2 Mendel’s Experiments Gregor Mendel Joined the Augustinian Monastery at the age of 21Taught in a secondary school, was fascinated with science and nature (physics, evolution, botany and natural sciences)Attended the University of Vienna to study physics and biologyReturned to the monastery and began his experiments with the common garden plant (Pisum sativum)
3 Mendel’s Experimental, Quantitative Approach Advantages of pea plants for genetic study:Many varieties with distinct heritable features, or characters (such as flower color); character variants (such as purple or white flowers) are called traitsMating of plants can be controlledEach pea plant has sperm-producing organs (stamens) and egg-producing organs (carpels)Cross-pollination (fertilization between different plants) can be achieved by dusting one plant with pollen from another
5 Mendel’s Experiments Mendel’s Experimental Design Crossed only peas with desired traits, two methodsSelf-fertilization – pollen from anther falls onto the stigma of the same flower before it opensCross-fertilization – pollen from one plant fertilizes anotherMendel opened the keel and removed the anther before self-fertilization could occurSelected 7 discrete, nonoverlapping characteristicsFlower color – purple or whiteNonoverlapping characteristics: heights may vary so he chose the dwarf variety to compare to the tall. Tall simply refers to the nondwarf type.
7 Mendel’s Experiments Mendel’s Experimental Design Grew the plants for two years to identify homogenous, pure-breeding characteristicsStarted by crossing two pure-breeding plants (the parent generation, P), one purple, one whiteThe offspring (first filial generation, F1), were referred to as hybrids (mixture of both parents)Monohybrids – hybrid of only one characteristicAll the F1 plants were purple – purple flower color is a dominant traitThus the white flower color trait is recessiveNonoverlapping characteristics: heights may vary so he chose the dwarf variety to compare to the tall. Tall simply refers to the nondwarf type.
8 Mendel’s Experiments Mendel’s Experimental Design He allowed the F1 to self-fertilizeThe resulting second filial generation (F2) showed both dwarf and tall characteristics705 purple, 224 white, a ratio of 3:1Mendel did not recognize that the traits were controlled by genes (term coined in 1909)He did propose that each trait relied on two related, but different, determinants.Alleles represent different forms of a geneNonoverlapping characteristics: heights may vary so he chose the dwarf variety to compare to the tall. Tall simply refers to the nondwarf type.
9 Mendel’s Experiments Mendel’s Experimental Design Phenotype – observable characteristicsAllele for purple flower plants (P) is dominant over white flower plants (p)Genotype - combination of alleles an organism possessesHomozygous purple– PP (Both alleles are the same)Heterozygous purple– Pp (Two alleles are different)Nonoverlapping characteristics: heights may vary so he chose the dwarf variety to compare to the tall. Tall simply refers to the nondwarf type.
10 EXPERIMENT P Generation (true-breeding parents) Purple flowers White FigEXPERIMENTP Generation(true-breedingparents)PurpleflowersWhiteflowersF1 Generation(hybrids)All plants hadpurple flowersF2 Generation705 purple-floweredplants224 white-floweredplants
11 Fig. 14-5-3 P Generation Appearance: Purple flowers White flowers Genetic makeup:PPppGametes:PpF1 GenerationAppearance:Purple flowersGenetic makeup:PpGametes:1/2P1/2pSpermFigure 14.5 Mendel’s law of segregationF2 GenerationPpPPPPpEggspPppp31
12 Segregation Law of Segregation During gamete formation the alleles will separate randomlyFertilization is the fusion of two gametes, reestablishing the two copies of a geneAllele for purple flowersHomologouspair ofchromosomesLocus for flower-color geneAllele for white flowers
13 SegregationLaw of segregation explains Mendel’s experiments with four related concepts:1. Alternate versions of a gene accounts for variations in inherited characters2. For each character, an organism inherits two alleles, one from each parent
14 SegregationLaw of segregation explains Mendel’s experiments with four related concepts:3. If two alleles at a locus differ, then one, the dominant allele, determines the organisms appearance, the other, the recessive allele, has no noticeable effect on the organisms appearance.4. The two alleles for a heritable character separate (segregate) during gamete formation and end up in different gametes.
15 How are each of the 4 parts of Mendel’s law of segregation portrayed in his experiment with crossing peas plants with different flower colors?
