chi-Square (Χ2 ) Test 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. • 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. 120 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. 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. 3 Chi-Square (Χ2 ) Test • Procedure: Chi-Square (Χ2 ) Test • 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. 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. Chi-Square (Χ2 ) Test eg. 1 • The allele c causes albinism in mice (C causes mice to be black). A student tested the hypothesis that a black mouse is a heterozygote. • Procedure: 1.State the simple hypothesis that gives the precise expectation: the null hypothesis 1.H0: based on Mendel’s first law of equal segregation the observed progeny is 50% C/c and 50% c/c. – in our example df = 2-1=1 2. Calculate the chi-square, (using observed values not percentages) O E (O-E)2 (O-E)2/E Class O E (O-E)2 (O-E)2/E 55 60 25 25/60 = 0.42 Albino 55 60 25 25/60 = 0.42 Black 65 60 25 25/60 = 0.42 Black 65 60 25 Total= Χ2 = 0.84 • in our example Χ2 =0.84 Degrees of freedom (df) = number of independent variables minus 1 (in this case phenotypic classes -1) Chi-Square (Χ2 ) Test Albino Chi-Square (Χ2 ) Test • 1.H0: based on Mendel’s first law of equal segregation the observed progeny is 50% C/c and 50% c/c. Class 3. Estimate the probability of obtaining a deviation from the expected at least this large by chance alone Χ2 = Σ (O-E)2 /E 1.State the simple hypothesis that gives the precise expectation: the null hypothesis 2. Calculate the chi-square, (using observed values not percentages) Chi-Square (Χ2 ) Test – • The allele c causes albinism in mice (C causes mice to be black). A student tested the hypothesis that a black mouse is a heterozygote. • Procedure: Chi-Square (Χ2 ) Test 3. Estimate the probability of obtaining a deviation from the expected at least this large by chance alone • Chi-Square (Χ2 ) Test eg. 1 Χ2 = Σ (O-E)2 /E – • in our example Χ2 =0.84 Degrees of freedom (df) = number of independent variables minus 1 (in this case phenotypic classes -1) – in our example df = 2-1=1 25/60 = 0.42 Total= Χ2 = 0.84 Chi-Square (Χ2 ) Test 3. Estimate the probability of obtaining a deviation from the expected at least this large by chance alone • • • Χ2 = 0.84 Degrees of freedom (df) = 2-1=1 0.5>p>0.1 4 4. Report the chi-square value, the degrees of freedom, and the probability and the rejection level. 4. Report the chi-square value, the degrees of freedom, and the probability and the rejection level. 4. Report the chi-square value, the degrees of freedom, and the probability and the rejection level. • Χ2 =0.84 • Degrees of freedom (df) = 2-1 = 1 • Probability of observing a deviation from the expected results at least this large on the basis of chance alone, in our example 0.5>p>0.1 • Rejection level is p=0.05 • Conclusion: At the 5% rejection level we fail to reject the null hypothesis that the observed progeny is 50% C/c and 50% c/c, a 1:1 ratio of pigmented to albino mice. We conclude that the black mouse used in the test cross was a heterozygote. • Χ2 =0.84 • Degrees of freedom (df) = 2-1 = 1 • Probability of observing a deviation from the expected results at least this large on the basis of chance alone, in our example 0.5>p>0.1 • Rejection level is p=0.05 • Conclusion: At the 5% rejection level we fail to reject the null hypothesis that the observed progeny is 50% C/c and 50% c/c, a 1:1 ratio of pigmented to albino mice. We conclude that the black mouse used in the test cross was a heterozygote. • Χ2 =0.84 • Degrees of freedom (df) = 2-1 = 1 • Probability of observing a deviation from the expected results at least this large on the basis of chance alone, in our example 0.5>p>0.1 • Rejection level is p=0.05 • Conclusion: At the 5% rejection level we fail to reject the null hypothesis that the observed progeny is 50% C/c and 50% c/c, a 1:1 ratio of pigmented to albino mice. We conclude that the black mouse used in the test cross was a heterozygote. The dihybrid situation The dihybrid situation The dihybrid situation • In dogs dark coat colour (C) is dominant to albino and short hair (S) is dominant over long hair. The ratio of progeny from a backcross to the homozygous recessive is – – – – 140 CS 135 cs 110 