Lisa McDonnell & Jennifer Klenz

Lisa McDonnell & Jennifer Klenz
Teaching Genetic Linkage and Recombination through Mapping with Molecular Markers Lisa McDonnell & Jennifer Klenz Note for instructors: Please see the notes field for comments that might be useful to you as an instructor when running this lesson.

Mapping with Molecular Markers
Where is the drought tolerance locus? Insert this cartoon for context:

Lesson Plan Part 1 Assessing genetic linkage between a trait of interest and a molecular marker if we do a testcross (dihybrid F1 x homozygous recessive). Part 2 Predicting and assessing genetic linkage when we do a dihybrid x dihybrid cross (F1 x F1) Notes: In this lesson we are defining a testcross as a cross where our heterozygote (F1) is crossed with a homozygous recessive genotype. The homozygous recessive plant is the tester strain.

Learning Objectives Explain how molecular markers (such as SNPs, microsatellites, VNTRs) can be used to map the location of genes/loci, including what crosses you would do, and why. From analysis of data from a cross (such as gel banding patterns) determine if crosses involve linked genes. Be able to justify your analysis by describing the information in the data that allows you to determine genes are linked. Use linkage information between multiple loci to construct a genetic map

Brainstorm in your groups
Genes closer together on a chromosome are more often inherited together If we know the location of some genes in a genome, how can we use that information to determine what DNA sequences cause specific traits of interest? What are some examples of traits for which you (or someone) might want to know the location of a gene in the genome that controls that trait? Why would you want to know the location of a gene? Notes: 1 – we expect genes that are linked to be inherited together. Therefore alleles of those genes will be inherited together – so we’ll expect to see traits of linked genes segregate together more often than they would if they were not linked. 2 – e.g.: genes involved in disease, or genes that give a desired trait (e.g. in plant or crop breeding) 3- once we know the location we can do more testing: sequence the gene, identify promoters, clone the gene to do further manipulations and experiments Write down your group answers and hand them in

Using genetic linkage to map the location of genes for traits of interest
“Drought Gene” Notes: The “Drought Gene” is one that was described by Monsanto ( The sequence codes for the CSPB protein which has been shown to bind to a broad array of RNA, allowing them to adopt the correct conformation under stress conditions and improve cellular function in the plant. Corn image: By Illustratedjc (Own work) [CC-BY-SA-3.0 ( via Wikimedia Commons Drought conditions: (available for non-profit reuse) Lecture Script: “Climate predictions indicate temperatures in various regions will continue to rise and, in some areas, precipitation will decrease resulting in drought conditions in larger areas of the world that last for more months of the year.” (During the lesson we like to emphasize possible local impacts of drought to provide an opportunity for students to connect to the scenario.) “For example, many states (such as California) are experiencing extreme drought, leading to reduced crop yields and rising food prices in places that depend on California food crops. Not surprisingly, to address the impact of these future droughts on humans, there is great interest in determining the mechanisms of drought tolerance in many species of crop plants (such as corn11 and rice12), including any genes, and their alleles, that are involved in drought tolerance. You might think that just knowing the genome is sufficient to understand these genes. In fact, for many plants, the entire genome sequence is already available.” “However, knowing the genome sequence does NOT mean that we know all of the genes in the genome, much less the identity of the gene(s) for the trait in which we are interested. So, if a random mutant plant has a phenotype of interest, such as drought tolerance (as presented on slide 7), researchers may need to study this mutant further, to determine which gene or genes is/are responsible for the phenotype. This information would not only help us understand more about the basic biology of plants, but also provide targets that can be used to identify or create other drought tolerant plants, including crops. Traditional genetic mapping analyzes the segregation of two (or more) phenotypes of interest over multiple generations. Obviously, for some plants, that mapping process can take years or even decades. Luckily, scientists have developed alternatives! We will explore one of them: genetic mapping using known molecular markers to rapidly identify the genes that may produce the drought resistant phenotype.” National Center for Atmospheric Research | University Corporation for Atmospheric Research

Example: a drought tolerant plant.
Most plants do not thrive in drought conditions. They may still reproduce, but they have lower yield. Notes: A drought tolerant plant appears in a crop experiencing drought conditions. Instructor asks the class: 1) What could be the reason for this plant being drought tolerant? Answer: genetic cause (i.e. mutation creates new allele), or environmental (something about where that plant is growing is making it more tolerant, or not experience drought the same as the other plants) Lecture Script: Describe the scenario: “This crop plant is normally drought sensitive, so in times of drought the crop plants wilt, are less healthy, and produce lower yields. In one field among these wilting plants, a healthy plant is observed, one that is much more tolerant of drought compared to the normal, drought sensitive plants. What might cause this drought tolerant phenotype?” The instructor could choose, at this point, to solicit ideas from the class about what could be causing this particular plant to be more tolerant of drought. “Until we do further experiments, we do not know if it is a genetic or environmental cause and, if it is genetic, which genes are responsible for the phenotype.”

Is the drought tolerant phenotype heritable?
Goals: Is the drought tolerant phenotype heritable? Is it caused by a single gene? Mapping – where in the plant’s genome is the mutation/locus involved in drought tolerance? Notes: A drought tolerant plant appears in a crop experiencing drought conditions. Instructor asks the class: 1) What could be the reason for this plant being drought tolerant? Answer: genetic cause (i.e. mutation creates new allele), or environmental (something about where that plant is growing is making it more tolerant, or not experience drought the same as the other plants)

What can you conclude based on these results?
Plants are grown under drought conditions x Cross #1: Drought tolerant plants selfed: x Cross #2: Drought tolerant x true-breeding wild-type (drought sensitive) What can you conclude based on these results? Wild type is dominant Drought tolerance is dominant Drought tolerance is heritable A and C B and C Cross #3: offspring of cross #2 crossed to each other x Notes: Reveal cross #1 and cross #2, but do not draw conclusions from them just yet. A is true (cross #2) C is true (cross #1 & #3) Ask: “Is there another cross we should do?” Cross #3: We should allow the F1 from the first cross to self, and if you see drought tolerance in the F2 that would indicate it is a heritable, recessive trait. A 3:1 ratio in the offspring (25% drought tolerant, 75% sensitive) also indicates that it is the result of a single gene. Lecture Script: “Before we start the mapping protocol (slide 11), we first want to answer the following questions: 1) Is the drought-tolerant phenotype due to a stable, heritable genetic change (i.e. mutation); and 2) Is the phenotype is due to a mutation in a one gene (slide 9). Alternatively, the drought tolerant plant could be the result of environmental factors or the result of multiple genes. So, we need to determine heritability and number of genes by crossing plants and screening offspring for the drought tolerant phenotype.” Slide 9 has animations to walk through the crosses. Present the crosses to the students but do not tell them what we conclude from each cross. Let them analyze the results and after showing Cross #1 and #2 ask them the clicker question: “What can we conclude based on these results?” Cross #1: The drought tolerant plant is self-fertilized and the offspring screened for drought tolerance by growing under drought conditions both in the field and in the lab. If all of the offspring continue to show the drought tolerant phenotype, the trait is heritable, and now we have a true-breeding (homozygous) drought tolerant population to use for subsequent experiments. To determine if the drought tolerant phenotype is the result of a dominant or recessive allele, we need to do another cross (Cross #2). Cross #2: We cross one of the true-breeding drought tolerant plants generated from Cross #1 with a true-breeding wild-type (drought-sensitive) plant and examine the phenotype of the offspring (F1s). All of the F1 plants are wild-type (drought-sensitive), indicating the drought tolerance allele is recessive to the drought sensitive allele. If all of the F1 plants were drought tolerant, we would conclude the drought tolerance allele was dominant to the wild-type/drought sensitive allele. Note, one could have crossed the original drought-tolerant plant with a true-breeding wild-type plant and examined the offspring. There are a variety of possible outcomes: 1) if the drought-tolerance allele is recessive to the wild-type allele, the original plant must have been true-breeding to exhibit the phenotype and therefore when crossed to a true-breeding wild-type plant we expect all the F1 to be wild-type; 2) if the drought-tolerance allele is dominant then it is possible that the original drought-tolerant plant is heterozygous, in which case we expect half of the F1 to be drought tolerant and half to be wild-type. Alternatively, if the drought tolerant plant was homozygous and the tolerance allele was dominant, we expect all the F1 to be drought tolerant. Ask the class “Is there another cross we should do?” Solicit their response, and then show them Cross #3. Cross #3: We would then allow the F1s from Cross #2 to self-fertilize (F1 x F1) to generate an F2 and observe the ratio of phenotypes to determine if drought tolerance is caused by a single gene. Show students the results of Cross #3, and ask “what do you conclude based on these results?” If we see 3 drought sensitive plants for every 1 drought tolerant plant, i.e. a 3:1 ratio, we can assume the phenotype is due to a single gene. (Remember, the F1s will be heterozygous for the drought-tolerance gene. Work through a Punnett Square if you need to remember why you get this 3:1 ratio of wild-type (drought sensitive) to mutant (drought resistant) progeny. “From these three crosses, our hypotheses that the drought tolerant phenotype is heritable and recessive to drought sensitive are confirmed.”

