2How traits are inherited: The debate in the 1800’s The Blended Hypothesis: the traits from 2 parents are mixed to become a third trait.The Particulate Hypothesis: the traits from 2 parents are joined but remain discrete, and can be separated again to their original forms.Which hypothesis seems more logical to you?-Write your response down in your notes & giveone supporting argument sentence.
3What does “true-breeding” mean? In order to test these 2 hypotheses, a scientist must have “true breeding parents”What does “true-breeding” mean?The parent organism produces only one type of offspring for a particular characteristic
4The blended hypothesis Parental generationF1 generationTrue breeding black coffeeTrue breeding white creamThe traits from two parents are mixed to become a third trait.
5F1 generation F2 generation What type of offspring can the parents in the testcross make?Only mocha colored; one phenotypeWhat type of offspring can the F1 generation create?Mocha-colored offspring; only one phenotypeCan the 2 traits for the characteristic of color be separated out again in this example?No
6Do we see examples of this type of character blending in real life?
7The particulate hypothesis True breeding yellowTrue breeding redParental generationF1 generationThe traits from the two parents are joined but remain discrete, and can be separated again to their original forms.
8F1 generation What type of offspring can the parental generation make? Only two-colored offspring; one phenotype possibleWhat type of offspring can the F1 generation create?Yellow, red, or a mixture of red and yellow; 3 phenotypes are possibleCan the 2 traits for color be separated out again in this example?Yes
9Do we see examples of this type of character mixing in real life?
10Which hypothesis, do you think, is more likely to be accurate? The particulate hypothesis could give just one type of offspring, which could account for why we see some blended traits, but with the blended hypothesis, traits could never separate out to yield variation in subsequent generations, as with discrete traits.
11PREDICTING PROBABILITY OF POTENTIAL OFFSPRING MONOHYBRID CROSSESPREDICTING PROBABILITY OF POTENTIAL OFFSPRING
12All organisms pass on inherited information using haploid gametes. Chromosomes are discrete packages of genetic material that can be traced back to the two parents that produced the zygote
13The marbles in this example represent certain traits or alleles The marbles in this example represent certain traits or alleles. Where are these alleles located (in us)?They are located at specific places on paired chromosomes, calledgenes.One chromosome in a pair has Mom’s allele at a gene locus & theother chromosome in the pair has Dad’s allele at the same gene locus.Because the alleles come from different parents, they may not be thesame.
14The zygote grows by mitosis to become a multi-cellular organism How do inherited alleles from your parents show themselves in physical traits all over your body?Because DNA replicates before cell division, the alleles from each parent are passed on to every cell that makes up your body and thus the chromosomes can express any trait that is needed in any part of the body.
15How do these inherited traits get passed on to your offspring? One of these cells with all of Dad’s chromosomes and all of Mom’s chromosomes will become a gamete-producing cell in the gonads and begin making sperm and eggs with half the number of required chromosomes.
16Which chromosomes will your children get- the ones from your dad or the ones from your mom? They will get a random mixture of both as a consequence of crossing-over and the law of segregation.
17Do you know the following terms? GenotypePhenotypeDominantRecessiveTest crossThe alleles of a given gene locus; genetic make-upThe expression of gene locus; observable traitAn allele that is fully expressed in a heterozygoteAn allele that isn’t expressed at all in a heterozygoteBreed an organism of unknown genotype with a recessive homozygote
18Crossing pea plants for the color of flower Name the phenotypes of the flowers in the P generationName the genotype of the flower in the F1 generationWhich of the traits are dominant and which are recessive?
19What are the possible gametes of parents of each of the following genotypes? AAAaaaA or AA or aa or a
21Let’s work out a problem with dogs… Heterozygous parents for black hair are crossed and their progeny are then test-crossed. Determine the expected genotypic and phenotypic ratios among the test cross progeny.BbParental generation: Bb x Bb.BBBbbbBTest cross of the F1 is to a white haired dog (bb).b¼ of the F1 will be black haired homozygous dominant. When crossed with bb the result will be all black (Bb).½ of the F1 will be black haired heterozygous (Bb). When crossed with bb the result will be ½ black and ½ white.¼ of the F1 will be white haired (bb). When crossed with bb all offspring will be white hairedSo… ½ Bb and ½ bb ratio= 1:1 for phenotype and 0:1:1 for genotype
24ProbabilityIf a coin is flipped once, what is the probability it will land on heads?If it lands on heads, what is the probability it will land on heads again?If I keep flipping the coin and it keeps coming up heads, will that affect the probability of the outcome of my next flip?
