mendelian genetics review simple dominance codominance incomplete dominance sex linked dihybrid...
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Mendelian Mendelian GeneticsGeneticsReviewReviewSimple DominanceSimple DominanceCodominanceCodominanceIncomplete DominanceIncomplete DominanceSex LinkedSex LinkedDihybrid CrossesDihybrid Crosses
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Simple DominanceSimple Dominance
PP11 cross b/t pure breeding dominant and cross b/t pure breeding dominant and recessive traits yields an Frecessive traits yields an F11 generation that is generation that is 100% dominant in phenotype.100% dominant in phenotype.
Heterozygotes express dominant phenotypeHeterozygotes express dominant phenotype A cross between the FA cross between the F11 hybrids (heterozygotes) hybrids (heterozygotes)
yields a predictable 1:2:1 ratio of homozygous yields a predictable 1:2:1 ratio of homozygous dominant: heterozygous: homozygous dominant: heterozygous: homozygous recessive in Frecessive in F22 offspring offspring
FF22 monohybrid phenotypic ratio = 3 dom:1 rec. monohybrid phenotypic ratio = 3 dom:1 rec.
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CodominanceCodominance
PP11 cross b/t pure breeding strains of two traits cross b/t pure breeding strains of two traits
yields an Fyields an F11 generation that exhibits both generation that exhibits both
phenotypes fully expressed (heterozygous)phenotypes fully expressed (heterozygous) A cross between the FA cross between the F11 hybrids (heterozygotes) hybrids (heterozygotes)
yields a predictable 1:2:1 ratio of homozygous yields a predictable 1:2:1 ratio of homozygous trait 1: heterozygous 1+2: homozygous trait 2 trait 1: heterozygous 1+2: homozygous trait 2 in Fin F22 offspring offspring
FF22 monohybrid phenotypic ratio = 1:2:1 monohybrid phenotypic ratio = 1:2:1
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Incomplete DominanceIncomplete Dominance
PP11 cross b/t pure breeding strains of two cross b/t pure breeding strains of two traits yields an Ftraits yields an F11 generation that exhibits generation that exhibits the intermediate trait (heterozygous)the intermediate trait (heterozygous)
A cross between the FA cross between the F11 hybrids hybrids (heterozygotes) yields a predictable 1:2:1 (heterozygotes) yields a predictable 1:2:1 ratio of homozygous trait 1: heterozygous ratio of homozygous trait 1: heterozygous ½ blend: homozygous trait 2 in F½ blend: homozygous trait 2 in F22 offspringoffspring
FF22 monohybrid phenotypic ratio = 1:2:1 monohybrid phenotypic ratio = 1:2:1
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Sex LinkageSex Linkage
Can be X or Y linked: carried only on that Can be X or Y linked: carried only on that chromosomechromosome
X-linked recessive: expressed more in X-linked recessive: expressed more in males. Why?males. Why?
Y-linked: Y-linked: onlyonly expressed in males expressed in males X-linked recessiveX-linked recessive
Female carriers (heterozygotes) X normal Female carriers (heterozygotes) X normal males yield Fmales yield F11 normal females and 50% normal females and 50% affected malesaffected males
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Dihybrid CrossDihybrid Cross
PP11 cross of organisms with two sets of cross of organisms with two sets of
seemingly opposite traits yield offspring seemingly opposite traits yield offspring that express only one of each of the traits that express only one of each of the traits = 100% dominant for both phenotypes.= 100% dominant for both phenotypes.
A cross of the FA cross of the F11 hybrid offspring yields a hybrid offspring yields a
predictable 9:3:3:1 phenotypic ratio for predictable 9:3:3:1 phenotypic ratio for the Fthe F22 generation. generation.
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What does it mean when What does it mean when these predictions DON’T these predictions DON’T occur?occur?There is something else controlling the There is something else controlling the
expression of the genes.expression of the genes.
Such as: gene linkage, epistasis, Such as: gene linkage, epistasis, dominance series, polygenes, etc.dominance series, polygenes, etc.
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alleleallele
Variable form of a geneVariable form of a gene
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genotypegenotype
Symbolic representation of a gene Symbolic representation of a gene combinationcombination
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phenotypephenotype
Expression of a gene combinationExpression of a gene combination
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Punnett squarePunnett square
Graphical representation of the probability Graphical representation of the probability of inheriting a particular traitof inheriting a particular trait
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More TermsMore Terms
Generations: PGenerations: P11, F, F11, F, F22 . . . . . .
