Ch. 5 Genetic Linkage and Mapping Worksheet

Goldberg - Genetics: From Genes to Genomes Ch. 5 Genetic Linkage and Mapping Goldberg - Genetics: From Genes to Genomes Ch. 5 Genetic Linkage and Mapping CONCEPT: OVERVIEW OF MAPPING Mapping is used to determine the position of gene loci on a chromosome □ Linked genes are genes that are inherited together - They exist on the ___________________ chromosome (the chromosome is the unit of inheritance) - Linked genes do not undergo independent assortment - Complete linkage describes genes that are always inherited together - F2 ratio from heterozygous cross is 1:2:1; test cross is 1:1 EXAMPLE: Linkage Group Linkage Group P D r □ Sometimes alleles on the same chromosomes _________________ inherited together - Crossing over is a physical breaking and rejoining of homologous chromosomes - Occurs during meiosis - Produces genetic recombination which leads to a new combination of alleles - Crossing over explains how genes on the same chromosome may not be inherited together EXAMPLE: Example of crossing-over Page 1 Goldberg - Genetics: From Genes to Genomes Ch. 5 Genetic Linkage and Mapping □ Chromosomal mapping is performed by looking at the _______________________ of recombination - The closer the genes are to each other, the less likely they are to cross over - The farther the genes are to each other, the more likely they are to cross over □ The recombination frequency can be used to determine the location of genes relative to each other EXAMPLE: Linkage Group P D 5 m.u. r 8 m.u 13 m.u Page 2 Goldberg - Genetics: From Genes to Genomes Ch. 5 Genetic Linkage and Mapping CONCEPT: CROSSING OVER AND RECOMBINANTS Gamete Genotypes ● Mendel’s law of independent assortment states that the alleles of two genes assort independently EXAMPLE: Genotype AaBb Phenotype Yellow Round What are the genotypes of the gametes? □ If the two genes are on the same chromosome, then they are physically linked and sorted into gametes EXAMPLE: A B a b Genotype Phenotype AB/ab Yellow Round What are the genotypes of the gametes? Page 3 Goldberg - Genetics: From Genes to Genomes Ch. 5 Genetic Linkage and Mapping □ Crossing over is the physical breaking and rejoining of sections of homologous chromosomes EXAMPLE: A B a b Genotype Phenotype AB/ab Yellow Round What are the genotypes of the gametes? Discovery of Crossing Over ● Many studies have supported the idea of crossing over between homologous chromosomes □ McClintock and Creighton studied two traits of maize (corn) - Their corn was heterozygous for color (Cc) and starchy (Wx) / waxy (wx) - Chromosomes were: C c wx Wx □ One of the plants homologous chromosomes had two added markers knob C c Foreign chromosome wx Wx □ They mated their chromosome with a colorless, starchy plant knob C Foreign chromosome wx X c Wx c Wx c wx Page 4 Goldberg - Genetics: From Genes to Genomes Ch. 5 Genetic Linkage and Mapping □ Many plants looked like the parental types, but some were recombinant - Recombinant describes a mixing between the two parents (genotypes or phenotypes) - The chromosomal markers identified the recombinant offspring knob C Foreign chromosome wx + X c Wx c wx cC c wx c colorless waxy c Wx c wx Wx wx colorless colored starchy Page 5 Goldberg - Genetics: From Genes to Genomes Ch. 5 Genetic Linkage and Mapping □ The second major study was from Thomas Hunt Morgan and Alfred Sturtevant - The studies fruit flies with two traits: Eye color, and wing length - Color is red if p+ or purple if pp - Wings are long if vg+ but short if vg vg (vestigial) □ Morgan’s cross went like: P: p+/p+ vg+/vg+ Gametes p+ x vg+ F1 p/p vg/vg p vg p+/p vg+/vg F1 x tester p+/p vg+/vg x p/p vg/vg Expected Offspring Ratios: _______ p+ (red) ________ p (purple) ___________ Long wings _____ red/long wings ___________ vestigial _____ red/vestigial ___________ long wings _______ purple/long ___________vestigial _______purple/vestigial Observed Offspring Ratios: Genotype p+ vg+ p vg p+ vg pr vg+ - Phenotype Red, Long wing Purple, vestigial Red, vestigial Purple, long wing Offspring total (2839) 1339 1195 151 154 Types Parental Parental Recombinant Recombinant There was 10.7% recombinant ( (151+154)/2839 ) instead of the predicted 50% Page 6 Goldberg - Genetics: From Genes to Genomes Ch. 5 Genetic Linkage and Mapping □ Alfred Sturtevant figured out why it wasn’t equal - It was because the two genes were on the same chromosome p+ vg+ p vg Genotype Phenotype p+p/vg+vg F1 Red eyes and long wings What are the genotypes of the