25What is the probability that I will get heads on 2 coin tosses in a row? This changes the probability because you are now asking for the probability of flipping heads and flipping heads again.When you have 2 separate probabilities that both must occur, the 2 separate probabilities must be multiplied to find the overall probability.So the probability of heads and heads occurring is ½ x ½ = ¼
26The rule of multiplication Used whenever there is a statement of one probability “and” another probabilityLet’s practice:What is the probability that a couple would have 3 boys?What is the probability that parents with the genotypes Aa and Aa would have a child with the genotype AA?½ x ½ x ½ = 1/8Probability of dad giving “A” is ½ and the same can be said for mom then… ½ x ½ = ¼
27Let’s try another!!!Black wool in sheep is recessive (w) to white wool which is dominant (W). A white ram is crossed with a white ewe. Both parents carry the allele for black (w).They produce a white ram lamb which is then back crossed to the female parent.Determine the probability of the back cross offspring being black.The white ram lamb (F1) is either WW or Ww.1/3 of the white offspring (WW) crossed with parent Ww will result in zero black lambsWwWWWwwwW2/3 of the white offspring (Ww) crossed with parent Ww will result in ¼ black lambs.wSo… 1/4 x 2/3 = 1/6 or 0.16
28Let’s use a deck of cards (52 cards) If the probability of getting dealt an ace from a deck of cards is 4 out of 52, what is the probability of being dealt an ace or a king?The probability of getting an ace is 4/52 and the probability of getting a king is 4/52. You must add the probabilities together to get the total probability.4/52 + 4/52 = 8/52
29Rule of additionUsed whenever there is a statement of one probability “or” another probability.Let’s practice:What is the probability of being dealt an ace or a face card from one deck of cards?The probability of getting an ace is 4/52 & the probability of getting a face card is 12/52So… 4/ /52 = 16/52 = 4/13
31IIGG x iigg IiGg “IG” “Ig” “iG” and “ig” I = inflated i= constricted G= green g= yellow How many traits or genes are represented by these genotypes? When gametes are made in meiosis, how many alleles does each gamete get of a single gene? What gametes can the 1st parent make? What gametes can the 2nd parent make? What type(s) of offspring will be produced? If the F1 generation was allowed to cross, what types of gametes would result?2 – pod texture & pod colorOne allele of each geneIGigIiGg= F1 generation“IG” “Ig” “iG” and “ig”
32If both parents are heterozygous (IiGg x IiGg) How can we predict the outcome?Creating a dihybrid cross!!!
33Recall the law of segregation: Do both dominant alleles have to go to the same gamete?Can a gamete have a mixture of dominant and recessive alleles?Do you have a mixture of dominant and recessive traits from your parents?How is it that each gene can segregate independently when they are all on a limited number of chromosomes?NoYesYesCrossing over during prophase 1 allows alleles to mix randomly as the gene loci are not located too near one another on the chromosome
35How does a genotype result in a phenotype? The genotype is an abbreviation for the allele found at a specific gene.The gene is a sequence of DNA that codes for the “recipe” to make a particular protein.For example” melanin gene locus can have different sequences of DNA (different alleles) that either make melanin – a dark skin pigment that absorbs UV or don’t.
36If the genotype is “MM” then the person has 2 alleles, or strands of DNA, that make the protein melanin.If the genotype is “Mm” then they have one allele that makes the protein & one that doesn’tWhich genotype would make the darker skin?
37THIS IS AN EXAMPLE OF INCOMPLETE DOMINANCE Let’s look at Snap DragonsCR CRCW CWWhat color would a snap dragon be with a genotype of CR CW ?THIS IS AN EXAMPLE OF INCOMPLETE DOMINANCE
38Gene expressionAlleles do not have to be completely expressed or unexpressed.An allele can make small amounts of a protein in its recessive form and greater amounts in its dominant form.There is a wide range of expression for any allele termed dominant or recessive.
39CODOMINANCE- neither of the 2 alleles of the same gene totally masks the other. The result is a combination of both dominant traits.
