Genetics Extensions of Mendelian Principles: Genetic Interactions ...
September 23rd, 2013
Biology 2C03: Genetics Extensions of Mendelian Principles: Genetic Interactions
Module 3 Outline: Cats are not Peas - At the end of this module, you will be able to: solve probles dealing with multiple alleles 1. Define, recognize, describe and apply Mendel’s second principle 2. Solve genetic problems relating to dihybrid and trihybrid crosses 3. Use the rules of probability to determine genotypic and phenotypic proportions 4. Know how and when to use a testcross 5. Recognize modified dihybrid Mendelian ratios 6. Test your theory that a Mendelian ratio is consistent with your data using a Chi-squared test Multiple Alleles - Mutations anywhere along the gene can still give a loss of function, but are considered different alleles - Population: multiple alleles may exist - Individual (diploid): only two alleles coexist in each cell
Example of the Multiple Alleles Contributing to Fur Coloration in Cats - Five alleles: C: full colour cb-: Burmese (mutation at a particular locus) cs: Siamese (mutation at a different allele) c: white, blue eyes ca: albino, pink eyes - C>cb=cs>c>ca (dominant/allelic series) If each individual has two alleles, the dominant/allelic series tells you which one will be dominant - Where: > indicates
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= indicates incomplete dominance or codominance cb/cs = mink Ex, for a cat to be Siamese they must have the cs allele with either a c or ca allele This represents a dominance series or allelic series. Important is the relationship between pairs of alleles
The C Gene Codes for an Enzyme Called Tyrosinase - The albino allele (the most recessive allele in this series) contains a cytosine deletion in Tyrosinase at position 975 in exon 2 - Causes a frame shift resulting in a premature stop codon residues downstream from the mutation – loss of a single nucleotide that alters the reading frame - This is a non-functional protein and the allele is described as a loss of function allele - Definition of frame-shift mutation: the loss of a single nucleotide causes the reading frame, the series of three neucleotide codons, to be read differently The Number of Different Genotypes is Determined by the Number of Alleles in the Population - Number of genotypes is determined by the number of different possible allele combinations in the population - c, ca. cs, cb, C (wildtype) - This gives, 15 genotypes of which there are five different homozygous genotypes and 10 different heterozygotes
Biochemical Definition of Wildtype - Wildtype allele: a functional enzyme or other protein is produced in the cell - Loss of function allele: an enzyme or other protein is no longer being produced, is produced at lower levels, or is nonfuncitonal, or at least subfunctional - Sometimes the term wildtype is used to refer to the most common phenotype (or genotype) found in a natural population - Often the wildtype allele is dominant over the loss of function allele - Half as much protein is synthesized yet this is often sufficient to achieve the wildtype phenotype (heterozygotes) - This is called haplosufficiency
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For fur colour, in the C/c genotype, half as much Tyrosinase is synthesized, but this is sufficient to achieve full colouration
Because there are Always Exceptions to the Rule - It is not always the case that the dominant allele is ‘normal’ and the recessive allele is a mutation - Dominant alleles can be gain of function mutations, in which the mutant allele produces a protein that has increased (detrimental) function - E.g. Huntington’s disease is an autosomal dominant disease which manifests the phenotype later in life; 50% chance of passing on the allele to your offspring - Dominant allele can be a loss of function alleles, but in the heterozygote, half as much protein is synthesized and this is not sufficient for a normal phenotype; haploinsufficient – inheriting both copies of the mutation is a lethal genotype - E.g. tailless cats (Manx)
Gregor Mendel Defined the Principle of Segragation - Mendel’s principle of segregation: 1) Each individual organism possesses two alleles encoding a trait 2) Alleles separate when gametes are formed 3) Alleles separate in equal proportions - He did this using monohybrid crosses, e.g. Aa x Aa - What would happen if he followed two traits at the same time? Mendel’s Law of Independent Assortment - The inheritance pattern of one trait will not affect the inheritance pattern of another trait - Examined dihybrid crosses e.g. Aa Bb x Aa Bb Taking true breeding round yellow seeds with true breeding wrinkled green seeds producing heterozygotes Four possible phenotypes
