2 Gregor Mendel 1822 – 1884 Austrian monk Experimented with garden peas Provided a basis for understanding heredity
3 Mendel cont’dPublished a paper in 1866 stating that parents pass discrete heritable factors on to their offspringFactors retain individuality generation after generationIdentified that each trait is inherited by a pair of factors, one from each parentOne form of a factor may be dominant over an alternative formReasoned that each egg and sperm must contain only 1 copy of a factor for each trait
4 Mendel cont’d. Mendel’s law of segregation Each individual has two factors (genes) for each traitThe factors segregate (separate) during the formation of gametesEach gamete contains only one factor from each pair of factorsFertilization gives each new individual 2 factors for each trait
5 Modern Genetics – Genes Sections of chromosomes which give instructions for one characteristic or protein.Located at the same point or locus, on each member of a homologous pairAll together make up the organism’s genome.Controls the physical characteristics of a species
6 Modern Genetics - Alleles Alternative forms of the same gene on each chromosomeOne allele comes from each parentDominant alleleMasks other traits presentOnly 1 dominant allele needs to be present for a certain trait to be expressedRepresented by a capital letterRecessive Allele2 copies of the recessive allele need to be present for trait to be expressedRepresented by a lower case letter
7 Modern Genetics – Alleles cont’d Gene locusFig. 23.2
8 Modern Genetics – Alleles cont’d GenotypeGenetic composition of a specific traitHomozygous dominant – 2 dominant allelesHeterozygous – 1 dominant allele and 1 recessive alleleHomozygous recessive 2 recessive allelePhenotypePhysical expression of a specific traitHomozygous dominant or heterozygous Dominant TraitHomozygous recessive recessive trait
9 Single Gene Inheritance Simplest situationOne gene carries all the information responsible for one traitWidow’s PeakAlternative forms of alleles for hairline shapeWidow’s peak is dominant to straight hair lineW=allele for widow’s peakw= allele for straight hairline
12 Single Gene Inheritance cont’d. Monohybrid crossLooks at inheritance of one trait onlyA punnett squares used to find all possible combinations of alleles.
13 Single Gene Inheritance cont’d. Example 1:If a homozygous woman with a widow’s peak reproduces with a man with a straight hairline, what kind of hairline will their children have?
14 Single Gene Inheritance cont’d. Example:If two heterozygous parents reproduce what kind of hairline will their children have?WwWWWwww
15 Single Gene Inheritance cont’d. Genetic RatiosExpress ratio’s of possible outcomesGenotypic ratioHomozygous Dominant: Heterozygous: Homozygous RecessivePhenotypic ratioDominant trait : recessive traitOften expressed as probabilityThe probability resets and is the same for each pregnancy!Having one child with a trait has no effect on future children.
16 Single Gene Inheritance cont’d. Look back to example 1What are the genotypic and phenotypic ratio’s?WW x wwGenotypic Ratio0 : 4 : 0Phenotypic Ratio1:0100% Widow’s peakWw x WwGenotypic Ratio1: 2: 1Phenotypic Ratio3 : 1Probability is ¾ Widow’s Peak75%Probability ¼ Straight25%
17 Single Gene Inheritance cont’d Determining GenotypeNo way to distinguish between a homozygous dominant individual and a heterozygous individual just by lookingThey are phenotypically the sameTest cross may help us determineUsed in breeders of plants and animalsCross unknown with a recessive individualWe know one parent genotype, this will help us determine the other genotypeIf there are any offspring produced with the recessive phenotype, then the dominant parent must be heterozygous
18 Single Gene Inheritance cont’d Example : In rats, large ears is dominant to small ears. A rat breeder has a female rat with large ears, which she breeds with a male rat with small ears. In the first litter, all rats are born with large ears.What is the genotype of the female rat?In a second litter from the same parents, 4 baby rats have large ears, one has small ears.
19 Single Gene Inheritance cont’d Practice problemsBoth a man and a woman are heterozygous for freckles. Freckles are dominant over no freckles. What is the chance that their child will have freckles?Both you and your sibling have attached ear lobes, but your parents have unattached lobes. Unattached earlobes (E) are dominant over attached (e). What are the genotypes of your parents?A father has dimples, the mother of his children does not, and all 5 of their children have dimples. Dimples (D) are dominant over no dimples (d). Give the probable genotypes of all persons concerned.