What do we not know? recessive to wildtype (sensitive)
We have a true-breeding drought tolerant population, and we know drought tolerance is: recessive to wildtype (sensitive) caused by a single gene Notes: Give the students some time to think about this, it will take them a few minutes to realize we don’t know where the gene and allele is that causes drought tolerance! Lecture Script: Pose this question to the class: “The results of these crosses tell us a lot about the drought tolerant phenotype. But, what do we not know?” Give the students a minute to think about their answer, since it may take them some time to remember that we do not know the locus/gene that causes drought tolerance! And for that matter, we do not know how the drought tolerance allele differs from the wild-type allele. At this point we may ask the class “Why might we want to know what gene is involved in the drought tolerant phenotype? What can we do with the gene/locus, if we identify it?” Students should think about the answer to these questions and briefly (1 minute) discuss their ideas with their neighbors. Soliciting student answers then prompts a brief class discussion of how we might want to understand the gene product and how it contributes to drought tolerance, identify the gene in other crop species, and determine if we can manipulate expression of the gene for biotechnology and GMO applications13. The instructor should help the class realize: If we don’t know the location of the gene, we cannot use PCR to amplify the DNA sequence and identify the type of mutation that underlies the drought-tolerant allele. Hence, to further our investigation of drought tolerance, it would be very beneficial to know the location of the drought tolerance locus in the genome. After this brief discussion, transition to slide 11 which articulates the mapping goal. What do we not know?

Goal: Map (identify) the location of the locus/gene/mutation that causes the drought tolerant phenotype. Wildtype is drought sensitive, DRS New mutant is drought tolerant, DRT DRS is dominant to DRT We only know the phenotype and hypothetical genotypes assigned based on the results of crosses: Notes: Instructor, ask the class: Why might we want to know what locus is involved in the drought tolerance phenotype? Sequence the gene – what protein/product does it code for? Do we know anything about this kind of gene/protein from other species? Clone the gene – can then express it in other species to further study the role of this gene in drought tolerance. Lecture Script: “Our goal is to map the location of the locus/gene/mutation that is associated with the drought tolerant phenotype. Some plant geneticists name a locus based on the mutant phenotype, which in our case is drought tolerance. For now we will call our unknown locus DR, for drought. We will use a non-standard nomenclature for naming the alleles, whereby a superscript letter beside the locus symbol will represent the allele. The two alleles we are working with are wild-type - drought sensitive (DRS), and the newly discovered drought tolerant phenotype, DRT. We do not know the location of the drought tolerance locus; we only have the phenotype and hypothetical genotypes that we tentatively assigned based on the results of crosses. We are going to use microsatellite markers to map the location of the locus/gene involved in drought tolerance”. F1: X DRT / DRS DRT / DRT DRS /DRS

Here is a map of known microsatellite markers in our plant’s genome
Here is a map of known microsatellite markers in our plant’s genome. How can we use these markers to find the locus involved in drought tolerance? Chromosome 1 2 3 4 5 Notes: We will use microsatellite markers in the genome to determine the position of the locus involved in drought tolerance. What we are looking for is whether or not our drought tolerance phenotype segregates with one of these microsatellite markers. That is, is the drought tolerance allele more often inherited with one of these markers than the others? If so, the DR locus might be located near the marker it segregates with. A note about the microsatellite loci names: This map shows real microsatellites in the Arabidopsis genome. You can tell your students this is an Arabidopsis map or pretend it is for the plant you are using in this hypothetical drought tolerance experiment. Microsatellite loci names vary widely depending on the species. For our purposes we are simplifying the names of the microsatellites to A-G (instructor – show the next slide which replaces the loci labels with A-G). Note that these are not genes, nor do microsatellite alleles have dominance or recessiveness. In subsequent slides we will describe how the alleles of each locus differ. Image use and sharing via Creative Commons License 4.0 Lecture Script: On slide 12 we show a detailed map of known microsatellite molecular markers in the genome of a plant (the map is from Arabidopsis, which you may choose to tell your students or not). The map has dozens of molecular markers with varying names. We explain that names of molecular markers vary widely, depending on the species and when/by whom they were identified. The animation circles seven microsatellites that we will use for our hypothetical mapping experiment. Ask the class “how can we use these molecular markers to find the drought tolerance locus?” “Ultimately, what we are asking is: In the offspring of various crosses, does the drought tolerant phenotype segregate with a particular microsatellite more often than we would predict if the drought tolerance locus and the microsatellite marker were assorting independently (i.e. the drought tolerance locus and the microsatellite are located on different chromosomes or far apart on the same chromosome).” Cosson et al. Plant Methods 2014 10:2; CC 4.0

How do we find the locus involved in drought tolerance (DR)?
Here is a map of known microsatellite markers in our plant’s genome: Chromosome 1 2 3 4 5 F D DR locus? DR locus? A C E G Notes: Ultimately we are using these molecular markers to determine the location of the DR locus. At the start of a mapping experiment we don’t know where the DR locus is – we would have to perform multiple crosses and PCRs (hundreds) to determine which microsatellite allele segregates with the drought tolerance phenotype. The DR locus could be anywhere in the genome – use the sequence of animations to show this. We will start out mapping experiment using the G microsatellite locus. Lecture Script: On Slide 13, we have simplified the names of the selected microsatellites to A-G. It is important to emphasize that the molecular marker loci are not genes themselves; microsatellites are sequences of repetitive DNA that tend to be located in non-coding DNA. In addition, since the microsatellite marker alleles do not affect the phenotype of the plants, microsatellite alleles are not dominant or recessive. For example, the A microsatellite may have two alleles in a population: A300 and A100 (where the 300 and 100 refer to the number of nucleotides in the PCR product and the band we see on a gel). The animation on slide 13 shows how we do not yet know where the drought tolerance (DR) locus is within the genome. At this point, I tell the students that “for simplicity we are starting our mapping experiment with just the G microsatellite, which was chosen arbitrarily. In reality, we would be testing hundreds of microsatellites simultaneously to try to find one that co-segregated with the DR locus.” DR locus? B DR locus? -