9:3:3:1 ratio in the F2 phenotype - Each trait is inherited by the Mendelian ratio, 3:1, but together we see a phenotypic ratio of 9:3:3:1
Another Example: Cat Colour and Tail Length - 9 brown fur, short tail - 3 brown fur, long tail - 3 white fur, short tail - 1 white fur, long tail - 9:3:3:1 - 9 = two dominant traits - 1 = two recessive traits
More Complex Problems: Tribybrid Cross - You can use a similar procedure for determining the proportion of genotypes and phenotypes if you are following more than two alleles: E.g. trihybrid crosses (and more…) or why I learned to stop using Punnett squares Tryhibrid: 64 combinations of gametes (an 8x8 Punnett square) 27 different F2 genotypes 8 different F2 phenotypes ratio: 27:9:9:9:3:3:3:1
Take Home Problem - In cats, white patches are caused by the dominant allele P, while p/p individuals are solid coloured - Short hair is caused by a dominant allele, S, while s/s cats have long hair - A long-haired cat with patches whose mother was solid-coloured and shorthaired mates with a short-haired, solid coloured cat whose mother was longhaired and solid-coloured - What kinds of kittens can arise from this mating and in what proportions
Extensions of Mendelian Principles or Cats are Not Peas - How do multiple genes interact to determine a single phenotype? What Happens When Multiple Genes often Contribute to a Single Characteristic - What happens when many genes contribute to a single phenotype: interaction between genes - F1 cross: A/a B/b x A/a B/b, this is still a dihybrid cross so we still have to consider Mendelian principles - If there is complete dominance at two distinct traits: phenotypes = 9:3:3:1 - Four potential outcomes - Gene interactions may produce novel phenotypes that will modify the 9:3:3:1 F2 phenotypic ratio - But will still be based upon it – total 16
Complete Dominance - D/D = this cat does not carry the dilute gene - D/d = this cat is a carrier of the dilute gene - d/d = this cat carries two copies of the dilute gene. The coat colour is diluted. - Black dilutes to blue, chocolate dilutes to lilac, cinnamon dilutes to fawn, and red dilutes to cream Molecular Explanation for Complete Dominance of Cat Colouration - The dilute gene (or melanophilin) affects the distribution of melanin granules within the cats’ hair - Dominant or dense allele (D): produces dense pigmentation - Recessive dilute (d) allele: results in pigment clumping, forming areas in the hair with no granules. This causes the cat coat to dilute or lighten - The dilute effect is autosomal recessive, hence a cat requires two copies of the d allele for the coat to dilute Complementation - White x white = white - White x white = white - White offspring mate = black - F2 phenotypic ratio = 9:7 Complementation – Two Different Albinos
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White (c/c) x white (c/c) = white (c/c) White (a/a) x white (a/a) = white (a/a) White (c/c) x white (a/a) = black Mutations in two different genes produce the same mutant phenotype F2 phenotypic ratio = 9:7 P: white (c/c) x white (c/c) and white (a/a) x white (a/a) F1: white (c/c, A/A) x white (C/C, a/a) F2 phenotypic ratio= 9:7 black (C/c, A/a)
We Can Observe Complementation in Human Pedigrees - About 50 genes have recessive mutant alleles that can cause deafness in humans - A heterogenous trait: a mutation in any one of a number of genes can give rise to the same phenotype - Complementation: 9:7 - Non complementation:
Types of Natural Selection
September 24th, 2013
What Happens when Multiple Genes Contribute to a Single Characteristic? - Examples from last class Complete dominance Complementation - P: A/A b/b x a/a B/B; F1 A/a B/b - What gametes are produced? - Complete dominance (9:3:3:1)
Complementation - Complementation occurs when two strains of an organism with different homozygous recessive mutations that produce the same phenotype, produce offspring of the wild-type phenotype when mated or crossed - Complementation will only occur if the mutations are in different genes - The other genome supplies the wild-type allele to “complement” the mutated allele - Complementation will not occur if the mutations are in the same gene What Happens When Multiple Genes Contribute to a Single Characteristic? - What gametes are produced? - 9:7