16 SegregationMendel’s segregation model accounts for the 3:1 ratio he observed in the F2 generation of his numerous crossesThe possible combinations of sperm and egg can be shown using a Punnett square, a diagram for predicting the results of a genetic cross between individuals of known genetic makeupA capital letter represents a dominant allele, and a lowercase letter represents a recessive allele
17 Segregation Testing the Law of Segregation Along with the 3:1 phenotype ratio, there should also be a 1:2:1 genotypic ratio1 PP, 2 Pp, 1 ppThis can be tested by self-fertilizing the F2 to create an F3The resulting whites should be homozygousThe purple F2 plants should be 1/3 homozygous and 2/3 heterozygousThe homozygous should produce only purple flower plantsThe heterozygous should produce 3 purple flower plants for each 1 white flower plant (3:1)
19 Segregation Testing the Law of Segregation Another way to test is by using a testcross – cross any organism with a homozygous recessiveIf the organism in question is homozygous dominant, then all progeny will have the dominant phenotypeIf the organism is heterozygous, then the progeny will be 50% phenotypically dominant, and 50% phenotypically recessive
20 Practicing a Test Cross Draw punnett square crossing homozygous white flower pea plant with heterozygous purple flower pea plant.What are the possible offspring?Can you determine the genotypes of the offspring?Draw punnett square crossing homozygous white flower pea plant with homozygous purple flower pea plant.
21 TECHNIQUE RESULTS Dominant phenotype, unknown genotype: PP or Pp? Fig. 14-7TECHNIQUEDominant phenotype,unknown genotype:PP or Pp?Recessive phenotype,known genotype:ppPredictionsIf PPIf PporSpermSpermppppPPPpPpPpPpEggsEggsFigure 14.7 The testcrossPpPpPpppppRESULTSorAll offspring purple1/2 offspring purple and1/2 offspring white
22 Independent Assortment Mendel analyzed the inheritance of two different traitsHomozygous round yellow seeds (YY) crossed with homozygous wrinkled(rr) green seedsThe F1 (dihybrid) were all round yellow seedsDihybrid-individuals that are heterozygous for two charactersWhen the F1 was self-fertilized, the resulting F2 had all four combinations of characteristicsThe ratio is very close to 9:3:3:1
23 EXPERIMENT RESULTS Fig. 14-8 P Generation F1 Generation Hypothesis of YYRRyyrrGametesYRyrF1 GenerationYyRrHypothesis ofdependentassortmentHypothesis ofindependentassortmentPredictionsSpermorPredictedoffspring ofF2 generation1/4YR1/4Yr1/4yR1/4yrSperm1/2YR1/2yr1/4YRYYRRYYRrYyRRYyRr1/2YRYYRRYyRr1/4YrEggsYYRrYYrrYyRrYyrrEggsFigure 14.8 Do the alleles for one character assort into gametes dependently or independently of the alleles for a different character?1/2yrYyRryyrr1/4yRYyRRYyRryyRRyyRr3/41/41/4yrPhenotypic ratio 3:1YyRrYyrryyRryyrr9/163/163/161/16Phenotypic ratio 9:3:3:1RESULTS31510810132Phenotypic ratio approximately 9:3:3:1
24 Independent Assortment Punnett squares are used to visualize possible gamete fusionsAssumption: Four types of gametes from each dihybrid parent will be produced in equal numbersWGWgwGwgWWGGWWGgWwGGWwGgWWggWwggwwGGwwGgwwgg
25 Independent Assortment Law of Independent AssortmentAlleles for one gene can segregate independently of alleles for other genesEach phenotypic class is made of several different genotypesExcept the homozygous recessiveThe genotypic ratio is 1:2:1:2:4:2:1:2:1WGWgwGwgWWGGWWGgWwGGWwGgWWggWwggwwGGwwGgwwgg
26 Independent Assortment Testing the Law of Independent AssortmentThis law can be tested by doing a dihybrid testcrossReview: monohybrid heterozygous testcross resulted in a 1:1 phenotypic ratioTestcross WwGg with wwggThe result is a 1:1:1:1 phenotypic ratio
27 Independent Assortment Strictly speaking, this law applies only to genes on different, nonhomologous chromosomesGenes located near each other on the same chromosome tend to be inherited together
29 Laws of ProbabilityMendel’s laws of segregation and independent assortment reflect the rules of probability
30 Probability Types of Probability Probability (P) = Number of times an event is observed (a) / the total number of possible cases (n)P=a/nExamples:Probability of rolling a 4 on a six-sided diceP=1/6Probability of drawing a 7 of clubs from a card deckP=1/52