21 Independent Assortment Mendel reasoned from the results of his pea plant crosses that each pair of factors assorts independently into gametesEach trait is passed down individually. The allele you receive for any one gene is not related to any other alleles you receive.We can now explain this through independent alignment and crossing over in meiosis. (Fig 23.6)
23 Independent Assortment cont’d Called the Law of Independent AssortmentEach pairs of factors assorts independently (without regard to how the others separate)All possible combinations of factors can occur in the gametes
24 Independent Assortment cont’d Practice problemsFor each of the following genotypes, give all possible gametesWWWWSsTtTtggAaBbFor each of the following, state whether the genotype or a gamete is representedDLlPwLlGg
25 Dihybrid CrossPunnet squares, considering two-trait crosses at one time.Example:The traits for hairline and finger length are both single gene traits. As before, widows peak is dominant over straight hairline. Having short fingers is considered dominant over long fingers.Two parents who are both heterozygous for both traits have children.Determine the genotypic and phenotypic ratios for their children.
26 Dihybrid cross cont’d Possible Gametes? Genotypic Ratio? Phenotypic Ratio?
27 Dihybrid Crosses cont’d Determining Ratio’sProduct rule of probabilityThe chance of 2 or more independent events occurring together is the product of their chance of occurring separatelyIn our example:Probability of widow’s peak = ¾Probability of short fingers= ¾What is the probability of widow’s peak AND short fingers?¾ x ¾ = 9/16
28 Dihybrid Crosses cont’d Recall from our single trait crossesProbability of widow’s peak = ¾ Probability of short fingers= ¾Probability of straight hairline= ¼ Probability of long fingers= ¼Using the product ruleProbability of widow’s peak and short fingers = X =Probability of widow’s peak and long fingers = X = 3/16Probability of straight hairline and short fingers = ¼ X ¾ = 3/16Probability of straight hairline and long fingers = ¼ X ¼ = 1/16These values are standard for all heterozygous crosses! You don’t need to memorize them, but should be able to figure them out in your head!
29 Dihybrid Crosses cont’d Using the product ruleProbability of widow’s peak and short fingersProbability of widow’s peak and long fingersProbability of straight hairline and short fingersProbability of straight hairline and long fingers
30 Dihybrid Crosses cont’d Two-trait test crossCross an individual with the dominant phenotype for each trait with an individual with the recessive phenotype of both traitsW?S? x wwssWSW??S??wwssWwSsWs?s?wSs?w?s
31 Dihybrid Cross cont’dAttached earlobes are recessive, What genotype do children have if one parent is homozygous for earlobes and homozygous dominant for hairline, and the other is homozygous dominant for unattached earlobes and homozygous recessive for hairline?If an individual from this cross reproduces with another of the same genotype, what are the chances that they will have a child with a straight hairline and attached earlobes?A child who does not have dimples or freckles is born to a man who has dimples and freckles (both dominant traits) and a woman who does not. What are the genotypes of all persons concerned?