Does the drought tolerance phenotype segregate with a the G microsatellite locus?
To measure segregation/linkage between 2 loci what do we look for? Frequency of parental and recombinant combinations of genotypes/phenotypes. For our experiment, this means segregation of drought tolerance with a G microsatellite allele (a specific banding pattern). We need two “parental” combinations of alleles: Drought tolerance with a G microsatellite allele that gives one banding pattern Drought sensitive with a different G microsatellite allele that gives a different banding pattern G microsatellite Notes: This slide is optional, and could be review depending on the students (it could also be assigned as part of their pre-class reading). This information can be used if you want to review with your students why we need two different strains of plant that carry different microsatellite alleles. Gel image modified from Cosson et al. Plant Methods 2014 10:2; CC 4.0 Use and sharing via Creative Commons License 4.0 (You are free to: Share — copy and redistribute the material in any medium or format Adapt — remix, transform, and build upon the material for any purpose, even commercially). Lecture Script: Ask the class, “To measure segregation/linkage between two loci, what do we look for?” “What we measure is the frequency of parental and recombinant combinations of genotypes and/or phenotypes in a population. Therefore, we need to measure the frequency at which the drought tolerant phenotype segregates with a particular G microsatellite allele (i.e. a specific banding pattern). To determine this frequency, we need two “parental” combinations of alleles: 1) Drought tolerant with a G microsatellite allele that gives one banding pattern; and 2) Drought sensitive with a different G microsatellite allele that gives a different banding pattern. The data we will be looking at are bands of microsatellite DNA (produced by PCR) and the drought tolerant or drought sensitive phenotypes of the plants whose DNA was used as a template in the PCR reaction. If the drought sensitive parent plant had the same microsatellite alleles as the drought tolerant parent plant, all offspring of these plants (and subsequent generations) would have the same microsatellite banding pattern, so no segregation of microsatellite alleles would be occurring for us to measure.” Image adapted from Cosson et al. Plant Methods 2014 10:2; CC 4.0

Is the drought tolerance locus linked to the G microsatellite?
Experimental plan: a testcross to measure linkage 1. Identify a true-breeding drought sensitive strain that carries different microsatellite alleles (depending on the species these are available for purchase form research groups) 2. Generate a heterozygous plant, DRT G200 /DRS G Set up a testcross: heterozygote x homozygous recessive 4. Analyze offspring of the testcross, look for segregation of the drought tolerance phenotype with G200 and G400 alleles. Notes: You can either have students determine this for themselves, or share it with them. In our experience developing an experimental plan is very challenging for the students. Having them determine the plan could add an extra 10 minutes to the lesson. Point #1: many different varieties of a given species exist, and if you are doing research in a popular/model species there are likely varieties available with known polymorphisms for the microsatellite loci. If not, you would have to screen natural populations for a variety that has different alleles – a very time consuming step! Lecture Script: On slide 15, we remind students that we are trying to map the location of the DR locus and outline the steps of our experimental plan: “1) Identify a true-breeding drought sensitive (wild-type) strain that carries microsatellite alleles that differ from the alleles in the true-breeding drought tolerant strain. Depending on the species, such strains are available for purchase from companies or can be obtained for free from research groups; 2) Generate a doubly heterozygous plant, DRT G200 /DRS G400; 3) Set up a testcross: heterozygote x homozygous recessive, and 4) Analyze the offspring of the testcross to determine the frequency with which the drought tolerant phenotype segregates together with the G200 and G400 alleles.”

Let’s start with the G microsatellite.
Is the DR locus genetically linked to G? Microsatellite G DR locus? You have homozygous drought tolerant plants that have the 200 allele of the G microsatellite, and the drought sensitive plants are homozygous for the 400 locus: DRT G200/ DRT G200 DRS G400/ DRS G400 true-breeding: drought tolerant G200 true-breeding: drought sensitive G400 Notes: Point out that the picture shows the DR locus downstream of the G microsatellite, but it could also be on the other side. We are not yet sure of the position of DR. Note about nomenclature: we use the “/” to distinguish between the homologous chromosomes, and the “;” to distinguish between different chromosomes. Because we’re testing for linkage we will write the genotypes in a way that suggests the two loci could be on the same chromosome – but the at this time we still don’t know for sure if they are. Lecture Script: On slide 16, we remind the class we are starting with the G microsatellite marker. “We do not know if the DR locus is on the same chromosome as the G locus. In addition, if it is on the same chromosome, we do not know the location of the DR locus relative to the G locus, i.e. whether it is 5’ or 3’ to the G locus.” The picture on slide 16 shows the DR locus downstream of the G microsatellite, but it could also be on the other side (shown with animation). Introduce the genotypes of the two parental plants we are using: “our true-breeding drought tolerant plant carries two copies of both the drought tolerance allele and the G200 allele (i.e. has the genotype: DRT G200/ DRT G200). The other plant we are working with differs at both loci that we care about right now: it is homozygous for the drought sensitive allele and for the 400 allele of the G microsatellite (i.e. has the genotype: DRS G400/DRS G400). The 200 and 400 refer to the number of nucleotides in the microsatellite, which is important information when we amplify the microsatellite locus via PCR and analyze the gel banding patterns to determine microsatellite genotypes.” A note about the nomenclature used for the genotypes: we use “/” to distinguish between homologous chromosomes. Here we have written the genotypes in a way that suggests the two loci are on the same chromosome – but it is important to remind the students that at this time we do not know if that is true, we need to use the data from our crosses to determine if the DR and G loci are linked.

What crosses will we do next?
Is the drought tolerance locus linked to the G microsatellite? Parent plants: Drought tolerant DRTG200 / DRTG200 Drought sensitive DRS G400 / DRSG400 Ultimately we need to look for recombination (or independent assortment) between the DR and G loci. Notes: Walk through the animations of this slide by posing Qs to the class, soliciting answers, then revealing the next animation. Answer: Cross the two homozygous plants we already have. Lecture Script: We have our two parent plants (DRTG200 / DRT G200 and DRS G400/DRS G400). Ask the class “what crosses will you do next?” as a way to have the class recall the experimental plan. The next step is to generate a heterozygote that can be used in a testcross. We can do this by crossing the two parent plants to generate a heterozygous F1 population (DRT G200/ DRS G400). What crosses will we do next?