31 Probability Types of Probability An event that is certain has a probability of 1P=1/1An event that is impossible has a probability of 0An event has a probability of P, the likelihood of the alternative event is Q=1-PProbability of rolling a 4 on a six-sided diceP=1/6Probability of rolling anything elseQ=1-1/6= 5/6The probability of all possible events must equal 1P+Q=1
32 Probability Types of Probability Mutually exclusive outcomes Event in which the occurrence of one possibility excludes all other possibilitiesRolling a dice, only one side can face upIndependent outcomesEvents that do not influence one anotherRolling two dice, the face value of one does not influence the otherTwo rules of probability that affect geneticsSum rule and product rule
33 Probability Sum Rule When events are mutually exclusive The probability that one of several mutually exclusive events will occur is the sum of the probabilitiesProbability of a dice showing either a 4 or 6P=1/6 + 1/6 = 2/6 = 1/3The probability increases as the number of possible outcomes increaseProbability of a dice showing any numberP=1/1Not used for traits expressed on a continuum (human heights)
34 Probability Product Rule When one event is independent of other events The probability that two events will both occur is the product of their separate probabilitiesProbability of throwing a die two times and getting a 4 and then a 6P=1/6 x 1/6 = 1/32The probability will decrease as you increase the number of independent eventsMuch like the lottery, the more numbers you need the less likely you are to win
35 Probability Using Probabilities In a monohybrid cross, Dd X Dd (1:2:1), what is the probability of getting either a homozygous dominant or heterozygous?P= ¼ + ½ = ¾In a dihybrid cross WwGg X WwGg, what is the probability of getting a round and green seed?P=3/4 x 1/4 = 3/16
36 Solving Complex Genetics Problems We can apply the rules of probabilityTo predict the outcome of crosses involving multiple charactersA dihybrid or other multicharacter crossIs equivalent to two or more independent monohybrid crosses occurring simultaneouslyIn calculating the chances for various genotypes from such crossesEach character first is considered separately and then the individual probabilities are multiplied together
37 Probability Branch-Line Approach to Calculate Probabilities Punnett squares require 16 squares for a dihybrid cross, 64 squares for a trihybridPunnett squares are useful, but hard to use with more complex crossesThe branch-line approach is based on the law of independent assortmentIn branch-line each trait is examined independently
38 Probability Branch-Line Approach to Calculate Probabilities AaBb x AaBbTo determine the probability of this dihybdrid cross calculate the probability of each trait and apply the product ruleA phenotype 3/4 (either AA or Aa)B phenotype ¾ (either BB or Bb)AB phenotype 9/16 (A-B- genotype)b phenotype 1/4 (bb genotype)Ab phenotype 3/16 (A-bb genotype)Here you can see the 9:3:3:1 ratioa phenotype 1/4 (aa genotype)B phenotype ¾ (either BB or Bb)aB phenotype 3/16 (aaB- genotype)b phenotype 1/4 (bb genotype)ab phenotype 1/16 (aabb genotype)
39 Probability Branch-Line Approach to Calculate Probabilities Use branch-line to determine the probabilities of this trihybrid cross: AaBbCc x AabbCcA phenotype 3/4a phenotype 1/4B phenotype 1/2b phenotype 1/2C phenotype 3/4c phenotype 1/4ABC phenotype 9/32ABc phenotype 3/32AbC phenotype 9/32Abc phenotype 3/32aBC phenotype 3/32aBc phenotype 1/32abC phenotype 3/32abc phenotype 1/32The key is to look at each cross and what the probabilities are going to be. Aa x Aa has ¾ probability to express the A allele and ¼ the a allele. Then look at the next cross, Bb x bb. This yields a ½ B and ½ b phenotype expression. Do the same with the Cc cross.
40 Probability Branch-Line Approach to Calculate Probabilities The branch-line approach can also be used to determine genotypesThe key is to look at each cross and what the probabilities are going to be. Aa x Aa has ¾ probability to express the A allele and ¼ the a allele. Then look at the next cross, Bb x bb. This yields a ½ B and ½ b phenotype expression. Do the same with the Cc cross.
41 StatisticsWhen dealing with probabilistic events, there is a chance the data will cause us to support a bad hypothesisMendel’s F2 heterozygous pea plants yielded 788 tall and 277 dwarf.2.84:1 not exactly 3:1Is 2.84:1 close enough to represent 3:1?