Take Home Problem - In cats, white patches are caused by the dominant allele P, while p/p individuals are solid coloured - Short hair is caused by a dominant allele, S, while s/s cats have long hair - A long-haired cat with patches whose mother was solid-coloured and shorthaired mates with a short-haired, solid coloured cat whose mother was longhaired and solid-coloured - What kinds of kittens can arise from this mating and in what proportion - Ppss x ppSs Answer to Take Home Problem - ¼ PpSs: patches, short hair - ¼ ppSs: solid, short hair - ¼ Ppss: patches, long hair - ¼ ppss: solid, long hair - Does this agree with your answer Genetic Epistasis - 9:7 – observing complementation, assume that two alleles are acting on the genotype - Epistasis: the masking of the expression of one gene by another. No new phenotypes are produced - The epistatic gene does the masking - The hypostatic gene is masked - F2 phenotypic ratio = 9:3:4 – recessive epistasis Homozygous recessives at one gene pair mask expression from the other gene A/-b/b and a/ab/b have the same phenotype bb gives recessive phenotype - F2 phenotypic ratio = 12:3:1 dominant epistasis One dominant allele at one gene masks expression from the other gene B locus does not matter, as long as there is a dominant A allele it will be the same phenotype
A/-B/- and A/-b/b have the same phenotype
Recessive Epistasis Example - F2 phenotypic ratio = 9:3:4 recessive epistasis - If the genotype is c/c, the mouse is white, regardless of the genotype at the other locus
Molecular Mechanism of Recessive Epistasis - c/c: no pigment is synthesized; therefore mice are white regardless of genotype at the A locus; this is recessive epistasis - The genotype at the A locus determines how the pigment is deposited: either black (a/a) or agouti (A/-) - Figure 5.9 in text has additional example
Duplicate Recessive Epistasis - Look back at the model of the two genes that produced white cats where F2 phenotypic ratio = 9:7 - This is an example of duplicate recessive epistasis because c/c masks expression from the A locus and a/a masks expression from the C locus, both giving white fur colour Dominant Epistasis
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F2 phenotypic ratio = 12:3:1 If the genotype is W/-, the cat is white, regardless of the genotype at the other locus Only in the w/w genotype, B/- is black and b/b is brown
Recessive Lethal Alleles - Essential genes, when mutated lead to a lethal genotype - Y/y x Y/y gives ratio 2 yellow:1 non yellow - Lethal allele (Y) is recessive: causes death only in homozygotes - But the effect of the allele on colour is dominant - Can you think of an example of a dominant lethal allele Recessive Lethal Alleles: Example #2 - t = dominant allele for absence of tail - T = recessive allele for normal tail - TT = normal tail - Tt = no tail - Tt = lethal - Manx cats are tailless - So, t is also recessive lethal allele Summary of F2 Ratios
In Class Question - The tailless Manx phenotype in cats is caused by a dominant autosomal allele at the T locus. A blank Manx cat was crossed to a brown cat. One quarter of their progeny (the F1) were black Manx, one quarter of their progeny were brown Manx, one quarter were black normal-tailed, and one quarter were brown normal-tailed. You know from previous work that the black coat colour is dominant to the brown coat colour at the B locus. a) What are the most likely genotypes of the black Manx cat and brown cat used in the initial cross? Most likely is BbMm (black) x bbmm (brown). This accounts for the 1:1 ratio of both traits independently
b) How could you test your hypothesis about the likely genotype of the black Manx cat used in the initial cross? Use a test cross: take a cat you know is recessive for both the B and M loci (bbmm) and see if you get the same ratios as above when you cross it to the black Manx cat. Take Home Question - Trihybrid cross of agouti, brown and coat colour genes: the overall interplay of these three genes causes a complex pattern of inheritance of coat colour that can be analyzed readily through use of the product rule, and diagrammed through use of the branching line approach - Remember that cc is totally epistatic over any other effects (giving white), and that in animals that have a C/- genotype, the A and B genes interact to yield agouti (A/-B/- phenotype), black (a/a/B/- phenotype), cinnamon (a/b/b phenotype), and brown (a/ab/b phenotype) - What are the expected phenotypic combinations from A/aB/bC/c x A/aB/bC/c - Choose either a Punnett square or branching line approach and bring your answer to the beginning of next class How do you Test a Theory of Inheritance? 