40CODOMINANCE RR x WW All offspring will be …..? RW RW RED & WHITE !! RW R = red colored coatW= white colored coatRR x WWRRAll offspring will be …..?RWRWWRED & WHITE !!WRWRW
41Blood Types There are 6 genotypes. They make up 4 phenotypes (blood types).A and B are codominant, and O is recessive.GenotypePhenotype(Blood Type)IAIA or AAIAi or AOIBIB or BBIBi or BOIAIB or ABii or OOAABBABO
42What are the possible blood types of the potential children of an AB (IAIB) male and an B (IB i) female?What % chance will the offspring be type B?Hint: use a punnett square
43IAIB x IBi IA IB TYPES: A, B, or AB IAIB IBIB IAi IBi IB i 50% What are the possible blood types of the potential children of an AB (IAIB) male and an B (IB i) female?IAIB x IBiIAIBTYPES: A, B, or ABIAIBIBIBIAiIBiIBWhat % chance will the offspring be type B?i50%
44PEDIGREESCollecting information about a family ‘s history for a particular trait and creating a family tree that displays how the trait is passed down from generation to generation.
45One type of pedigree shows only the phenotypes of individuals.
46Another type of pedigree chart gives information about all of the individuals’ genotypes for a trait.
47Addams FamilyBbbbbbBbbbBbBbbbB_What is the percent chance the new Addams’ baby will have web feet?
48Using pedigrees to trace inheritance patterns WS When you are done with the WS, get together with your lab groups and come up with 3-4 rules that apply to recessive allele inheritance patterns and 3-4 rules that apply to dominant allele inheritance patterns.
49Possible Rules for Inheritance Patterns RecessiveDominantUnaffected parents can have affected offspringThe phenotype can skip a generationIndividuals with no sign of the allele can be carriersAffected offspring must have at least one affected parentThe phenotype appears in every generation without skippingTwo unaffected parents have no affected offspring
51Chromosomal Basis of Sex Short segments at either end of the Y chromosome are homologous with the same regions on the X chromsomeAllows the X and Y chromosomes to pair up during meiosis.XX= female and XY= maleAnatomical distinction of sex in an embryo occurs at about 2 monthsSRY gene on the Y chromosome is a trigger for testes to form.If not present, ovaries form.
52Hemophilia: A Sex-Linked Trait Hemophilia is an inherited condition in which the blood clots slowly or not at allTwo genes that encode blood-clotting proteins reside on the X chromosomeHemophilia is an X-linked recessive disorderMales develop hemophilia if they inherit one mutant allele from their motherFor females to develop hemophilia, they have to inherit two mutant alleles, one from each parent
53Royal hemophiliaStarted by a mutant allele in Queen Victoria of EnglandThree of her nine children received the defective alleleThey transferred it by marriage to other royal familiesFig. 8.28Queen Victoria
54Fig. 8.28 In all, 10 of Victoria’s male descendants had hemophilia Escaped the disorder
55Sickle-Cell Anemia: Recessive Trait Sickle-cell anemia is an autosomal recessive trait in which the protein hemoglobin is defectiveAffected individuals cannot properly transport oxygen to their tissuesFig. 8.29
56Huntington’s Disease: Dominant Trait Huntington’s disease is an autosomal dominant trait that causes progressive deterioration of brain cellsFig. 8.33It is a fatal diseaseHowever, it persists in human populations because it has a late onset
57Examples of sex linked traits: hemophilia and colorblindness THESE ARE TRAITS LOCATED ON THE SEX CHROMOSOMES.MALES PASS GENES LOCATED ON THE “X” CHROMOSOME TO ALL OF THEIR DAUGHTERS & NONE OF THEIR SONS.WHATEVER MOM HAS ON HER “X” CHROMOSOME WILL BE EXPRESSED IN HER SONS EVEN IF THE TRAIT IS RECESSIVE.Examples of sex linked traits: hemophilia and colorblindness
58SEX LINKED ( X-LINKED) XhXh x XHY Xh Xh XH Y 100% XHXh XhY H= non hemophiliah= hemophiliaXhXh x XHYXhXhXHXhXhYXHYIF THESE PARENTS HAVE A BOY, WHAT ARE HIS CHANCES THAT HE WILL HAVE HEMOPHILIA?100%
59X inactivation in Female Mammals One of the two X chromosomes in each cell is randomly inactivated during embryonic development
60If a female is heterozygous for a particular gene located on the X chromosome She will be a mosaic for that character due to the turning off or on of the gene in a cellTwo cell populationsin adult cat:Active XOrangefurInactive XEarly embryo:X chromosomesAllele forblack furCell divisionand XchromosomeinactivationBlackFigure 15.11Allele for orange fur
62Linked genes tend to be inherited together because they are located near each other on the same chromosomeEach chromosomeHas hundreds or thousands of genesGenes that are close together on the same chromosome are linked and do not assort independentlyUnlinked genes are either on separate chromosomes of are far apart on the same chromosome and assort independently with crossing over.