42 Statistics Hypothesis Testing Statistics are used by scientists to summarize data and test their hypothesis by comparing data predicted resultsTo determine if the data is consistent with the hypothesis we generate a null hypothesisThe null hypothesis assumes the difference between the observed and expected results are due to chance
43 Statistics Chi-Square (O – E)2 χ2=∑ E This test is used when data is distributed among discrete categories (tall and dwarf plants)Formula for the chi-squareχ is the Greek letter chi, O is the observed number for a category, E is the expected number for the category, and ∑ means to sum the calculations for all categoriesχ2=∑(O – E)2E
44 Statistics Chi-Square (787 – 798)2 = 0.15 798 (277 – 266)2 = 0.45 266 Example: Mendel observed 787 tall plants and 277 dwarf. Total of The expected results for a 3:1 ratio are 798 tall and 266 dwarf.Start with figuring the chi-square for tall plantsThen figure it for the dwarf plantsNow sum the two results(787 – 798)2798= 0.15(277 – 266)2266= 0.45= 0.60
45 Statistics Chi-Square (787 – 532)2 = 122.23 532 (277 – 532)2 = 122.23 Example: Use the same data, but assume we were testing for a 1:1 ratio. How does the chi-square change? Mendel observed 787 tall plants and 277 dwarf. Total of 1064.Start with figuring the chi-square for tall plantsThen figure it for the dwarf plantsNow sum the two results(787 – 532)2532=(277 – 532)2532==
46 Statistics Chi-Square To properly use chi-square values we need to convert them to a probability value (p)As the number of categories increases so will the chi-square value (because we sum each category)To help solve this we use degrees of freedom (# of categories minus 1)Example: two categories would have 1 degree 0f freedomConsult the chi-square table
48 Statistics Chi-Square Table We will look a p value of 0.05 We have only 1 degree of freedomProbabilitiesDegrees of freedom0.990.950.800.500.200.050.0110.0000.0040.0640.4551.6423.8416.63520.0200.1030.4461.3863.2195.9919.21030.1150.3521.0052.3664.6427.81511.34540.2970.7111.6493.3575.9899.48813.27750.5541.1452.3434.3517.28911.07015.08660.8721.6353.0705.3488.55812.59216.812
49 Statistics Chi-Square Table There is a 0.05 probability of getting a χ2 value of or larger by chance alone, given the hypothesis is correctProbabilitiesDegrees of freedom0.990.950.800.500.200.050.0110.0000.0040.0640.4551.6423.8416.63520.0200.1030.4461.3863.2195.9919.21030.1150.3521.0052.3664.6427.81511.34540.2970.7111.6493.3575.9899.48813.27750.5541.1452.3434.3517.28911.07015.08660.8721.6353.0705.3488.55812.59216.812
51 Inheritance PatternsInheritance patterns are often more complex than predicted by simple Mendelian geneticsThe relationship between genotype and phenotype is rarely simple
52 Extending Mendelian Genetics for a Single Gene The inheritance of characters by a single gene may deviate from simple Mendelian patternsThe Spectrum of DominanceComplete dominanceOccurs when the phenotypes of the heterozygote and dominant homozygote are identicalCodominanceTwo dominant alleles affect the phenotype in separate, distinguishable waysEx. The human blood group MN is an example of codominance
53 The Spectrum of Dominance Incomplete dominanceThe phenotype of F1 hybrids is somewhere between the phenotypes of the two parental varietiesP GenerationF1 GenerationF2 GenerationRedCRCRGametesCRCWWhiteCWCWPinkCRCWSpermCw1⁄2EggsCR CRCR CWCW CW
54 The Relation Between Dominance and Phenotype Dominant and recessive alleles do not really interact; it is in the pathway from genotype to phenotype that dominance and recessiveness come into playLead to synthesis of different proteins that produce a phenotypeExamples: Mendel’s Pea Shape, Tay-Sachs diseaseDominant alleles are not always the most common in a populationExample: polydactyly (+5 digits)
55 Multiple Alleles Most genes exist in populations Table 14.2Most genes exist in populationsIn more than two allelic formsThe ABO blood group in humansIs determined by multiple alleles
56 Blood TypesGlycoproteins on surface determine blood type; Important in transfusions/transplants IA and IB are codominant ii (type O) is recessive to A or B Type O = universal donor Type AB= universal recipient Differences in Rh factor (Mom Rh- and baby Rh+) can result in erythroblastosis fetalis
57 Pleiotropy Pleiotropy-a gene has multiple phenotypic effects Ex. Hereditary diseases such as cystic fibrosis & sickle cell disease
58 Extending Mendelian Genetics for Two or More Genes Some traits may be determined by two or more genesepistasis- a gene at one locus alters the phenotypic expression of a gene at a second locuspolygenic inheritance- an additive effect of two or more genes on a single phenotype
60 An example of epistasis BCbCBcbc1⁄4BBCcBbCcBBccBbccbbccbbCcBbCCbbCCBBCC9⁄163⁄164⁄16SpermEggsEX: Coat color in miceB = Black b = brownC = color deposited in coatc = color NOT depositedcc-mouse looks white even though it has color genes
61 Polygenic Inheritance Many human charactersVary in the population along a continuum and are called quantitative charactersQuantitative variation usually indicates polygenic inheritanceAn additive effect of two or more genes on a single phenotype
62 POLYGENIC traits are recognizable by their expression as a gradation of small differences (a continuous variation).The results form a bell shaped curve.