1) Develop a theory about a pattern of inheritance Example: you propose that two traits are segregating independently of one another and show a phenotypic ratio of 1:1:1:1 2) Frame your Null Hypothesis Null hypothesis: no significant difference between the observed results and the predicted results In other words chance alone is responsible for deviations from predicted results 3) Statistical analysis assays whether any difference is due to chance. If not, the null hypothesis is rejected Chi-square test, χ2 Statistical test that provides information about how well observed values fit expected values In Class Example - A brown cat with short tail mates with a cinnamon cat with a long tail. A series of matings yields the following progeny: 154 brown, short tail 124 brown, long tail 144 cinnamon, short tail 146 cinnamon, long tail total: 568 - Theory: parent 1 is a heterozygote. Assuming Mendel’s Law, you should see a phenotypic ratio of 1:1:1:1
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B/b1S/s x b1/b1s/s Null hypothesis: there is no significant difference between observed and predicted data Total number of progeny is 568. Expected number of each phenotypic class is 568/4 = 142 ∑ ⁄ df = (number of independent classes) -1, 4-1 = 3 χ2 = 3.43 Only differences due to chance
Interpreting your Results - Options: Reject the hypothesis or fail to reject the hypothesis - Generally, if the probability of obtaining a χ2 value is greater that 5/100 (P>0.05), then you fail to reject your null hypothesis that there is no difference between the two sets of data - In this example, we fail to reject the null hypothesis, that is, there is no statistical reason for rejecting the hypothesis. The assumptions of inheritance (1:1:1:1 pattern based upon simple dominant-recessive relationships) are consistent with the results that you have obtained In Class Example - A brown cat with a short tail mates with a cinnamon cat with a long tail. A series of matings yields the following progeny: 165 brown, short tail 120 brown, long tail 165 cinnamon, short tail 118 cinnamon, long tail Total: 568 - Theory: parent 1 is a heterozygote. Assuming Mendel’s Law, you should see a phenotypic ratio 1:1:1:1, B/b1S/s x b1/b1s/s - Null hypothesis: there is no significant difference between observed and predicted data - χ2 = 14.93 - The P value is
Biology 2C03: Genetics Extensions of Mendelian Principles: Genetic Interactions
Module 3 Outline: Cats are not Peas - At the end of this module, you will be able to: solve probles dealing with multiple alleles 1. Define, recognize, describe and apply Mendel’s second principle 2. Solve genetic problems relating to dihybrid and trihybrid crosses 3. Use the rules of probability to determine genotypic and phenotypic proportions 4. Know how and when to use a testcross 5. Recognize modified dihybrid Mendelian ratios 6. Test your theory that a Mendelian ratio is consistent with your data using a Chi-squared test Multiple Alleles - Mutations anywhere along the gene can still give a loss of function, but are considered different alleles - Population: multiple alleles may exist - Individual (diploid): only two alleles coexist in each cell
Example of the Multiple Alleles Contributing to Fur Coloration in Cats - Five alleles: C: full colour cb-: Burmese (mutation at a particular locus) cs: Siamese (mutation at a different allele) c: white, blue eyes ca: albino, pink eyes - C>cb=cs>c>ca (dominant/allelic series) If each individual has two alleles, the dominant/allelic series tells you which one will be dominant - Where: > indicates
-
= indicates incomplete dominance or codominance cb/cs = mink Ex, for a cat to be Siamese they must have the cs allele with either a c or ca allele This represents a dominance series or allelic series. Important is the relationship between pairs of alleles
The C Gene Codes for an Enzyme Called Tyrosinase - The albino allele (the most recessive allele in this series) contains a cytosine deletion in Tyrosinase at position 975 in exon 2 - Causes a frame shift resulting in a premature stop codon residues downstream from the mutation – loss of a single nucleotide that alters the reading frame - This is a non-functional protein and the allele is described as a loss of function allele - Definition of frame-shift mutation: the loss of a single nucleotide causes the reading frame, the series of three neucleotide codons, to be read differently The Number of Different Genotypes is Determined by the Number of Alleles in the Population - Number of genotypes is determined by the number of different possible allele combinations in the population - c, ca. cs, cb, C (wildtype) - This gives, 15 genotypes of which there are five different homozygous genotypes and 10 different heterozygotes