32 Polygenic Inheritance Not fully understood by geneticists.Generally:One trait controlled by 2 or more genes at different lociThe higher the number of dominant alleles you possess, the stronger the expression of the trait.Result is a continuous range of phenotypesDistribution resembles a bell curveThe more gene pairs involved, the more continuous the pattern of variationEx: human height, skin pigmentation, eye colour
34 Polygenic Inheritance cont’d Skin colorControlled by many gene pairs and many allelesLet’s assuming a simple model of two alleles at 2 lociA and BIf two heterozygous parents have children, children can range from very light to very darkGenotypePhenotypeAABBVery Dark SkinAABb or AaBBDark SkinAaBb, AAbb, or aaBBMedium brown skinAabb, or aaBbLight SkinaabbVery light skin
35 Polygenic Inheritance cont’d Eye colour is controlled by 3 genes we have identifiedWe suspect there are moreNot a clear dominant and recessiveBrown allele
36 Environmental Influences on Inheritance Environment can influence gene expression and therefore phenotypeEx: sunlight exposure on skin; coat color in Himalayan rabbitsHuman twin studiesPolygenic traits are most influenced“nature vs. nurture”Identical twins separated at birth are studiedIf they share a trait in common even though raised in different environments, it is likely genetic
38 Incomplete Dominance Incomplete dominance Heterozygous individual has a phenotype intermediate to the two homozygous individualsEx: Curly-haired Caucasian woman and a straight- haired Caucasian man produce wavy-haired childrenWhen 2 wavy-haired people have children, the phenotypic ratio is 1 curly: 2 wavy: 1 straight
40 Codominance Multiple allele inheritance The gene exists in several allelic forms, but each person still has only 2 of the possible allelesOccurs when both alleles are equally expressedEx: type AB blood has both A antigens and B antigens on red blood cells
41 Codominance cont’d. ABO blood types IA = A antigens on RBCs IB = B antigens on RBCsi = has neither A nor B antigens on RBCsBoth IA and IB are dominant over I, IA and IB are codominantPhenotypeGenotypeAIAIA or IAiBIBIB or IbiABIAIBOii
42 Codominance cont’d. Paternity testing- ABO blood groups often used Can disprove paternity but not prove itRh factor- another antigen on RBCsRh positive people have the antigenRh negative people lack itThere are multiple alleles for Rh negative, but all are recessive to Rh positive
44 Practice ProblemsA polygenic trait is controlled by three pairs of alleles. What are the two extreme genotypes for this trait?What is the genotype of the lightest child that could result from a mating between two medium-brown individuals?A child with type O blood is born to a mother with type A blood. What is the genotype of the child? The mother? ‘what are the possible genotypes of the father?From the following blood types determine which baby belongs to which parents:Baby 1 type O Mrs. Doe type A Mrs. Jones type ABaby 2 type B Mr. Doe type A Mr. Jones type AB
45 Sex-linked inheritance Sex chromosomes22 pairs of autosomes, 1 pair of sex chromosomesX and yIn females, the sex chromosomes are XXIn males, the sex chromosomes are XYNote that in males the sex chromosomes are not homologousTraits controlled by genes in the sex chromosomes are called sex-linked traitsX chromosome has many genes, the Y chromosome does not
46 Sex-linked inheritance cont’d. X-linked traitsRed-green colorblindness is X- linkedThe X chromosome has genes for normal color visionXB = normal visionXb – colorblindnessGenotypesPhenotypesXBXBfemale with normal visionXBXbcarrier female, normal visionXbXbcolorblind femaleXBYmale with normal visionXbYcolorblind male
48 Practice ProblemsBoth the mother and the father of a colorblind male appear to be normal. From whom did the son inherit the allele for colorblindness? What are the genotypes of the mother, father, and the son?A woman is colorblind. What are the chances that her son will be colorblind? If she is married to a man with normal vision, what are the chances that her daughters will be colorblind? Will be carriers?Both the husband and the wife have normal vision. The wife gives birth to a colorblind daughter. Is it more likely the father had normal vision or was colorblind? What does this lead you to deduce about the girl’s parentage?What is the genotype of a colorblind male with long fingers is s=long fingers? If all his children have normal vision and short fingers, what is the likely genotype of the mother?
49 Inheritance of linked genes The sequence of individual genes on a chromosome is fixed because each allele has a specific locusAll genes on a single chromosome form a linkage groupWhen linkage is complete, a dihybrid produces only 2 types of gametesAny time traits are inherited together, a linkage group is suspectedIf very few recombined phenotypes appear in offspring, linkage is also suspected
50 Inheritance of linked genes Crossing over between 2 alleles of interest, can result in 4 types of gametesOccurrence of crossing over can indicate the sequence of genes on a chromosomeMore frequent between distant genesFig
51 Practice ProblemsWhen AaBb individuals reproduce, the phenotypic ratio is about 3:1. What ratio was expected? What may have caused the observed ratio?The genes for ABO blood type and for fingernails are on the same homologous pair of chromosomes. In an actual family, 45% of offspring have type B blood and no fingernails, and 45% have type O blood and fingernails; 5% have type B blood and fingernails, and 5% have type O blood and no fingernails. What process accounts for the recombinant phenotypes?