X F1 Draw chromosomes of the parent plants and the resulting F1.
Generating a heterozygous plant X F1 DRT G200 / DRT G200 DRS G400 / DRS G400 Draw chromosomes of the parent plants and the resulting F1. What phenotypes will we see in the F1? What G microsatellite banding patterns will we see from the parent and F1 plants? F1 plant: Parent: 200 400 Notes: Explain that we would harvest some tissue from parent and F1 plants (e.g. a leave) and from that we would extract DNA and use the DNA as a template for PCR, to amplify the G microsatellite locus (as shown in the picture). Have students predict phenotype and banding patterns. After a few minutes or two I would solicit student ideas and draw the banding patterns they expect to see using a tablet or document camera (or show a student drawing and review it as a class to determine if the prediction is correct). The next slides is a pre-made feedback slide if real-time drawing is not ideal. We now have a population of heterozygous plants (the F1) that we can use for the testcross. Lecture Script: “We crossed the two true-breeding parent plants and generated a heterozygous F1” (DRT G200/DRT G200 x DRS G400/ DRS G400  DRT G200 /DRS G400). “We can extract DNA from both parent plants and the F1 plants, use the DNA as the template for amplifying the G microsatellite locus using PCR, and visualize the bands using gel electrophoresis.” We then give the students just a few minutes to individually work on these tasks: Draw chromosomes of the parents of this cross and the resulting F1. What phenotypes will we see in the F1? What G microsatellite banding patterns will we see from the parent and F1 plants? To provide feedback on student work, we solicit their ideas, drawing the chromosomes in real time as students provide answers, labelling the alleles, and diagraming the expected banding patterns. Alternatively, you could show a volunteered student drawing and review it with the class to determine if the student’s drawings and predictions are correct and why or why not. Students should label the gel with the phenotypes of the plants (drought tolerant or sensitive). We expect to see a single band of 200 nucleotides from the homozygous drought tolerant parent plant, a single band of 400 bases from the homozygous drought sensitive plant, and two bands in the F1 because it is heterozygous (produces both 200 nucleotide and 400 nucleotide bands). The bands for the parent plants are thicker because we assume the homozygote contains two copies of only one template (the 200 or 400) and therefore will produce twice as much PCR product as the heterozygote.

Expected F1 phenotypes and banding patterns
P generation: DRT G200 / DRT G200 DRS G400 / DRS G400 DRS G400/ DRT G200 F1 plant: drought sensitive Parent: drought tolerant Parent: Drought sensitive 200 400 Notes: This is a pre-done feedback slide for the predictions they were asked to make on slide 18. Lecture Script: Slide 19 is a pre-made feedback slide, showing the expected banding patterns, which could be used in lieu of having students draw their predicted gels. “We now have a population of heterozygous plants (the F1) that we can use for the testcross.”

F2 Assessing linkage: using a testcross
F x homozygous recessive (“tester”) DRS G400 / DRT G200 DRT G200 / DRT G200 Drought sensitive Drought tolerant F2 What gamete genotypes will each plant produce? (draw the chromosomes!) Which are parental? Recombinant? What phenotype (drought tolerant or sensitive) and what banding patterns will F2 have? (draw a Punnett Square to make your predictions)? What F2 results will suggest linkage between the DR locus and G microsatellite locus? Notes: I give students about 5 minutes to work on this, encouraging them to draw the chromosomes labeled with alleles for the gametes produced by the F1 and the tester plant. If the class is struggling you could first ask them to predict the gametes the F1 will produce and the homozygous recessive will produce. Then, label the gametes as parental and recombinant, and then task them to complete Q3 and Q4. Once they have had some time to answer the questions alone/in groups, I pose the question on the next slide to the class. Lecture Script: Here, we show students the testcross, in which we cross the heterozygous F1 generated from the previous cross to a homozygous recessive plant (the “tester” plant): DRT G200 / DRT G20 0x DRT G200 / DRS G40 0 F2. “Our next task is to predict the F2 phenotypes and banding patterns, as well as how genetic linkage between the DR and G loci would affect the F2 phenotypes and ratios. As a researcher, it is important to predict what the results could be and what they would mean, so that you can compare your actual results to your original predictions.” Give students 5 or so minutes to: 1) Determine the gamete genotypes that will be produced by the F1 and tester plant; 2) Identify which gamete genotypes are parental versus recombinant; 3) Predict what phenotypes (drought tolerant or sensitive) and what banding patterns the F2 population will have; and 4) predict what F2 results will suggest linkage between the DR and G loci. Encourage them to draw both a Punnett Square to make predictions and a gel showing the predicted banding patterns. Solicit their answers to question 1 and 2 on slide 20 and, in response to their answers, we recommend the instructor draws the gamete chromosomes that make the different F2 plants. The homozygous recessive plant can produce only one gamete genotype with respect to the DR and G microsatellite locus (DRT G200). The heterozygous F1 plant can produce four gamete types: DRT G200 and DRS G400 (both parental), and DRT G400 and DRS G200 (both recombinant). Half of the F2 are expected to be drought tolerant and half drought sensitive (Figure 3). All of the plants should have a 200 band, inherited from the homozygous recessive tester plant. The other band for each plant is inherited from the F1 (either a recombinant or a parental band).

Is the drought tolerance locus linked to the G microsatellite?
Which combination will go up if the DR and G loci are genetically linked? 1 and 3 1 and 4 2 and 3 2 and 4 F2 200 400 Drought tolerant Drought tolerant Drought sensitive Drought sensitive Notes: A. 1 and 3. In the tolerant plants the 200 allele is parental, in the sensitive plants the parental allele is 400. Linkage results in more parental allele combinations and less recombinant combinations (compared to independent assortment) Lecture Script: At this point, we transition to slide 21 and ask the clicker question: “If the DR and G loci are genetically linked, which combination of phenotype and banding pattern will be higher than predicted?” This answer requires students to have correctly identified which bands are parental and which are recombinant. If students struggled to get the correct answer on this question it is recommended that the instructor: 1) draw the chromosomes of the doubly heterozygous F1, labelled with loci and alleles, and illustrate how the crossing-over between the DR and G loci results in the recombinant genotypes; 2) draw all the resulting gamete chromosomes produced by the F1, labelling them as recombinant and parental; and 3) diagramming the banding patterns that would be associated with each gamete type. CLICKER Question

F2 Recombinant Parental One of these is parental
200 400 Drought tolerant Drought tolerant Drought sensitive Drought sensitive Notes: Pre-made feedback slide for the question on slide 21. Lecture Script: Slide 22 is a pre-made feedback slide. You can solicit responses from the class, and then, if needed, walk them through the logic: “We know the tester plant can only donate DRT G200, so we expect all plants to have the 200 nucleotide band. The other band represents what microsatellite allele was inherited from the F1 parent. Recall that DRT is recessive to DRS; therefore, all the drought tolerant plants must be DRT/DRT. The parental combination of alleles is DRT G200 and DRS G400, so if a drought tolerant F2 plant has a 400 nucleotide band, it must have inherited a recombinant genotype (DRT G400) from the F1. Thus, a 200/400 heterozygous banding pattern in a drought tolerant plant is recombinant. If the DR and G loci are on the same chromosome, the recombinant genotype results from a recombination event during meiosis I in the F1, with crossing over between the DR and G loci of homologous chromosomes. Considering the drought tolerant plants, if the DR and G loci assort independently (i.e. are unlinked), we expect an equal proportion of the F2 plants will have 200/200 banding patterns (single 200 band on the gel) and 200/400 banding patterns (two bands on the gel). If the DR and G loci are genetically linked, we expect that the F2 population will have fewer than 50% recombinant types (200/400) and greater that 50% parental types (200/200).” “Now let us consider the drought sensitive plants. Again, we expect all plants to have the 200 nucleotide band, inherited from the homozygous tester plant. The other band represents what microsatellite allele was inherited from the F1 parent. If an F2 plant is drought sensitive, it had to have inherited the DRS allele from the F1 plant. Recall that, of the DRS-containing gametes generated by the F1, DRS G400 is a parental genotype and DRS G400 is a recombinant genotype. Thus, a drought sensitive F2 plant that has a 200/400 banding pattern represents the parental type: the 200 band came from the homozygous tester plant and the 400 from the F1. If a drought sensitive F2 plant has a 200/200 banding pattern, represented by a single 200 nucleotide band on the gel, it is a recombinant type. If the DR and G microsatellite loci are genetically linked we expect the 200/400 banding pattern to be observed more often in the drought sensitive F2 plants (e.g. in greater than 50% of the drought sensitive F2 plants), and the 200/200 banding pattern to be observed less often among the drought sensitive plants.” Recombinant Parental One of these is parental One of these is recombinant