Biochemical Definition of Wildtype - Wildtype allele: a functional enzyme or other protein is produced in the cell - Loss of function allele: an enzyme or other protein is no longer being produced, is produced at lower levels, or is nonfuncitonal, or at least subfunctional - Sometimes the term wildtype is used to refer to the most common phenotype (or genotype) found in a natural population - Often the wildtype allele is dominant over the loss of function allele - Half as much protein is synthesized yet this is often sufficient to achieve the wildtype phenotype (heterozygotes) - This is called haplosufficiency
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For fur colour, in the C/c genotype, half as much Tyrosinase is synthesized, but this is sufficient to achieve full colouration
Because there are Always Exceptions to the Rule - It is not always the case that the dominant allele is ‘normal’ and the recessive allele is a mutation - Dominant alleles can be gain of function mutations, in which the mutant allele produces a protein that has increased (detrimental) function - E.g. Huntington’s disease is an autosomal dominant disease which manifests the phenotype later in life; 50% chance of passing on the allele to your offspring - Dominant allele can be a loss of function alleles, but in the heterozygote, half as much protein is synthesized and this is not sufficient for a normal phenotype; haploinsufficient – inheriting both copies of the mutation is a lethal genotype - E.g. tailless cats (Manx)
Gregor Mendel Defined the Principle of Segragation - Mendel’s principle of segregation: 1) Each individual organism possesses two alleles encoding a trait 2) Alleles separate when gametes are formed 3) Alleles separate in equal proportions - He did this using monohybrid crosses, e.g. Aa x Aa - What would happen if he followed two traits at the same time? Mendel’s Law of Independent Assortment - The inheritance pattern of one trait will not affect the inheritance pattern of another trait - Examined dihybrid crosses e.g. Aa Bb x Aa Bb Taking true breeding round yellow seeds with true breeding wrinkled green seeds producing heterozygotes Four possible phenotypes
9:3:3:1 ratio in the F2 phenotype - Each trait is inherited by the Mendelian ratio, 3:1, but together we see a phenotypic ratio of 9:3:3:1
Another Example: Cat Colour and Tail Length - 9 brown fur, short tail - 3 brown fur, long tail - 3 white fur, short tail - 1 white fur, long tail - 9:3:3:1 - 9 = two dominant traits - 1 = two recessive traits
More Complex Problems: Tribybrid Cross - You can use a similar procedure for determining the proportion of genotypes and phenotypes if you are following more than two alleles: E.g. trihybrid crosses (and more…) or why I learned to stop using Punnett squares Tryhibrid: 64 combinations of gametes (an 8x8 Punnett square) 27 different F2 genotypes 8 different F2 phenotypes ratio: 27:9:9:9:3:3:3:1
Take Home Problem - In cats, white patches are caused by the dominant allele P, while p/p individuals are solid coloured - Short hair is caused by a dominant allele, S, while s/s cats have long hair - A long-haired cat with patches whose mother was solid-coloured and shorthaired mates with a short-haired, solid coloured cat whose mother was longhaired and solid-coloured - What kinds of kittens can arise from this mating and in what proportions
Extensions of Mendelian Principles or Cats are Not Peas - How do multiple genes interact to determine a single phenotype? What Happens When Multiple Genes often Contribute to a Single Characteristic - What happens when many genes contribute to a single phenotype: interaction between genes - F1 cross: A/a B/b x A/a B/b, this is still a dihybrid cross so we still have to consider Mendelian principles - If there is complete dominance at two distinct traits: phenotypes = 9:3:3:1 - Four potential outcomes - Gene interactions may produce novel phenotypes that will modify the 9:3:3:1 F2 phenotypic ratio - But will still be based upon it – total 16