gametes? - But what is the significance of the 10.7% recombination? - It is because the area between the two genes is 10.7% of the length of the chromosome - So we say these two genes are 10.7 map units apart Page 7 Goldberg - Genetics: From Genes to Genomes Ch. 5 Genetic Linkage and Mapping Crossing Over Terminology ● Crossing over occurs between two ________________________ □ Crossing over occurs between two chromatids - Sister chromatids are two copies of the same chromosome - Non-sister chromatids are two copies, one from each homologous pair □ During Meiosis, homologous pairs line up and undergo crossing over - Tetrads is the term describing the four paired chromatids - Dyads is a pair of two chromatids - Bivalent refers to the pair of homologous chromosomes - Chiasmata is the structure that forms between dyads during crossing over - Usually between non-sister chromatids, but can also occur between sister EXAMPLE: Page 8 Goldberg - Genetics: From Genes to Genomes Ch. 5 Genetic Linkage and Mapping □ Linked genes have certain terminology - Cis conformations means that dominant alleles of two genes are on the same chromosome - AB/ab or ++/ab - Trans conformation means that two different alleles of two genes are on the same chromosome - Ab/aB or +b/+a □ Linked alleles are written differently - Alleles on the same homolog have no punctuation between them (Ab instead of A/b) - The “/” separates to homologs, instead of two genes - If linkage is unknown you write like A/a  B/b PRACTICE: 1. Which of the following gametes can be formed from the genotype AaBb if AB and ab are linked? a. AB, ab b. Ab, aB c. Aa, Bb 2. An experiment that was performed found the recombination frequency between two genes was 12.5%. What is the distance (in mapping units) between two genes? a. 25 b. 6.25 c. 12.5 d. 10 Page 9 Goldberg - Genetics: From Genes to Genomes Ch. 5 Genetic Linkage and Mapping 3. Which of the following terms describes two copies of the same chromosome? a. Non-sister chromatids b. Sister-chromatids c. Bivalent d. Dyads Page 10 Goldberg - Genetics: From Genes to Genomes Ch. 5 Genetic Linkage and Mapping CONCEPT: MAPPING GENES ● Measuring recombination is traditionally the best way to map gene loci on a chromosome □ Recombination frequencies is the frequency of recombinant offspring produced in a cross Number of recombinant offspring Map Distance = X 100 Total number of off spring □ Morgan’s cross went like: P: p+/p+ vg+/vg+ Gametes p+ x vg+ F1 p/p vg/vg p vg p+/p vg+/vg F1 x tester p+/p vg+/vg x p/p vg/vg Observed Offspring Ratios: Genotype p+ vg+ p vg p+ vg pr vg+ Phenotype Red, Long wing Purple, vestigial Red, vestigial Purple, long wing 151+ 154 Offspring total (2839) 1339 1195 151 154 X 100 = Types Parental Parental Recombinant Recombinant 10.7% 2839 □ What is the significance of the 10.7% recombination? - The area between the two genes is 10.7% of the length of the chromosome - So we say these two genes are 10.7 map units (m.u.) apart - Therefore, physical distance is directly correlated with recombination frequencies - The closer two genes are, the less likely they are to cross over and recombine p+ vg+ 10.7 m.u Page 11 Goldberg - Genetics: From Genes to Genomes Ch. 5 Genetic Linkage and Mapping □ Recombination frequencies can also determine __________________ a gene is linked - Linkage is likely occurring if the recombination frequencies are less than 50% - Linkage is likely not occurring if the recombination frequencies are close or equal to 50% □ Recombination frequencies are never greater than 50% because: - Independent assortment equally assorts alleles, but cannot cause more than 50% recombination EXAMPLE: Modern Mapping ● There are ____________________ types of gene loci maps □ Recombination maps use recombination frequencies to determine gene loci - Can be used to map 2+ gene loci □ Physical maps use the action genomic sequence to determine gene loci - Involves sequencing the entire chromosome, or genome of an organism □ Mapping via certain genomic markers can also be used - Single Nucleotide Polymorphisms are single nucleotide changes that can be markers - Restriction fragment length polymorphisms (RFLPs) are sequences that restriction enzymes cut - Restriction enzymes are proteins that cut DNA at one specific sequence - Microsatellites are short repetitive sequences found throughout the genome Page 12 Goldberg - Genetics: From Genes to Genomes Ch. 5 Genetic Linkage and Mapping PRACTICE 1. The genetic distances between three genes (ab, nm, kf) were determined using a two-point mapping cross. Determine the sequence of the three genes. a. Ab – nm – kf b. Ab – kf – nm c. Kf – ab – nm Gene ab - nm ab – kf nm – kf Distance 45 3 42 2. True or False: Recombination frequencies are never greater than 50% a. True b. False Page 13 Goldberg - Genetics: From Genes to Genomes Ch. 5 Genetic Linkage and Mapping 3. Using the following data collected from a test cross, calculate the recombination frequency. a. b. c. d. 15.9% 10.3% 40.5% 32.7% Phenotype Parental Parental Recombinant Recombinant Offspring 1400 1200 150 150 Page 14 Goldberg - Genetics: From Genes to Genomes Ch. 5 Genetic Linkage and Mapping CONCEPT: TRI-HYBRID CROSS FOR MAPPING ● A tri-hybrid cross, is a mating to look at the inheritance of _________ traits in the offspring □ For unlinked genes, this should follow normal Mendelian inheritance □ For linked genes, recombination frequencies in the offspring can be used to map all three gene loci EXAMPLE: In a cross of fruit flies, there are three traits of interest. 1. Eye color: v+ = red, v = vermillion (like orange/purpleish), 2. Wing veins: cv+= presence of wing crossvein, cv = absence of crossvein 3. Wing shape: ct+ = normal wing, ct = cut wing edges v+/v+  cv/cv  ct/ct P: Parental phenotypes: Red, crovsveinless,cut wings Gametes: v+  cv  ct F1 trihybrid v+/v  cv/cv+  ct/ct+ F1 test cross v+/v  cv/cv+  ct/ct+ F1 phenotypes: v+ cv ct v cv+ ct+ Offspring # ct+ ct ct+ ct ct ct+ ct Parental Parental Recombinant Recombinant Recombinant Recombinant Recombinant 580 592 45 40 89 94 3 v+  cv  ct+ Recombinant Total 5 1448 cv+  cv  cv  cv+  cv  cv+  cv+  x vermillion, crossvein, normal wing v  cv+  ct+ Phenotypes v  v+  v  v+  v  v+  v  v/v  cv+/cv+  ct+/ct+ x Red, crossvein, normal wing Parental gametes: Gametes x v/v  cv/cv  ct/ct vermillion, crossvienless, cut wings Recomb. v and cv Recomb v and ct Recomb cv and ct 268 191 93 Page 15 Goldberg - Genetics: From Genes to Genomes Ch. 5 Genetic Linkage and Mapping Now, you determine the recombination frequencies: 1. For v and cv: RF = 268/1448 = 18.5 2. For v and ct: RF = 191/1448 = 13.2 3. For cv and ct: RF = 93/1448 = 6.4 What can we determine from these ________________________ frequencies? 1. All three genes are linked, because their RFs are less than 50% 2. The orientation and distances of the gene loci □ Why does 13.2 + 6.4 = 19.6 and not the 18.5 (RF for v and cv)? - We should have counted the double cross overs twice, instead of once! Gametes v  ct+  v+  ct  v  ct+  v+  ct  v  ct  v+  ct+  v  ct  v+  ct+  cv+ cv cv cv+ cv cv+ cv+ cv v+ ct cv v ct+ cv+ Phenotypes Parental Parental Recombinant Recombinant Recombinant Recombinant Recombinant Recombinant Total Offspring # 580 592 45 40 89 94 3 5 1448 Recomb. v and cv R R R R R x2 R x2 284 RF = 284/1448 = 19.6% Page 16 Goldberg - Genetics: From Genes to Genomes Ch. 5 Genetic Linkage and Mapping PRACTICE 1. The following table shows data from a cross (ABC x abc) examining three genes (a, b, and c). Calculate the recombination frequency for A and B a. b. c. d. 20% 32% 37% 9.8% Genotype ABC abc aBC Abc AbC aBc ABc abC Offspring 320 276 145 152 43 34 9 12 AC AB BC 2. The following table shows data from a cross examining three genes (a, b, and c). Calculate the recombination frequency for A and C a. b. c. d. 32% 37% 20% 9.8% Genotype ABC abc aBC Abc AbC aBc ABc abC Offspring 320 276 145 152 43 34 9 12 AC AB BC Page 17 Goldberg - Genetics: From Genes to Genomes Ch. 5 Genetic Linkage and Mapping 3. The following table shows data from a cross examining three genes (a, b, and c). Calculate the recombination frequency for B and C a. b. c. d. 20% 32% 37% 9.8% Genotype ABC abc aBC Abc AbC aBc ABc abC Offspring 320 276 145 152 43 34 9 12 AC AB BC 4. The following table shows data from a cross examining three genes (a, b, and c). Determine the order of genes a. A B C Genotype Offspring A C AB BC b. B C A A B C 320 c. C A B abc 276 aBC 145 Abc 152 AbC 43 aBc 34 ABc 9 abC 12 Page 18 Goldberg - Genetics: From Genes to Genomes Ch. 5 Genetic Linkage and Mapping CONCEPT: MULTIPLE CROSS OVERS AND INTERFERENCE ● Mapping the loci of 3+ genes can be more difficult, because of multiple cross overs □ Multiple crossovers occur when more than one crossover causes 2+ changes in the gamete genotype EXAMPLE: v+ ct cv v ct cv+ v ct+ cv+ v+ ct+ cv □ Calculate information on double cross overs by: 1. Double counting the double crossovers when calculating the RF for the genes of farthest distance - This will help