Linkage: less than 50% recombinant types
If two loci assort independently we expect 50% parental types and 50% recombinant types. If two loci are linked we expect less than 50% recombinant types Frequency of recombinants = Number of recombinant types Total number of samples analyzed Lecture Script: Slide 23 contains a summary of the effects of linkage versus independent assortment and how to calculate the frequency of recombinants. If two loci assort independently, we expect 50% parental types and 50% recombinant types. If two loci are linked, we expect less than 50% recombinant types and more than 50% parental types. The frequency of recombinants can be used to determine the map distance between two linked loci, by dividing the number of recombinant types by the total number of samples analyzed and multiplying the result by 100%. “Consider that the bands we see on the gel represent two chromosomes: one inherited from the F1 plant and one from the tester plant. As discussed above, because the homozygous tester plant can only donate one gamete genotype, the banding patterns, in combination with the drought tolerant/sensitive phenotypes, are representative of the recombination (or lack thereof) that occurred in the F1 plant. That is to say, only one chromosome of each F2 plant can be classified as parental or recombinant. Hence, if an F2 plant contains one recombinant chromosome, the entire plant can be classified as recombinant. For example, if we looked at 10 drought tolerant F2 plants and there were 4 that had the 200/400 banding pattern we would say there are recombinants among the 10 sample: a recombination frequency of 40%, or 40 map units between the two linked.”

What is the map distance?
Is the drought tolerance locus linked to the G microsatellite? Now that we’ve made predictions, let’s analyze some data Our testcross scenario: F1 x homozygous recessive F2 Examine the four gels and determine which (if any) indicates the drought tolerance locus is genetically linked to the G microsatellite? (gels are on the next two slides) Notes: Project the gel pictures and ask them to record their answers in their notes. After a couple minutes of analysis put up the clicker question (slide 26) to solicit their answers. Lecture Script: “Now that we have made predictions about what results will indicate genetic linkage between the DR and G loci, we will analyze some banding patterns for drought tolerant and drought sensitive plants.” Students are asked to analyze the four gels (slide 25 and 26) and determine which, if any, of the gels contain inheritance patterns that indicate genetic linkage between the DR and the G microsatellite loci. We project all four gels at once using dual screens. If you do not have dual screens, you can project the first two gels (slide 25) and ask the students to analyze them and record their conclusions in their notes before moving on to the second two gels (slide 26). Alternatively, you can print these gels as a one-sheet handout, which has the advantage that students like having a handout to annotate. Analysis of all the gels takes 2-5 minutes. What is the map distance?

Which result indicates the drought tolerance locus is genetically linked to the G microsatellite? (these are F2 results from the testcross) Gel A 10 drought tolerant plants analyzed: 200 400 Gel B 10 drought sensitive plants analyzed: Notes: Independent assortment. To determine linkage, calculate the recombinant frequency in the population = number of recombinant types (banding patterns) divided by the total number of plants sampled. For the drought tolerant plants the recombinant type is the 400 band. Here we see 6/10 plants with the recombinant type. Likewise, in Gel B we count 5/10 plants with the recombinant type (G200 is recombinant in the drought sensitive plants). 200 400

Which result indicates the drought tolerance locus is genetically linked to the G microsatellite? (these are F2 results from the testcross) Gel C 10 drought tolerant plants analyzed: 200 400 Gel D 10 drought sensitive plants analyzed: Notes: C & D For tolerant plants (C) the 200 band is parental and the 400 band is recombinant. We see two 2 recombinant types/10 plants. For the drought sensitive plants, 200 is the recombinant type. We see 3 recombinant types/10 plants Combined, that is 5 recombinant plants/20 *100%= 25% Lecture Script: After a few minutes, pose the clicker question on slide 27 to determine if students are on track. 200 400

What is the map distance?
Is the drought tolerance locus linked to the G microsatellite? F1 x tester  F2 From your analysis of the four gels, which (if any) indicates the drought tolerance locus is genetically linked to the G microsatellite locus. Gels A & B Gels C & D Gels A & D Gels B & C None of the results indicate linkage Notes: C & D. See notes in slide 24 and 25 for explanations. “What is the map distance?” In Gel C, drought sensitive plants, we see two 2 recombinant types/10 plants. In Gel D we see 3 recombinant types/10 plants. Combined, we have 5 recombinant types/20 plants  5/20*100 = 25% recombinants or 25 map units. Lecture Script: To determine linkage, calculate the recombinant frequency in the population (number of recombinant type banding patterns divided by the total number of plants sampled). If the recombinant frequency is less than 50%, we can assume that the two loci are genetically linked. Recall that the 400 band is recombinant in the drought tolerant plants and the 200 band is recombinant in the drought sensitive plants. When we asked the question on slide 27, requiring students to indicate which of the four gels, if any indicates, that the drought tolerance locus is genetically linked to the G microsatellite locus, only 50% of students selected the correct answer, whereas the other 50% were selecting the two gels containing banding patterns that are indicative of independent assortment. Upon seeing this result, we cued the students by saying “Recall which banding patterns are recombinant and which are parental, and pay close attention to what kind of plants – drought tolerant or drought sensitive – are being tested”. Students were given another couple of minutes to discuss with one another and analyze the gels, and then revote. After this cued peer discussion, 94% of students selected the correct answer. You can tell the class to assume that Gels C & D (showing genetic linkage between the DR and G loci) are the results of our mapping experiment, and then ask them “What is the map distance between DR and G?” In Gel C, drought sensitive plants, we see two 2 recombinant types out of 10 plants. In Gel D, we see 3 recombinant types out of 10 plants. Combined in the population of plants we analyzed, we have 5 recombinant types/20 plants: 5/20 x100% = 25% recombinants or 25 map units. What is the map distance? CLICKER Question

Summary Always determine what allele combination is considered parental, and which is recombinant. Draw the chromosomes of the parents and F1 to be sure you keep track of information correctly. If you do an F1 testcross you can analyze the entire F2 population to determine linkage. You always know what gamete genotype the tester donates, so the F2 phenotype proportions are representative of the recombination events (and frequencies) that happened in the F1. This is a good place to stop. Part 2, analyzing linkage via an F1 x F1 dihybrid cross is the next lesson.