Cs 115 cS • In dogs dark coat colour (C) is dominant to albino and short hair (S) is dominant over long hair. ….a backcross to the homozygous recessive … • C/c;S/s X c/c;s/s – – – – • Is this ratio different from the expected 1:1:1:1 ratio? Chi-Square (Χ2 ) Test Χ2 (O-E)2 • = Σ /E • Degrees of freedom (df) = number of independent variables minus 1 (in this case phenotypic classes -1) • Probability of observing a deviation from the expected results at least this large on the basis of chance alone. • In dogs dark coat colour (C) is dominant to albino and short hair (S) is dominant over long hair. ….a backcross to the homozygous recessive … • C/c;S/s X c/c;s/s • The ratio of progeny from a backcross to the homozygous recessive is 140 CS 135 cs 110 Cs 115 cS • Parental vs recombinant Chi-Square (Χ2 ) Test eg. 2 1. H0: based on Mendel’s first law of equal segregation the observed progeny fits a 1:1:1:1 ratio of phenotypes CS:cs:Cs:cS (coat colour C= dark and hair length S= short). 2. Calculate chi-square Class CS cs CS cS df= 4-1 O 135 135 110 115 500 E 125 125 125 125 (O-E)2 (O-E)2/E 225 1.8 100 0.8 225 1.8 100 0.8 Chi-square = 5.2 Chi-Square (Χ2 ) Test eg. 2 1. H0: based on Mendel’s first law of equal segregation the observed progeny fits a 1:1:1:1 ratio of phenotypes CS:cs:Cs:cS (coat colour C= dark and hair length S= short). 2. Calculate chi-square Class CS cs CS cS df= 4-1 O 135 135 110 115 500 E 125 125 125 125 (O-E)2 (O-E)2/E 225 1.8 100 0.8 225 1.8 100 0.8 Chi-square = 5.2 5 Chi-Square (Χ2 ) Test eg. 2 Chi-Square (Χ2 ) Test eg. 2 Other 1:1 ratios? • A mouse cross: • A/a;B/b x a/a;b/b gave: – – – – • Χ2 =5.2 , df = 3, p>10%, rejection level =5% • Conclusion: given the chi-square =5.2 and df=3 a deviation from the expected 1:1:1:1 ratio at least this large would occur by chance alone 10% of the time so we fail to reject the null hypothesis that the observed ratio of CS:cs:Cs:cS is 1:1:1:1. • Χ2 =5.2 , df = 3, p>10%, rejection level =5% • Conclusion: given the chi-square =5.2 and df=3 a deviation from the expected 1:1:1:1 ratio at least this large would occur by chance alone 10% of the time so we fail to reject the null hypothesis that the observed ratio of CS:cs:Cs:cS is 1:1:1:1. Other 1:1 ratios? • A mouse cross: • A/a;B/b x a/a;b/b gave: – – – – 25% A/a;B/b 25% A/a;b/b 25% a/a;B/b 25% a/a;b/b • What is the probability of getting a female mouse that is a/a;b/b? • Another mouse cross: • C/c;D/d X c/c;d/d gave: – – – – 38% C/c;D/d 12% C/c;d/d 38% c/c;D/d 12% c/c;d/d • The student could not find a female cc;d/d mouse. In fact all the _;d/d mice were male! • Why? Other 1:1 ratios? • Another mouse cross: • C/c;D/d X c/c;d/d gave: – – – – 38% C/c; D/d 12% C/c; d/d 38% c/c; D/d 12% c/c; d/d • C/c;D/d X c/c;d/d gave: – – – – 38% C/c; D/d 12% C/c; d/Y 38% c/c; D/d 12% c/c; d/Y • The student could not find a female cc;d/d mouse. In fact all the _;d/d mice were male! Other 1:1 ratios? Other 1:1 ratios? • For autosomal traits males and females are affected in equal proportions • From the mouse cross that gave no _;d/d female progeny we would test whether the d is X-linked • The c allele is likely autosomal • In the P generation the homozygous recessive mice were female • For autosomal traits males and females are affected in equal proportions • From the mouse cross that gave no _;d/d female progeny we would test whether the d is X-linked • The c allele is likely autosomal • In the P generation the homozygous recessive mice were female • Another mouse cross: • C/c;D/d X c/c;d/d gave: 25% A/a;B/b 25% A/a;b/b 25% a/a;B/b 25% a/a;b/b • What is the probability of getting a female mouse that is a/a;b/b? – – – – 38% C/c;D/d 12% C/c;d/d 38% c/c;D/d 12% c/c;d/d • The student could not find a female cc;d/d mouse. In fact all the _;d/d mice were male! • Why? Other 1:1 ratios? • For autosomal traits males and females are affected in equal proportions • From the mouse cross that gave no _;d/d female progeny we would test whether the d is X-linked • The c allele is likely autosomal • In the P generation the homozygous recessive mice were female Other 1:1 ratios • What is the probability of 6 children being girls? • P (girl)= ½ – P (all 6 being girls)= (½)6 = 1/64. • Probability that of 6 children all will be the same sex? – P (all 6 being girls) + P (all 6 being boys) = (½ )6 + (½)6 = 1/32 The distribution of boys and girls in 240 families (O-E)2/E Sex O E (O-E)2 Girls 129 120 81 0.67 Boys 111 120 81 0.67 6 Other 1:1 ratios • What is the probability of 6 children being girls? • P (girl)= ½ – P (all 6 being girls)= (½)6 = 1/64. • Probability that of 6 children all will be the same sex? – P (all 6 being girls) + P (all 6 being boys) = (½ )6 + (½)6 = 1/32 Other 1:1 ratios • • • H0: based on Mendel’s first law of equal segregation the observed progeny fits a 1:1 ratio of girls : boys. Chi-square = 1.34, df= 1, p>0.1, rejection level is p=0.05 Therefore we conclude that for Χ2 =1.34, df=1 we would expect a deviation from the 1:1 ratio at least this large would occur by chance alone more than 1% of the time so we fail to reject the null hypothesis that the observed ratio of progeny fits a 1:1 ratio of girls to boys. Other 1:1 ratios • • • H0: based on Mendel’s first law of equal segregation the observed progeny fits a 1:1 ratio of girls : boys. Chi-square = 1.34, df= 1, p>0.1, rejection level is p=0.05 Therefore we conclude that for Χ2 =1.34, df=1 we would expect a deviation from the 1:1 ratio at least this large would occur by chance alone more than 1% of the time so we fail to reject the null hypothesis that the observed ratio of progeny fits a 1:1 ratio of girls to boys. The distribution of boys and girls in 240 families Sex O E (O-E)2 (O-E)2/E Girls 129 120 81 0.67 Boys 111 120 81 0.67 Other 1:1 ratios • • • Pedigree of an autosomal dominant condition H0: based on Mendel’s first law of equal segregation the observed progeny fits a 1:1 ratio of girls : boys. Chi-square = 1.34, df= 1, p>0.1, rejection level is p=0.05 Therefore we conclude that for Χ2 =1.34, df=1 we would expect a deviation from the 1:1 ratio at least this large would occur by chance alone more than 1% of the time so we fail to reject the null hypothesis that the observed ratio of progeny fits a 1:1 ratio of girls to boys. Polydactyly a rare dominant phenotype • The unaffected couple in generation IV have12 offspring. What is the probability of all of 12 being boys? • His brother is affected, if before having children the probability of having an affected son was P(son)*P(affected) = ½ * ½ = ¼ Is it strange that they have 2 affected sons? Polydactyly a rare dominant phenotype • The unaffected couple in generation IV have12 offspring. What is the probability of all of 12 being boys? • His brother is affected, if before having children the probability of having two affected sons was P(son)*P(affected) = ½ * ½ = ¼ Is it strange that they have 2 affected sons? Pedigree: X-linked recessive • For I_1and I-2 the probability of grandson inheriting red-green colour blindness is: Pedigree: X-linked recessive • For I_1and I-2 the probability of grandson inheriting red-green colour blindness is: – Their daughter is a heterozygote so • P(grandson)*P(grandson is afflicted)= (½ ) * (½)= ¼ 7 Sex linkage Sex linkage • Recessive genes showing X-linked inheritance can be detected in pedigree charts by • Recessive genes showing X-linked inheritance can be detected in pedigree charts by – More males than females are affected – None of the offspring of an affected male show the phenotype – More males than females are affected – None of the offspring of an affected male show the phenotype • but his daughters are carriers and half of their sons show the recessive phenotype • And none of his sons show the recessive phenotype nor will they pass it on to their offspring Pedigree: X-linked dominant • Hypophosphatemia • but his daughters are carriers and half of their sons show the recessive phenotype • And none of his sons show the recessive phenotype nor will they pass it on to their offspring Pedigree: X-linked recessive • Haemophilia a clotting disorder most commonly attributed to the absence or malfunction of factor VIII . Sex linkage • Dominant genes showing X-linked inheritance can be detected as follows: – Affected males pass on the gene to all of their daughters and to none of their sons – Affected females pass on the gene to half of their offspring Course Overview 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. 16) Ch. 6 Ch. 6 Ch. 6 (Ch. 16) Ch. 18 Ch. 18 Ch. 16 + papers 8