Complete Dominance - D/D = this cat does not carry the dilute gene - D/d = this cat is a carrier of the dilute gene - d/d = this cat carries two copies of the dilute gene. The coat colour is diluted. - Black dilutes to blue, chocolate dilutes to lilac, cinnamon dilutes to fawn, and red dilutes to cream Molecular Explanation for Complete Dominance of Cat Colouration - The dilute gene (or melanophilin) affects the distribution of melanin granules within the cats’ hair - Dominant or dense allele (D): produces dense pigmentation - Recessive dilute (d) allele: results in pigment clumping, forming areas in the hair with no granules. This causes the cat coat to dilute or lighten - The dilute effect is autosomal recessive, hence a cat requires two copies of the d allele for the coat to dilute Complementation - White x white = white - White x white = white - White offspring mate = black - F2 phenotypic ratio = 9:7 Complementation – Two Different Albinos
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White (c/c) x white (c/c) = white (c/c) White (a/a) x white (a/a) = white (a/a) White (c/c) x white (a/a) = black Mutations in two different genes produce the same mutant phenotype F2 phenotypic ratio = 9:7 P: white (c/c) x white (c/c) and white (a/a) x white (a/a) F1: white (c/c, A/A) x white (C/C, a/a) F2 phenotypic ratio= 9:7 black (C/c, A/a)
We Can Observe Complementation in Human Pedigrees - About 50 genes have recessive mutant alleles that can cause deafness in humans - A heterogenous trait: a mutation in any one of a number of genes can give rise to the same phenotype - Complementation: 9:7 - Non complementation:
Types of Natural Selection
September 24th, 2013
What Happens when Multiple Genes Contribute to a Single Characteristic? - Examples from last class Complete dominance Complementation - P: A/A b/b x a/a B/B; F1 A/a B/b - What gametes are produced? - Complete dominance (9:3:3:1)
Complementation - Complementation occurs when two strains of an organism with different homozygous recessive mutations that produce the same phenotype, produce offspring of the wild-type phenotype when mated or crossed - Complementation will only occur if the mutations are in different genes - The other genome supplies the wild-type allele to “complement” the mutated allele - Complementation will not occur if the mutations are in the same gene What Happens When Multiple Genes Contribute to a Single Characteristic? - What gametes are produced? - 9:7
Take Home Problem - In cats, white patches are caused by the dominant allele P, while p/p individuals are solid coloured - Short hair is caused by a dominant allele, S, while s/s cats have long hair - A long-haired cat with patches whose mother was solid-coloured and shorthaired mates with a short-haired, solid coloured cat whose mother was longhaired and solid-coloured - What kinds of kittens can arise from this mating and in what proportion - Ppss x ppSs Answer to Take Home Problem - ¼ PpSs: patches, short hair - ¼ ppSs: solid, short hair - ¼ Ppss: patches, long hair - ¼ ppss: solid, long hair - Does this agree with your answer Genetic Epistasis - 9:7 – observing complementation, assume that two alleles are acting on the genotype - Epistasis: the masking of the expression of one gene by another. No new phenotypes are produced - The epistatic gene does the masking - The hypostatic gene is masked - F2 phenotypic ratio = 9:3:4 – recessive epistasis Homozygous recessives at one gene pair mask expression from the other gene A/-b/b and a/ab/b have the same phenotype bb gives recessive phenotype - F2 phenotypic ratio = 12:3:1 dominant epistasis One dominant allele at one gene masks expression from the other gene B locus does not matter, as long as there is a dominant A allele it will be the same phenotype
A/-B/- and A/-b/b have the same phenotype
Recessive Epistasis Example - F2 phenotypic ratio = 9:3:4 recessive epistasis - If the genotype is c/c, the mouse is white, regardless of the genotype at the other locus
Molecular Mechanism of Recessive Epistasis - c/c: no pigment is synthesized; therefore mice are white regardless of genotype at the A locus; this is recessive epistasis - The genotype at the A locus determines how the pigment is deposited: either black (a/a) or agouti (A/-) - Figure 5.9 in text has additional example
Duplicate Recessive Epistasis - Look back at the model of the two genes that produced white cats where F2 phenotypic ratio = 9:7 - This is an example of duplicate recessive epistasis because c/c masks expression from the A locus and a/a masks expression from the C locus, both giving white fur colour Dominant Epistasis
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F2 phenotypic ratio = 12:3:1 If the genotype is W/-, the cat is white, regardless of the genotype at the other locus Only in the w/w genotype, B/- is black and b/b is brown