Part 2: Determining linkage when we do a dihybrid cross

What if we do not cross the F1 to a homozygous recessive plant, but instead we let the F1 self?
DRT G200 / DRS G400 DRT G200 / DRS G400 Drought sensitive Drought sensitive X DRS G400 DRS G400 DRT G200 Notes: During this class we will explore how to map the location of the drought tolerance locus if the F2 we analyze are the product of an F1 self. Practically speaking, one reason we might let the F1 self instead of doing at testcross is that it can be very laborious to force crosses between plants. The first image shows the F1 self (swooping X), the animation replaces that with another F1, because an F1 self is like crossing two F1 plants. Lecture Script: Provide the F1 x F1 scenario to students and then move on to the activity. Practically speaking, one reason we might let the F1 self instead of doing a testcross is that it can be very laborious to force crosses between plants. Thus, just allowing the F1s to self-fertilize can be easier. DRT G200

Plants are grown under drought conditions
Experimental Plan Plants are grown under drought conditions Selfed Many hundreds of F2 plants. Screen phenotypes & use PCR for the microsatellite marker DRT G200 / DRS G400 Notes: This (and the next slide) is a reminder of the experimental process. Collect seeds and plant them. The next sequence of activities will help them predict what proportion of plan phenotypes/molecular marker banding patterns they will observe if the DR locus is or is not genetically linked to the molecular marker. Lecture Script: Slide 31 is a reminder of the experimental process: Self-fertilize the F1, generate a large F2 population, screen the F2 population for drought tolerant and sensitive phenotypes, and extract DNA from plants to amplify the microsatellite locus by PCR.

Insert a picture of a growth chamber, such as the picture found at: Notes: A large number of plants might be grown in the field (top left)a greenhouse (bottom) growth chamber (top right) or in Crop in field image: Creative Commons Deed CC0, from Greenhouse: “Strawberry Greenhouse” Creative Commons, Joi Ito via Wikipedia Commons Lecture Script: Slide 32 is optional, to remind or introduce to students that our experimental plants could be grown in the field, a greenhouse, or climate-controlled growth chambers.

P generation: X DRS G400/DRS G40 DRTG200 / DRT G200 F1: DRT G200 / DRS G400 Self-fertilized Drought sensitive Notes: Students assemble into groups of 6 and then break into 2 mini-groups: 3 people in each mini-group, MAKE PREDICTIONS BASED ON WHICH SCENARIO YOU WERE ASSIGNED (LINKED OR UNLINKED). Come back together in your group of 6. COMPLETE HANDOUT B. Teach each other about your predictions and identify how they are different, and how they are different from when we did an F1 x tester cross. The next slide also illustrates this modified jigsaw activity (group – split – regroup Lecture Script: “Recall our parental plants were two true-breeding strains; one was drought tolerant (DRTG200 / DRT G200) and the other was drought sensitive (DRS G400/DRS G400). Crossing these parental strains generated a heterozygous drought sensitive F1 (DRT G200/ DRS G400). We will now explore mapping the unknown mutation/locus using the F2 generated when the doubly heterozygous F1s are self-fertilized (essentially an F1 x F1 cross). The next activity involves predicting the drought phenotypes and banding patterns of the F2 and which F2 plants we can use to determine if the DR and G loci are linked.” Students will engage in a jigsaw-type activity14. Students should form teams of six, and then the team of six breaks into two three-person mini-groups. One group of three will work on prediction #1: predict the F2 genotypes, phenotypes, and banding patterns if the DR locus is independently assorting from the G microsatellite locus. The other group of three will work on prediction #2: predict the F2 genotypes, phenotypes, and banding patterns if the DR locus is genetically linked to the G microsatellite locus. We recommend teams of six physically assemble in one spot. You introduce the mini-group goals, and then have the mini-groups physically split from their team of six (e.g. mini-groups working on prediction #1 could work on one side of the classroom and mini-groups working on prediction #2 on the other side of the classroom). Each group will work through a set of questions on Handout A and make their predictions (drawings, Punnett Squares, etc.) on a sheet of flip-chart paper (about 27” x 34”) which they will tape up on the wall to share with their team of six during the second phase of this activity. (If you do not have the space, you can have groups work directly on the handout, omitting the flip-chart component.) Provide each mini-group with a sheet of flip chart paper and a couple of colored flipchart markers and either Handout A1 (predicting results if the loci are linked, Supplemental Material S2a) or Handout A2 (predicting results if the loci are not linked, Supplemental Material S2b). Encourage them to draw the gametes, the chromosomes labeled with alleles, and the expected bands on a gel for the F2 population. Mini-groups can complete their predictions in approximately minutes. We recommend that instructors and/or teaching assistants circulate around the room, asking questions and soliciting explanations about the mini-groups’ drawings. Also encourage groups to complete the questions on Handout A, as these questions will be needed for the second phase of the activity, when mini-groups reassemble into their team of six to teach one another about their predictions. In our experience, many students have a hard time determining which banding patterns will increase and decrease if genetic linkage is occurring. The key is to look at the gametes produced by the F1 and identify which are parental and recombinant. We can then adjust the frequencies of each gamete type according to our scenario: if there is genetic linkage, the recombinant types will be lower than predicted for independent assortment (i.e. each recombinant gamete type is expected to occur at a frequency of 25% if independent assortment occurs), and that decrease in recombinant types will impact the frequency of each banding pattern observed. Figure 4 shows an example of a Punnett Square drawing used to make predictions. The answers to Handout A questions are described below for slides 36 and 37 and directly in slide of the Lesson Slides (Supplemental Material S1). Predict what the F2 data will look like if there is linkage Predict what the F2 data will look like if there is independent assortment

Prediction Activity You have 15 minutes!
Students assemble into groups of 6 and then break into 2 mini-groups: 3 people in each mini-group. You have 15 minutes! Each group, on the flip chart paper: draw gametes, chromosomes with alleles labelled, banding patterns on a gel. Put your work on the wall when you are done. COMPLETE HANDOUT A 3 people will “predict what the F2 will look like, and banding patterns, if marker and DR locus are linked” 3 people will predict what the F2 will look like, and banding patterns, if marker and DR locus are not linked (assort independently) Notes: Have students assemble into groups and then break into their mini groups. Give the mini groups a version of handout A, either predicting if independent assortment is occurring or predicting if linked. These questions are designed to help them make the predictions. They should Then ask a representative from each group come and pick up a flip chart page and markers (or have teaching assistants hand out if possible). We recommend giving them a time limit to encourage efficient group work. You can adjust if needed, but we recommend walking around the room while they work and encouraging them to get working right away. Come back together in your group of 6. Teach each other about your predictions and identify how they are different, and how they are different from when we did an F1 x tester cross. COMPLETE HANDBOUT B. Class Slide 1

Instructor answer slide
Notes: This slide is included in the lesson slides in the event that you feel the students should see this completed Punnett Square as feedback for their Handout A and B work. We recommend you do not provide it to students until after this entire lesson is complete, as post-lesson feedback. Lecture Script: In order to assess linkage, we must know the genotype of the plant at both loci. Recall the testcross scenario from Part 1 of this lesson: we could accurately infer the genotypes of all the plants because we knew that all of them inherited the DRT allele from the tester plant, therefore the tolerance/sensitive phenotypes observed in the F2 from the testcross indicated which DR allele the F2 plants inherited from the F1. However, in the F1 x F1 dihybrid cross scenario we can only accurately infer the genotypes of the drought sensitive F2 plants because tolerant is recessive to sensitive. For this reason, we can only use the drought tolerant F2 plants to assess linkage we would only know the genotypes of the drought tolerant plants because they have to be homozygous recessive to be tolerant. If there was no linkage, we would expect the frequency of each gamete type to be 0.25. If linkage is occurring, the frequency of parental gamete types will go up, and the recombinant types will go down.