63 Nature and Nurture: The Environmental Impact on Phenotype Another departure from simple Mendelian genetics arisesWhen the phenotype for a character depends on environment as well as on genotypeThe norm of reactionIs the phenotypic range of a particular genotype that is influenced by the environment
64 Environment influences Phenotype “Nature vs Nurture” Siamese cats and Himalayan rabbits have dark colored fur on their extremitiesAllele that controls pigment production is only able to function at the lower temperatures of those extremities.
65 Integrating a Mendelian View of Heredity and Variation Multifactorial charactersAre those that are influenced by both genetic and environmental factorsAn organism’s phenotypeIncludes its physical appearance, internal anatomy, physiology, and behaviorReflects its overall genotype and unique environmental historyEven in more complex inheritance patternsMendel’s fundamental laws of segregation and independent assortment still apply
72 recessive to the normal working allele. A mutation in an allele that causes a protein to be NON-FUNCTIONAL would appearrecessive to the normal working allele.Examples of autosomal recessive GENETIC DISORDERS:Phenylketonuria (PKU)Tay-Sachs DiseaseCystic Fibrosis
73 Phenylketonuria (PKU) CAUSE: Mutation in gene for an enzyme the breaks down an amino acid called phenylalanineBuild up causes mental retardation
74 Phenylketonuria (PKU) ALL babies are tested for PKU before they leave the hospital.Treatment: Need a diet low in phenylalanine to extend life and prevent mental retardation
76 Carrier Heterozygous individual That carries one recessive allele for agenetic disorderDoesn’t show thedisorder themselves,but can pass it on tooffspring
77 TAY-SACHS DISEASE Autosomal Recessive CAUSE: Mutation in gene for an enzyme the breaks down a kind of lipid in the developing brainAs these lipids build up in brain infant suffers seizures, blindness, loss of motor & mental function > > > leads to early death.Found more frequently in people with Jewish, Mediterranean, or Middle Eastern ancestry
78 Disorders caused by autosomal codominant alleles: SICKLE CELL DISEASE CAUSE: A changed to T in gene forHemoglobin(protein in red blood cellsthat carries oxygen in blood)
79 SICKLE CELL DISEASE SYMPTOMS: Red blood cells become sickle shaped under low oxygen condition in persons with two sickle cell alleles (ss)Ss=Sickle cell trait Normally healthy, but can suffer some sickle cell episodes
80 Cells stick in capillaries Loss of blood cells (anemia) SICKLE CELL DISEASECirculatory problemsCells stick in capillariesLoss of blood cells (anemia)Organ damage (brain, heart, spleen)Can lead to DEATH
81 HUNTINGTON’S DISEASE is AUTOSOMAL DOMINANT CAUSE: Extra CAG repeats at end of gene on chromosome 4The more repeats the more severe the symptoms.
82 HUNTINGTON’S DISEASE Begins in middle age Huntington’s brainBegins in middle ageCauses progressive loss of muscle control and mental function1 in 10,000 people in U.S.have Huntington’s diseaseNormal brain
83 A person withHuntington’s diseasehas a _____ chance ofpassing the disorder on totheir offspring.50%Problem: Symptoms of disorder usually don’t show until ____________ . . .so you don’t know you have it until ________ you have had children.MIDDLE AGEAFTER
84 ACHONDROPLASIA (One kind of Dwarfism) CAUSE: Autosomal Dominant gene1 in 25,000 birthsDD = lethalDd = dwarf phenotypedd= = normal height200,000 “little people” worldwideOne of oldest known disorders – seen in Egyptian art
85 ACHONDROPLASIA (One kind of Dwarfism) Normal size head and torso; short arms and legsProblem with way cartilage changes to bone as bones grow
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