Recessive Lethal Alleles - Essential genes, when mutated lead to a lethal genotype - Y/y x Y/y gives ratio 2 yellow:1 non yellow - Lethal allele (Y) is recessive: causes death only in homozygotes - But the effect of the allele on colour is dominant - Can you think of an example of a dominant lethal allele Recessive Lethal Alleles: Example #2 - t = dominant allele for absence of tail - T = recessive allele for normal tail - TT = normal tail - Tt = no tail - Tt = lethal - Manx cats are tailless - So, t is also recessive lethal allele Summary of F2 Ratios
In Class Question - The tailless Manx phenotype in cats is caused by a dominant autosomal allele at the T locus. A blank Manx cat was crossed to a brown cat. One quarter of their progeny (the F1) were black Manx, one quarter of their progeny were brown Manx, one quarter were black normal-tailed, and one quarter were brown normal-tailed. You know from previous work that the black coat colour is dominant to the brown coat colour at the B locus. a) What are the most likely genotypes of the black Manx cat and brown cat used in the initial cross? Most likely is BbMm (black) x bbmm (brown). This accounts for the 1:1 ratio of both traits independently
b) How could you test your hypothesis about the likely genotype of the black Manx cat used in the initial cross? Use a test cross: take a cat you know is recessive for both the B and M loci (bbmm) and see if you get the same ratios as above when you cross it to the black Manx cat. Take Home Question - Trihybrid cross of agouti, brown and coat colour genes: the overall interplay of these three genes causes a complex pattern of inheritance of coat colour that can be analyzed readily through use of the product rule, and diagrammed through use of the branching line approach - Remember that cc is totally epistatic over any other effects (giving white), and that in animals that have a C/- genotype, the A and B genes interact to yield agouti (A/-B/- phenotype), black (a/a/B/- phenotype), cinnamon (a/b/b phenotype), and brown (a/ab/b phenotype) - What are the expected phenotypic combinations from A/aB/bC/c x A/aB/bC/c - Choose either a Punnett square or branching line approach and bring your answer to the beginning of next class How do you Test a Theory of Inheritance? 1) Develop a theory about a pattern of inheritance Example: you propose that two traits are segregating independently of one another and show a phenotypic ratio of 1:1:1:1 2) Frame your Null Hypothesis Null hypothesis: no significant difference between the observed results and the predicted results In other words chance alone is responsible for deviations from predicted results 3) Statistical analysis assays whether any difference is due to chance. If not, the null hypothesis is rejected Chi-square test, χ2 Statistical test that provides information about how well observed values fit expected values In Class Example - A brown cat with short tail mates with a cinnamon cat with a long tail. A series of matings yields the following progeny: 154 brown, short tail 124 brown, long tail 144 cinnamon, short tail 146 cinnamon, long tail total: 568 - Theory: parent 1 is a heterozygote. Assuming Mendel’s Law, you should see a phenotypic ratio of 1:1:1:1
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B/b1S/s x b1/b1s/s Null hypothesis: there is no significant difference between observed and predicted data Total number of progeny is 568. Expected number of each phenotypic class is 568/4 = 142 ∑ ⁄ df = (number of independent classes) -1, 4-1 = 3 χ2 = 3.43 Only differences due to chance
Interpreting your Results - Options: Reject the hypothesis or fail to reject the hypothesis - Generally, if the probability of obtaining a χ2 value is greater that 5/100 (P>0.05), then you fail to reject your null hypothesis that there is no difference between the two sets of data - In this example, we fail to reject the null hypothesis, that is, there is no statistical reason for rejecting the hypothesis. The assumptions of inheritance (1:1:1:1 pattern based upon simple dominant-recessive relationships) are consistent with the results that you have obtained In Class Example - A brown cat with a short tail mates with a cinnamon cat with a long tail. A series of matings yields the following progeny: 165 brown, short tail 120 brown, long tail 165 cinnamon, short tail 118 cinnamon, long tail Total: 568 - Theory: parent 1 is a heterozygote. Assuming Mendel’s Law, you should see a phenotypic ratio 1:1:1:1, B/b1S/s x b1/b1s/s - Null hypothesis: there is no significant difference between observed and predicted data - χ2 = 14.93 - The P value is