F1x F1  F2 Consider the drought sensitive F2 plants
What is recombinant? 200 400 200 400 200 400 This is to review question A and B from Handout A, and question 1 from Handout B. Notes: E. Each can contain a recombinant band. Look at what gametes can make the PHENOTYPE of drought sensitive in the Punnett square. Remember we need to recall the original DR and G allele combinations in the parent plants (those that made the F1) to decide what is parental and what is recombinant. There are three drought sensitive plants that are homozygous G200: 2/3 of them are the result of a parental chromosome and a recombinant chromosome uniting, 1/3 are the result of two recombinant chromosomes. If the loci are linked we expect the parental types to go up and recombinant to go down. Expect het and homozygous G400 to go up. Lecture Script: When working through these clicker questions, we ask students to answer them alone for the first vote. If the responses indicate confusion warranting peer instruction (e.g. less than 70% of the class select the correct answer), students could be cued to discuss their answer choice, and reasons for their answer choice, with their peers before re-voting. Pose the question on slide 36, which asks students to identify which banding patterns in drought sensitive F2 plants is/are recombinant (question 1 from Handout B). For the drought sensitive F2 plants, each banding pattern could represent a recombinant genotype. In our experience, as only 33% of students selected the correct answer on the first vote (Figure 5A and 5B) students still struggle to select the correct answer, even after engaging in the group discussion activity with Handout B. Approximately 50% of students did not select the 200/200 banding pattern as one that could be represent a recombinant gamete genotype. In a drought sensitive F2 plant, the 200/200 band could be the result of a parental and recombinant gamete combining (DRT G200/ DRS G200), or it could be the result of two recombinant gametes combining (DRS G200/ DRS G200). When only 33% of students selected the correct answer on the first vote, we asked students to consider what combination of gametes (parental and recombinant) could make drought sensitive F2 plants and to discuss the answers with their neighbor to systematically decide if each banding pattern could represent a recombinant gamete. Students were given a few minutes to discuss and then revote, which resulted in 80% selecting the correct answer. After the second vote, we used the document camera to draw the crossing-over events in the F1 at the chromosomal level, indicate the resulting gamete genotypes, label the genotypes as parental and recombinant, and predict the resulting banding patterns generated from the union of various gamete types. B. A. C. B and C Each can represent a recombinant CLICKER Question

Each contains a recombinant band
F1 self. Drought tolerant F2 plants only What is recombinant? 200 400 200 400 200 400 B. This is to review question A and B from Handout A, and question 2 from Handout B. Notes: D. B and C. Remember the tolerant allele (DRT) was originally inherited with the G200 allele in the F1 plant (from the parent plant). We know the drought tolerant F2 plants must be DRT/DRT because the tolerance allele is recessive to the sensitive. Therefore, a gamete that has the DRT allele combined with a G400 allele is recombinant (the result of recombination between chromosomes if these loci are on the same chromosome). the The G400 band is a recombinant type. If linked, we expect the frequency of G200 bands to go up, so the homozygous G200, the G400 homozygous types and heterozygous types will go down. Lecture Script: We then pose the clicker question on slide 37, which asks students to identify recombinant bands in the drought tolerant F2 plants (Question 2 of Handout B). In our experience, after the discussion and feedback on the previous clicker question, students are fairly successful at identifying the correct answer in this question (74% correct, Figure 5C and 5D). A. C. B and C Each contains a recombinant band CLICKER Question

Which plants would you use to test for genetic linkage?
A. The entire F2 population B. The F2 drought tolerant plants only C. The F2 drought sensitive plants only This is review of question 3 from Handout B. Notes: B. They are homozygous recessive so you can be sure of their genotype Lecture Script: The clicker question on slide 38 reflects on their answers, by asking students to select which population they would choose for testing genetic linkage between the DR and G loci using F1 self-crossing (Question 3 of Handout B): the entire F2 population, drought tolerant plants only, or drought sensitive plants only. After discussions from the previous two questions, most students select the correct answer on the first vote (the drought tolerant F2 only). CLICKER Question

10 drought tolerant F2 plants analyzed:
Based on these PCR results (banding patterns), what is the linkage/map distance between the DR and G loci? 10 drought tolerant F2 plants analyzed: 200 400 Notes: Note that 10 is far too small a sample size, we’re just doing 10 here as a simple example. Ideally you would probably want 50 or more plants. 4 recombinant chromosomes/20 chromosomes total = 20% recombination, 20 map units If students are choosing D it is because they are calculating by counting each plant – not each chromosome. We cannot necessarily call an entire banding pattern “recombinant” or “parental” (Like we could with an F1 x homozygous recessive) because the F1 was selfed – so each gamete that made the F2 plant can carry a parental or recombinant combination of alleles. IN comparison, when we did an F1 x homozygous recessive cross the homozygous recessive (tester) plant can only donate gametes of one type, so we only have recombinant types coming from one plant – the F1. With an F1 self (or F1 x F1) recombinant types can come from each plant, so we need to count the number of recombinant chromosomes out of the total number of chromosomes that could represent a recombination. Since we know the genotype of the homozygous recessive drought tolerant plants we know if each band on the gel represents a parental combination of alleles or recombinant. So, to calculate recombination frequency we need to calculate the number of recombinant chromosomes/total number of chromosomes. Each plant has two chromosomes (if diploid) – so in this simplified example we have 4 recombinant chromosomes/20 chromosomes = 20% recombination, or 20 map units. Lecture Script: Here, we show the banding patterns for 10 drought tolerant F2 plants. Students are asked to “review the banding patterns and compare the patterns to what they learned by making the two predictions, linked vs. independent assortment”. Then, based on their analysis, they are asked to determine if these results indicate that the DR and G microsatellite are genetically linked and, if so, to calculate the map distance between the two loci (clicker question on slide 39). Allow students to vote, and then share an explanation with them. “In the case of drought tolerant F2 plants, the 400 band is recombinant; hence, we would expect to see an equal number of 400 bands and 200 bands if the DR and G loci were assorting independently. Fewer than 50% recombinant bands suggests the two loci are linked. The gel on slide 39 has banding patterns for 10 plants: 20 bands, representing a total of 20 G loci (or 20 chromosomes, or 20 gametes). Of these, there are four 400 bands (2 heterozygotes and one homozygous 400/400). If the DR and G loci were assorting independently, we would expect at least 10/20 bands to be 400; thus, the results suggest genetic linkage. To calculate map distance, we divided the number of recombinant bands by the total number of bands on the gel: 4/20*100% = 20% recombinants, or 20 map units.” Note that, at the point in the lesson when we ask the question on slide 39, we have intentionally not yet talked explicitly about how to calculate map distance in the F1 x F1 scenario, and that the method used here differs from that used in the testcross scenario. If students apply the “rules” of calculating recombinant frequency and map distance that we learned in the testcross scenario they will answer that the loci are linked and they are 30 map units apart, because there are 3/10 plants with the 400 band. In our experience, most students are likely to make this mistake, which we use as an opportunity to have a meaningful discussion about why we calculate map distance differently in the F1 x F1 scenario compared to the testcross. Not linked Linked but cannot determine distance Linked, 20 map units apart Linked, 30 map units apart. CLICKER Question