63The farther apart genes are on a chromosome The more likely they are to be separated during crossing overIf genes are located closer to one another, what happens to their recombination frequency?The percentage of recombination decreases toward 0% as distance between the genes decreases.
64A linkage mapIs the actual map of a chromosome based on recombination frequenciesRecombinationfrequencies9%9.5%17%bcnvgChromosomeThe b–vg recombination frequency is slightly less than the sum of the b–cn and cn–vg frequencies because doublecrossovers are fairly likely to occur between b and vg in matings tracking these two genes. A second crossoverwould “cancel out” the first and thus reduce the observed b–vg recombination frequency.In this example, the observed recombination frequencies between three Drosophila gene pairs(b–cn 9%, cn–vg 9.5%, and b–vg 17%) best fit a linear order in which cn is positioned about halfway betweenthe other two genes:RESULTSA linkage map shows the relative locations of genes along a chromosome.APPLICATIONTECHNIQUEA linkage map is based on the assumption that the probability of a crossover between two genetic loci is proportional to the distance separating the loci. The recombination frequencies used to construct a linkage map for a particular chromosome are obtained from experimental crosses, such as the cross depictedin Figure The distances between genes are expressed as map units (centimorgans), with one map unitequivalent to a 1% recombination frequency. Genes are arranged on the chromosome in the order that best fits the data.Figure 15.7
65If gene A recombines 12% of the time with gene B but 16% of the time with gene C, which gene is closer to gene A ?BIf gene B and C recombine with one another 28% of the time, what is the order of genes A, B, and C along the chromosome?B – A – C; B and A are 12 map units or centimorgans apart; A and C are 16 centimorgans apart; therefore, B and C are the furthest apart, with 28 centimorgans between them.
66Determine the sequence of genes along a chromosome based on the following recombination frequencies: A – B , 8 map unitsA – C , 28 map unitsA – D , 25 map unitsB – C , 20 map unitsB – D , 33 map unitsDABC
67A wild-type fruit fly (heterozygous for gray body color & normal wings) is mated with a black fly with vestigial wings. The offspring have the following phenotypic distribution:wild-typeblack-vestigial 785black-normalgray-vestigialWhat is the recombination frequency between these genes for body color and wing size?Take the total number of recombinant types & divide by the total number of offspring.17%
68A wild-type fruit fly (heterozygous for gray body color & red eyes) is mated with a black fruit fly with purple eyes. The following are their offspring:721 wild-type751 black-purple49 gray-purple45 black-redWhat is the recombination frequency between these genes for body color and eye color?6%
69Alterations of chromosome number or structure cause some genetic disorders Nondisjunction = abnormal chromosome number
70If one of the aberrant gametes unites with a normal one… AneuploidyFertilization of a gamete with a missing chromosome the zygote would be: monosomicIf the chromosome is in triplicate in the zygote: trisomic for that chromosome.Zygotes with more than 2 sets of chromosomes would be considered: polyploidy (fairly common in plant kingdom)TriploidyTetraploidyone extra or missing chromosome disrupts genetic balance more than does an entire set of extra chromosomes
71Chromosome alterations Deletion – chromosomal fragment is lostDuplication – the deleted fragment attaches to a sister chromatid.Inversion – deleted fragment reattaches in original location but in reverse orientation.Translocation – deleted fragment attaches to a nonhomologous chromosomeDuringmeiosis
72Interactions between genes Some genes may not just control a single characteristic or trait in the phenotype of an organism.Most genes probably have an effect on one or more phenotypic traitsPleiotropy
74Interactions between genes In some cases a single characteristic may be controlled by more than one gene. -Polygeny
75We all inherit a set of three Rhesus (Rh) genes from each parent called a haplotype. They are referred to as the c, d, e, C, D and E genes. The upper case letters denote Rh positive genes and the lower case, negative and we inherit either a positive or negative of each gene from each parent (eg. CDe/cde, cdE/cDe etc.). This means that we then possess two of each gene and can pass either to our offspring.If a person is tested Rh positive, their blood is said to contain the Rhesus factor - if they are tested negative it does not. A person possessing one or more positive Rh genes (C, D or E), anywhere in their inherited haplotypes, has inherited the Rh factor (eg. cdE/De, cde/cDe etc.) and they are tested Rh positive - only a person with a genotype of cde/cde is truly Rh negative.
