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1 Honors Biology chapter 23 Genetics John ReganCopyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
2 23.1 Mendel’s laws Heredity - Biological inheritance Passing of traits from parents to offspringGenetics- Study of heredityGregor MendelPublished 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 parentLaw of DominanceOne form of a factor may be dominant over an alternative formDominant (capital letter)Prevents the expression of the other form –recessive (small letter)
3 Alleles- alternate forms of a gene for the same trait Mendel’s laws cont’d.Alleles- alternate forms of a gene for the same traitCapital letter- represents the dominant alleleSmall letter- represents the recessive alleleDominant-a certain trait will result if the individual has at least 1 dominant alleleRecessive- for a recessive trait to result the individual must have 2 copies of the recessive alleleGenotype- genetic composition of an individual with regard to a specific trait2 copies of the dominant allele- homozygous dominant1 copy of the dominant allele and 1 of the recessive- heterozygous2 copies of the recessive allele- homozygous recessivePhenotype- physical characteristics of a trait- outward appearance of a trait
4 Mendel’s laws 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 The Inheritance of Traits The parent generation is also known as the P generation.The offspring of this P cross are called the first filial (F1) generation.The second filial (F2) generation is the offspring from the F1 cross.
6 Mendel’s laws cont’d. What we know now Genes are sections of chromosomesChromosomes come in pairs called homologous pairsHomologous pairs have genes controlling the same traitsGenes are located at the same point or locus, on each member of the pairInformation contained within the homologous genes is not necessarily the same (ex: blues vs. brown eyes)Alternative forms of a gene for a trait are called alleles
8 Mendel’s laws cont’d. The inheritance of a single trait cont’d. Phenotype- physical appearance of the individual with regard to a traitHomozygous dominant individual and heterozygous individual will have the same genotype in this instanceHomozygous recessive individual will have a different phenotypeExample- inheritance of a widow’s peak vs. a straight hairline in peopleAlternative forms of alleles for hairline shapeWidow’s peak is dominant to straightW=allele for widow’s peakw= allele for straight hairline
10 Mendel’s laws cont’d. Inheritance of a single trait cont’d. Gamete formationDuring meiosis, homologous chromosomes separate so there is only 1 member of each pair in a gameteThere is one allele for each trait, such as hairline, in each gameteNo two letters in a gamete can be the same letter of the alphabetIf genotype is Ww, then gametes from this individual will contain either a W or a wIf the genotype is WwLl (looking at 2 traits), gametes can contain any of the following combinationsWL, Wl, wL, or wl
11 Mendel’s laws cont’d. Practice problems For each of the following genotypes, give all possible gametesWWWWSsTtTtggAaBbFor each of the following, state whether the genotype or a gamete is representedDLlPwLlGg
13 Mendel’s laws cont’d. One trait crosses-monohybrid cross Let’s consider a specific crossIf a homozygous woman with a widow’s peak reproduces with a man with a straight hairline, what kind of hairline will their children have?Use W= widow’s peak, w=straight hairlinefigure out the genotype of each parentWoman has a widow’s peak and we are told she is homozygous so she is WWMan has a straight hairline so he must be wwDetermine their gametes: her eggs will all have W and his sperm will all have wPut together all possible combinations: all offspring will receive a W from her and a w from him so all Ww
14 Mendel’s laws cont’d. One-trait cross cont’d. In this example there was only 1 combination of eggs and sperm possibleWhat if the parents were each Ww?Then sperm could have either W or w, and eggs also could have W or wOne way to figure out the possible combinations of eggs and sperm that could occur is to use a Punnett squareThis is illustrated on the following slide
15 A punnett square is an easy way to figure out all possible combinations of eggs and sperm
16 Mendel’s laws cont’d. One-trait cross cont’d. After the genotypes and phenotypes of offspring are determined, we can determine the ratiosIn our previous example, in the Punnett square we had the following offspring: WW, Ww, Ww, and wwThe genotypic ratio is 1 WW:2 Ww:1 wwThe phenotypic ratio is 3 individuals with a widow’s peak to 1 individual with a straight hairlineAnother way to phrase the phenotypic ratio is in terms of probabilityThis couple has a 75% chance of producing a child with a widow’s peak and a 25% chance of producing a child with a straight hairlineThe probability will be the same for each pregnancy between this couple
17 Mendel’s laws cont’d. One-trait crosses and probability Product rule of probabilityThe chance of 2 or more independent events occurring together is the product of their chance of occurring separatelyIn the cross Ww X Ww, what is the chance of obtaining either a W or a w from a parent?Chance of W = ½ and the chance of w = ½Therefore the probability of having these genotypes is as followsChance of WW= ½ X ½ = ¼Chance of Ww = ½ X ½ = ¼Chance of wW= ½ X ½ = ¼Chance of ww = ½ X ½ = ¼
18 Mendel’s laws cont’d. One-trait crosses and probability cont’d. Sum rule of probability-the chance of an event that can occur in more than one way is the sum of the individual chancesTo calculate the chance of an offspring having a widow’s peak, add the chances of WW, Ww, or wW from the preceding slide¼ + ¼ + ¼ = ¾ or 75%
19 Mendel’s laws cont’d. The one-trait test cross You cannot distinguish between a homozygous dominant individual and a heterozygous individual just by lookingThey are phenotypically the sameBreeders of plants and animals may do a test cross to determine the likely genotype of an individual with the dominant phenotypeCross with a recessive individual-has a known genotypeIf there are any offspring produced with the recessive phenotype, then the dominant parent must be heterozygous