Heterozygous, HomozygousHeterozygous, HomozygousTestcrossTestcrossMonohybrid, dihybrid, multihybridMonohybrid, dihybrid, multihybrid3:1/1:2:1 monohybrid3:1/1:2:1 monohybrid9:3:3:1 Dihybrid9:3:3:1 DihybridLaw of segregation/Law of independent Law of segregation/Law of independent AssortmentAssortment
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Solving Multihybrid crossesSolving Multihybrid crossesCrossing the following :Crossing the following :
AABBCCDDEEFF x aabbccddeeff PAABBCCDDEEFF x aabbccddeeff P11
All AaBbCcDdEeFf FAll AaBbCcDdEeFf F11
– Just like monohybrid crosses!!Just like monohybrid crosses!!
Crossing FCrossing F11 offspring, you can predict number of offspring, you can predict number of
gametes and genotypes of offspring easilygametes and genotypes of offspring easily– Number of gametes = 2Number of gametes = 2nn where n=number of where n=number of
heterozygous allele combinationsheterozygous allele combinations– Probability of offspring with a “multiple” Probability of offspring with a “multiple”
genotype = product of individual genotype genotype = product of individual genotype probabilitiesprobabilities
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ExampleExample
Crossing AaBbCcDdEe hybrids Crossing AaBbCcDdEe hybrids There are 5 pairs of alleles that are heterozygousThere are 5 pairs of alleles that are heterozygous
2255 = number of possible gametes = number of possible gametes
AABBccDdEeFf gives how many possible gametes?AABBccDdEeFf gives how many possible gametes?
2233 = 8 possible gametes = 8 possible gametes
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Offspring Probability Examp.Offspring Probability Examp.
What is the probability of getting an offspring What is the probability of getting an offspring with the genotype AABBCcDdEe from a with the genotype AABBCcDdEe from a Multihybrid cross of AaBbCcDdEe?Multihybrid cross of AaBbCcDdEe?
Treat each pair of alleles INDEPENDENTLY.Treat each pair of alleles INDEPENDENTLY.
Product of probabilities = overall probabilityProduct of probabilities = overall probability
¼ AA x ¼ BB x ½ Cc x ½ Dd x ½ Ee = 1/128 ¼ AA x ¼ BB x ½ Cc x ½ Dd x ½ Ee = 1/128 probability of AABBCcDdEe genotypeprobability of AABBCcDdEe genotype
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DNADNA
Deoxyribonucleic acidDeoxyribonucleic acid
Deoxyribose (pentose sugar) + Phosphate Deoxyribose (pentose sugar) + Phosphate backbonebackbone
Nitrogenous bases: A, G, C, TNitrogenous bases: A, G, C, T
Sugar + Phosphate + base = nucleotideSugar + Phosphate + base = nucleotide
Cells contain approximately 3 meters of Cells contain approximately 3 meters of DNA eachDNA each
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ChromosomeChromosome
Supercoiled DNA/protein complex Supercoiled DNA/protein complex designed to “manage” the immense designed to “manage” the immense information packaged in cellular DNAinformation packaged in cellular DNA
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GenomeGenome
Full complement of DNA contained within Full complement of DNA contained within and organismand organism
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GeneGene
Segment of DNA encoding a protein Segment of DNA encoding a protein and/or regulating the expression of other and/or regulating the expression of other genesgenes
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Sample Multihybrid Gametes and Sample Multihybrid Gametes and ProbabilitiesProbabilities
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A pea plant is heterozygous for both seed shape and seed color. S is the allele for the dominant, spherical shape characteristic; s is the allele for the recessive, dented shape characteristic. Y is the allele for the dominant, yellow color characteristic; y is the allele for the recessive, green color characteristic. What will be the distribution of these two alleles in this plant's gametes?
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25% SY
25%Sy
25% sY
25% sy
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Law of Independent Assortment: Alleles of different genes are assorted independently of each other during the formation of gametes.
Law of Segregation: during meiosis, alleles separate from each other
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The ability to taste the chemical PTC is determined by a single gene in humans with the ability to taste given by the dominant allele T and inability to taste by the recessive allele t. Suppose two heterozygous tasters (Tt) have a large family. a. Predict the proportion of their children who will be tasters and nontasters. Use a Punnett square to illustrate how you make these predictions.
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What is the likelihood that their first child will be a taster? What is the likelihood that their fourth child will be a taster?
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1st child-- 3/44th child-- 3/4 (each child is independent)
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What is the likelihood that the first three children of this couple will be nontasters?
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1/4 for each child; 1/4 x 1/4 x 1/4 = 1/64 that all three will be nontasters
Product rule: AND is key word
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Tail or No Tail . . . that is the question.
Tail = expressed trait
P1
F1
F2
F3
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Mackerel/Blotched Tabby Stripes
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+
+1 2
2
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+ +
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Inheritance Modes
All white v. not all whiteAll white dominant to not all whiteSome white v. no whiteSome white dominant to no white with variable expressivityDilute v. not diluteDense color dominant to dilute
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continued
Agouti v. non-agouti
Agouti dominant epistatis
Makeral v. blotched
Mackeral dominant
No Tail dominant over Tail