76The baby of an Rh negative woman may inherit the Rh positive factor from his/her father. This would result in the mother and baby having different blood types. During pregnancy, some of the baby's Rh positive red blood cells may enter the mother's circulation. The cells are recognized as being "foreign" by the mother's immune system, and she may produce antibodies. Theseantibodies can be permanent, and are capable of crossing over into the baby's blood and break down his/her Rh positive red blood cells; they will not harm the mother. Antibodies are usually produced too late in the first pregnancy to affect the baby being carried. Future babies are at risk since the antibodies are already present when pregnancy occurs.
77EPISTASISINTERACTION BETWEEN THE PRODUCTS OF TWO GENES IN WHICH ONE OF THE GENES MODIFIES THE PHENOTYPIC EXPRESSION PRODUCED BY THE OTHER.EX: COAT COLOR FOR LABRADOR RETRIEVERS.
78The E gene determines if dark pigment will be deposited in the fur or not. If the dog has ee there is no pigment & dog will be yellow.The B gene determines how dark the pigment will be.In yellow labs the B gene indicates the color on their nose, lips, & eye rims.
81SEX INFLUENCED TRAITSTHE PRESENCE OF MALE OR FEMALE HORMONES INFLUENCES THE EXPRESSION OF CERTAIN TRAITS.EX: PATTERN BALDNESSIF FEMALE IS HETEROZYGOUS SHE WILL NOT BE BALD.-the gene is recessive in femalesIF A MALE IS HETEROZYGOUS HE WILL BE BALD.-the gene is dominant in males
82ENVIRONMENTAL EFFECTS Many alleles are expressed depending on the environment.Some are heat sensitiveEx:Arctic foxes make fur pigment only when the weather is warm.
83CAN YOU SEE WHY THIS TRAIT WOULD BE AN ADVANTAGE?
84Inheritance of Organelle Genes Mitochondria and chloroplasts have small circular DNA (aka extranuclear genes)Nonnuclear inheritance
85Chloroplasts & mitochondria are randomly assorted to gametes & daughter cells Traits determined by them do not follow simple Mendelian rules.Does not distribute genes to offspring through the same process as nuclear DNA by way of meiosis.In animals, mitochondrial DNA is transmitted by the egg & not the spermMitochondrial determined traits are maternally inherited.Help make up protein complexes of the ETCATP synthase
86Most susceptible to energy deprivation Nervous system and muscles.Mitochondrial myopathy = weakness, intolerance of exercise, and muscle deteriorationLeber’s hereditary optic neuropathy= produces sudden blindness in people as young as 20.
87Chi Square Testa statistical test commonly used to compare observed data with data we would expect to obtain according to a specific hypothesis
88An ExampleIf, according to Mendel's laws, you expected 10 of 20 offspring from a cross to be male and the actual observed number was 8 males, then you might want to know about the "goodness to fit" between the observed and expected.
89The Question is…Were the deviations (differences between observed and expected) the result of chance, or were they due to other factors?How much deviation can occur before you, the investigator, must conclude that something other than chance is at work, causing the observed to differ from the expected?
90The answer is…The chi-square test is always testing what scientists call the null hypothesis, which states that there is no significant difference between the expected and observed result.The formula for calculating chi square is:X2 = ƩThat is, chi-square is the sum of the squared difference between observed (o) and the expected (e) data (or the deviation, d), divided by the expected data in all possible categories.(o-e)2e
91M & M CHI SQUARED ANALYSIS!!! It’s time for …M & M CHI SQUARED ANALYSIS!!!Are the numbers of different colors of M & M’s in a package really different from one package to the next? Or does the Mars Company do something to ensure that each package gets a certain number of each M & M color?