21 Mendel’s laws cont’d. Practice problems Both 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.
22 Mendel’s laws cont’d. The inheritance of two traits Mendel reasoned from the results of his pea plant crosses that each pair of factors assorts independently into gametesIf one pair of homologous chromosomes has a W on one and w on the other, and a second pair has an R on one and an r on the other, when gametes form all possible combinations of these letters are possible: WR, Wr, wR, and wrCalled the Law of Independent Assortment
23 Segregation and independent assortment Fig. 23.6
24 Mendel’s laws cont’d. The law of independent assortment states : Each pairs of factors assorts independently (without regard to how the others separate)All possible combinations of factors can occur in the gametes
25 Mendel’s laws cont’d. Two-trait crosses-dihybrid cross Let’s consider two traits, hairline and finger lengthWe’ll use the W and w as before, and now S=short fingers and s=long fingersA person who is WwSs widow’s peak and short fingers and a person who is also WwSs have childrenFigure the gametes for each parentWS, Ws, wS, and ws for bothMake a big Punnett square as illustrated on the following slideFigure the genotypes of the offspringPhenotypic ratio: 9 widow’s peak, short fingers: 3 widow’s peak, long fingers; 3 straight hairline, short fingers; 1 straight hairline, long fingers
27 Mendel’s laws cont’d. Two-trait crosses and probability Probability lawsProbability of widow’s peak = ¾Probability of short fingers= ¾Probability of straight hairline= ¼Probability of long fingers= ¼Using the product ruleProbability of widow’s peak and short fingers = ¾ X ¾ = 9/16Probability of widow’s peak and long fingers = ¾ X ¼ = 3/16Probability of straight hairline and short fingers = ¼ X ¾ = 3/16Probability of straight hairline and long fingers = ¼ X ¼ = 1/16
28 Mendel’s laws cont’d. The two-trait testcross Cross an individual with the dominant phenotype for each trait with an individual with the recessive phenotype of both traitsIf the dominant individual is homozygous for both traits, all offspring will show the dominant phenotypesIf the dominant individual is heterozygous, then a Punnett square shows the following phenotypic ratio of offspring1 dominant for both: 1 dominant for first trait recessive for second: 1 recessive for first trait dominant for second: and 1 recessive for both
30 Mendel’s laws Practice problems Attached 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) and a woman who does not. What are the genotypes of all persons concerned?
31 23.2 Beyond simple inheritance patterns Polygenic inheritanceControlled by 2 or more sets of allelesEach dominant allele codes for a product and effects are additiveResult 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
33 Beyond simple inheritance patterns cont’d. Polygenic inheritance cont’d.Skin colorControlled by many gene pairs and many allelesAssuming the simplest model, 2 alleles at lociA and BWhen an AaBb person has children with another AaBb person, children can range from very light to very darkAABBVery darkAABb or AaBBDarkAaBb, AAbb, aaBBMedium brownAabb or aaBbLightaabbVery light
34 Beyond simple inheritance patterns cont’d. Environmental influencesEnvironment can influence gene expression and therefore phenotypeEx: sunlight exposure on skin-suntan;coat color in Himalayan rabbits only make black pigment where body temp. is colder- extremitiesHuman 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
36 Beyond simple inheritance patterns cont’d. Incomplete dominance and codominanceIncomplete dominanceHeterozygous individual has a phenotype intermediate to the two homozygous individualsEx: 4 o’clock flowersred incompletely dominant with white; heterozygote is pinkEx: 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 straightCodominanceOccurs when both alleles are equally expressedEx: horses:chestnut codominant w/cream; heterozygote is palominoEx: cattle: red codominant w/ white; heterozygote is roanEx: type AB blood has both A antigens and B antigens on red blood cells
38 Beyond simple inheritance patterns cont’d. Multiple allele inheritanceThe gene exists in several allelic forms, but each person still has only 2 of the possible allelesABO blood typesIA = A antigens on RBCsIB = B antigens on RBCsi = has neither A nor B antigens on RBCsBoth IA and IB are dominant over I, IA and IB are codominantPhenotype GenotypeA IAIA or IAiB IBIB or IBiAB IAIBO ii
39 Beyond simple inheritance patterns cont’d. ABO blood types cont’d.Paternity testing- ABO blood groups often usedCan 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
41 Beyond simple inheritance patterns cont’d. 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
42 23.3 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
43 Sex-linked inheritance cont’d. X-linked traitsRed-green colorblindness is X-linkedThe X chromosome has genes for normal color visionXB = normal visionXb – colorblindnessGenotypes PhenotypesXBXB female with normal color visionXBXb carrier female with normal color visionXbXb colorblind femaleXBY male with normal color visionXbY colorblind male
45 Sex-linked inheritance-practice problems Both 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?
46 23.4 Inheritance of linked genes A single chromosome has many genesThe sequence 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 suspectedOr, if very few recombined phenotypes appear in offspring, linkage is also suspectedRecall that during meiosis crossing over sometimes occursIf crossing over occurs between 2 alleles of interest, then 4 types of gametes are formed instead of 2Fig on the following slide illustrates linkage
48 Inheritance of linked genes cont’d. Ocurrence of crossing over can indicate the sequence of genes on a chromosomeCrossing over occurs more frequently between distant genes than between those close together on a chromosomePractice 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?
49 Chapter 23 ObjectivesKnow vocabulary terms like allele, homozygous, heterozygous, genotype, phenotype.Know Mendel’s laws.Be able to solve a 1-trait genetics problem for simple dominance using a Punnett square and use it to determine chances of a trait appearing in the offspring.Be able to determine genotype of members of a family from phenotypic information.Be able to solve incomplete dominance and codominace problems.Be able to solve a 2-trait genetics problem.Know that the phenotypic ratio for a dihybrid cross is 9:3:3:1.Be able to solve ABO blood type problems.Know how multiple alleles and polygenic traits affect inheritance.Know that environmental factors can affect phenotypes.Be able to solve problems for sex-linked traits.
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