you accurately map the genes (Trihybrid cross) ● Interference is when crossovers in one region of the chromosome affects the chance of crossover in an adjacent region □ Independent crossovers (meaning, no interference) can be calculated using the double recombination frequency EXAMPLE: Double cross over calculation with no interference 1. If the frequency of crossing over is 20% for genes A and B, and 30% for genes B and C, what is the frequency of a double crossover between A and C? - (0.2 x 0.3) x 100 = 6% Page 19 Goldberg - Genetics: From Genes to Genomes Ch. 5 Genetic Linkage and Mapping □ If the calculated value does not equal the observed results, then that means interference is impacting the data - Uses the coefficient of coincidence Observed frequency or # of double recombinants I= 1Expected frequency or # of double recombinants EXAMPLE: Calculating the interference Observed frequency: 4 Expected frequency: 6 1-(4/6) = 0.33 x 100 = 33% □ Therefore, there were 33% fewer double crossovers occurred than expected - Crossing over at one location, partially decreased the chance of crossover at an adjacent location Page 20 Goldberg - Genetics: From Genes to Genomes Ch. 5 Genetic Linkage and Mapping PRACTICE: 1. A female with the following genotype can produce a number of different gametes. Choose the gamete produced if no crossovers have occurred. Genotype = ab+ a. a + c ++c b. + b c c. a b + d. + b + 2. A female with the following genotype can produce a number of different gametes. Choose the gamete produced if a single crossover has occurred. Genotype = ab+ a. + + c ++c b. + b c c. a b + d. a b c Page 21 Goldberg - Genetics: From Genes to Genomes Ch. 5 Genetic Linkage and Mapping CONCEPT: CHI-SQUARE TEST FOR LINKAGE ● A chi-square test can be used to identify the likelihood of gene linkage □ Chi-square test is used to evaluate if your experimental values are different from the predicted values EXAMPLE: I want to know if genes A and B are linked, so I do an experiment where I cross two heterozygous organisms. I get 50 offspring, 31 parental types, and 19 recombinant types. Is it likely that A and B are linked? 1. What are the expected numbers? - If two genes are not linked then the recombination frequency is 50%. Therefore, there would be 50% parental types and 50% recombinant types. So out of 50 offspring, 50/2 = 25. There would be 25 parental and 25 recombinants. 2. Calculate the chi-square value Phenotype Parental Recombinant Observed 31 19 Expected 25 25 Total (o-e)2/e 1.44 1.44 2.88 3. Determine the P value. a. Calculate degrees of freedom. DF=1 b. Find chi-square value on row 1. It is between 2.71 and 3.84 c. The appropriate p value is between 0.10 and 0.05 (10% to 5%) Page 22 Goldberg - Genetics: From Genes to Genomes Ch. 5 Genetic Linkage and Mapping 4. Do we accept or reject the null hypothesis? a. The null hypothesis states that the expected and observed values are not different. (This would mean that the genes would NOT be linked) b. Because the p values are greater than 0.05 (5%), so we accept the null hypothesis. 5. Therefore, we can say with 95% confidence the genes are NOT linked. a. Remember: This does not confirm linkage, it just states the likelihood PRACTICE: Black(B) rabbit coat colors are dominant to white(b) coat colors. Long hair (H) is dominant to short hair (h). A breeder crosses a rabbit homozygous for white, short hair with a black rabbit with long hair. The F1 is backcrossed to the rabbit with white, short hair and the following progeny are produced. Use the chi-square test to answer the following questions. Phenotype Black, Long Black, Short White, Long White, Short Total 1. Offspring 40 20 24 62 146 What are the expected offspring numbers if the two genes are not linked, and therefore assort independently? a. 73 parental, 73 recombinant b. 146 parental, 0 recombinant c. 60 parental, 86 recombinant d. 102 parental, 44 recombinant Page 23 Goldberg - Genetics: From Genes to Genomes Ch. 5 Genetic Linkage and Mapping 2. Calculate the chi-square value for the above problem. a. b. c. d. 23.00 5.89 0.02 0.467 3. In this example, how many degrees of freedom should be used? a. 1 b. 2 c. 3 d. 4 Page 24 Goldberg - Genetics: From Genes to Genomes Ch. 5 Genetic Linkage and Mapping 4. Using the appropriate chi-square value and degrees of freedom, do the coat color and hair length genes assort independently? a. Yes, both genes assort independently b. No, both genes do not assort independently (meaning they are linked) Page 25