92An example…Suppose that a cross between two pea plants yields a population of 880 plants, 639 with green seeds and 241 with yellow seeds. You are asked to propose the genotypes of the parents.Your hypothesis is that the allele for green is dominant to the allele for yellow and that the parent plants were both heterozygous for this trait.
93If your hypothesis is true, then the predicted ratio of offspring from this cross would be 3:1 (based on Mendel's laws) as predicted from the results of the Punnett squarePredicted offspring from cross between green and yellow-seeded plants. Green (G) is dominant (3/4 green; 1/4 yellow).
94To calculate X2 , first determine the number expected in each category To calculate X2 , first determine the number expected in each category. If the ratio is 3:1 and the total number of observed individuals is 880, then the expected numerical values should be 660 green and 220 yellow.X2 = Ʃ(o-e)2e
95Calculating Chi-Square GreenYellowObserved (o)639241Expected (e)660220Deviation (o - e)-2121Deviation2 (d2)(o-e)2441d2/e0.6682Ʃ (d2/e) = X2 Your final answer should be:2.668
96So what does this answer mean? Here's how to interpret the X2 value:Determine degrees of freedom (df). Degrees of freedom can be calculated as the number of categories in the problem minus 1. In our example, there are two categories (green and yellow); therefore, there is I degree of freedom.
97Determine a relative standard to serve as the basis for accepting or rejecting the hypothesis. The relative standard commonly used in biological research is p > 0.05.The p value is the probability that the deviation of the observed from that expected is due to chance alone (no other forces acting).In this case, using p > 0.05, you would expect any deviation to be due to chance alone 5% of the time or less.
98Using the appropriate degrees of 'freedom, locate the value closest to your calculated chi-square in the table.Determine the closest p (probability) value associated with your chi-square and degrees of freedom.
99Chi-Square Distribution Degrees ofFreedom(df)Probability (p)0.950.900.800.700.500.300.200.100.050.010.00110.0040.020.060.150.461.071.642.713.846.6410.8320.210.450.711.392.413.224.605.999.2113.8230.350.581.011.422.373.664.646.257.8211.3416.2741.061.652.203.364.887.789.4913.2818.4751.141.612.343.004.356.067.299.2411.0715.0920.5261.633.073.835.357.238.5610.6412.5916.8122.4672.172.833.824.676.358.389.8012.0214.0718.4824.3282.733.494.595.537.349.5211.0313.3615.5120.0926.1293.324.175.386.398.3410.6612.2414.6816.9221.6727.88103.944.866.187.279.3411.7813.4415.9918.3123.2129.59NonsignificantSignificant
100In this case (X2=2.668), the p value is about 0.10. Means that there is a 10% probability that any deviation from expected results is due to chance only.Based on our standard p > 0.05, this is within the range of acceptable deviation.In terms of your hypothesis for this example, the observed chi-square is not significantly different from expected.The observed numbers are consistent with those expected under Mendel's law.
101Step-by-Step Procedure for Testing Your Hypothesis and Calculating Chi-Square State the hypothesis being tested and the predicted results.Gather the data by conducting the proper experiment (or, if working genetics problems, use the data provided in the problem).Determine the expected numbers for each observational class.Remember to use numbers, not percentages.
102Calculate X2 using the formula. Complete all calculations to three significant digits.Round off your answer to two significant digits.Use the chi-square distribution table to determine significance of the value.Determine degrees of freedom and locate the value in the appropriate column.Locate the value closest to your calculated 2 on that degrees of freedom df row.Move up the column to determine the p value.
103State your conclusion in terms of your hypothesis. If the p value for the calculated X2 is p > 0.05, accept your hypothesis. 'The deviation is small enough that chance alone accounts for it. A p value of 0.6, for example, means that there is a 60% probability that any deviation from expected is due to chance only. This is within the range of acceptable deviation.If the p value for the calculated X2 is p < 0.05, reject your hypothesis, and conclude that some factor other than chance is operating for the deviation to be so great. For example, a p value of 0.01 means that there is only a 1% chance that this deviation is due to chance alone. Therefore, other factors must be involved.