F1 x x How do we calculate map distance? F1 self or F1 x F1
homozygous recessive DRT G200 / DRS G400 DRT G200/ DRT G200 Only gametes from the F1 can contain recombinant genotypes (chromosomes) Each F2 plant receives one gamete from the F1, so only one “band” in each F2 plant can be parental ore recombinant. Count the number of recombinant bands/total number of plants Gametes from each F1 can be parental or recombinant. Both gametes that make the F 2plant come form the F1, so both “band” in an F2 need to be classified as either parental or recombinant Count the number of recombinant bands/total number bands Notes: This is a summary of the different methods to calculate map distance depending on which cross was performed. Animations are used to walk through the comparison between the two methods. The emphasis of this discussion should be the gametes, and gamete genotypes, that make up the F2 plants When looking at an individual F2 plant, how many of the gametes that combined to make that plant could represent be a parental or recombinant type (genotype/chromosome/combination of alleles)? If two loci are genetically linked, then a recombinant gamete genotype represents a recombination event between the two parental chromosomes. In our example, the 400 band amplified from drought tolerant F2 plant DNA indicates that the plant carries a recombinant chromosome. In the F1 x homozygous recessive testcross scenario it is only the gamete genotypes from the F1 plant (and not the tester plant) that could be parental or recombinant, so an F2 plant can only represent one parental genotype or one recombinant genotype: hence we can classify a plant as either parental or recombinant. In comparison, in the F1 x F1 cross both gametes that make an F2 plant could be parental or recombinant, so each plant classified as either parental or recombinant. Instead we have to consider each gamete contribution (which we do by assessing the plant drought phenotype and G microsatellite alleles by analyzing the bands on the gel). Therefore, each plant represents two gamete contributions, so we count every band and divide by the total number of bands. Lecture Script: Slide 40 walks through the logic behind why simply counting plants that contain at least one recombinant chromosome is not the correct way to calculate map distance in this situation. Slide 40, comparing the use of a testcross and the F1 x F1 cross to assess genetic linkage: The emphasis of this discussion should be on the gamete genotypes that make up the F2 plants. “When we look at an individual F2 plant, how many of the gametes that combined to make that plant could represent a parental or recombinant type (genotype/chromosome/combination of alleles)? If two loci are genetically linked, then a recombinant gamete genotype represents a recombination event between the two homologous parental chromosomes. In the F1 x homozygous recessive testcross scenario we did during the first part of this lesson (DRT G200/ DRS G400 x DRT G200/ DRT G200  F2), only the gamete genotypes from the F1 plant (and not the tester plant) can be identified as parental or recombinant. In other words, only one of the DR-G allele combinations on one chromosome in these F2 plants can be classified as parental or recombinant. Hence, we can classify the entire F2 plant as either a parental organism or a recombinant organism with respect to the two loci we are testing. To determine map distance we calculate the frequency of recombinant organisms in the total population analyzed.” “In comparison, in the F1 x F1 cross (DRT G200/ DRS G400 x DRT G200/ DRS G400  F2), both gametes that make an F2 plant can be either parental or recombinant. As a result, we have to classify the DR-G allele combination on each chromosome that the F2 plant inherited as either parental or recombinant. With respect to the DR-G loci, one F2 plant may have no recombinant chromosomes, one recombinant chromosome, or two recombinant chromosomes. To determine recombination frequency we only analyze the plants that are homozygous recessive (i.e. DRT/ DRT, drought tolerant). So, in this F1 X F1 cross, to determine recombinant frequency (and map distance), we must count every recombinant band and divide by the total number of bands.”

F1 self-fertilized and F2 drought tolerant plants analyzed.
X G200 /G200 G400 /G400 G200 /G400 Drought tolerant Sensitive F1: sensitive Drought tol. Wilting F1 (parent)) (parent) Drought tolerant F2 plants only CONTROLS The F2 banding pattern for all of the drought tolerant plants is the same as the banding pattern shown above. Which is the best explanation for these F2 results? The drought tolerance locus and the molecular marker are: Genetically linked and some cross-overs occur between them. Genetically linked and no cross-overs occur Assort independently This result is impossible. Notes: Answer is B. All of the plants are 200/200 indicating they only inherited DRTG200 (and not the recombinant genotype DRTG400). This suggests that the F1 is only producing parental gametes (or a very small proportion of recombinant genotypes), suggesting that the DR and G loci are genetically linked and so close together that little or no crossing over occurs. You could also ask: If this were true, no crossing over, what would you expect for the F2 that were drought sensitive? Draw Punnett square. ¼ will be DRT200/DRT200, ½ will be DRT 200/DRS400, ¼ will be DRS400/DRS400. So, of the sensitive phenotype plants 2/3 of them will be 200/400, 1/3 will be 400/400. Lecture Script: This question can be used if time permits. The students are asked to analyze the banding patterns for drought tolerant F2 plants produced by an F1 x F1 cross. All of the plants we analyze have the same homozygous 200/200 banding pattern. After the students vote, as feedback, you can tell them: “in this case, all of the F2 plants have the same banding pattern (200/200), which indicates that all drought tolerant F2 plants inherited DRTG200 and not the recombinant genotype DRTG400. One reason for no recombinant genotypes showing up is that the DR and G loci are so tightly linked (i.e. very close to each other on the same chromosome), that little to no recombination by crossing-over occurs between them. It is possible that recombination does occur, but at such a low frequency that we do not see it in our sample size.” You could extend this idea by asking the class “If no recombination occurs between the two loci, what genotypes and banding patterns do you expect in the drought sensitive F2?” In this scenario, we predict the F2 of the F1 x F1 cross to be: ¼ DRT G200/DRT G200, ½ will be DRT G200/DRS G400, ¼ DRS G400/DRS G400. Among the drought sensitive plants 2/3 of them will be 200/400, 1/3 will be 400/400. CLICKER Question

Practice Assessing Linkage by analyzing results of an F1 x F1 cross
New trait of interest: potatoes that are resistant to the Colorado Potato Beetle. a pest that destroys millions of dollars of potato crops each year (and the beetle is resistant to most insecticides) You have found a potato plant that potato beetles do not attack. Is the plant producing something that deters attack from the beetle (or kills the beetle)? Notes: Image from: Creative Commons licence 3.0 Lecture Script: This portion of the lesson involves students assessing genetic linkage between molecular markers and a new trait of interest. Here, we present the example of resistance of potato plants to a common pest, the potato beetle. “We have identified a true-breeding potato plant that is resistant to potato beetle attacks: either the beetles do not attack this plant or they die when they eat it. Is the plant producing something that deters beetle attack (or kills the beetle)? Our goal is to locate the gene and mutation involved in resistance so we can better understand the mechanisms of resistance in the plant.” © 2008, Pilise Gábor. Creative Commons 3.0

E C F D A B Goal: Find the locus involved in beetle resistance (BR)
Chromosome 1 2 4 5 3 E C F D A B Notes: Remember our original goal – identify the location of the BR locus in the genome. Note that potatoes are actually 4n = 48, but I’ve only shown 5 chromosomes here for simplicity. Lecture Script: Slide 43 and 44 introduces students to the molecular markers we will use, the alleles we have assigned to the beetle resistant and susceptible phenotypes, and the crossing scheme. Students are asked to work in groups of three, and each group receives a copy of Handout C (Supplemental Material S4), which has six gels (one gel for each molecular marker). Tell the class that the first group to determine the correct genetic linkage, including the correct map distance, for all six markers wins the challenge. In the past we have given the winning group some chocolate bars, but we have also simply awarded the winning team “bragging rights”. The group that finishes first can be encouraged to either raise their hand or come to the front of the room and explain their answer to us, or to write their answer on a page projected by the document camera. We’ll call the locus involved in potato beetle resistance BR BRR is the resistance allele BRS is the wildtype allele (potatoes are susceptible to beetle attack) BRS is dominant to BRR We have six molecular markers we are using: A-F

Which F2 plants will we select for our analysis? - Beetle resistant
HANDOUT C In your teams of 3, analyze the banding patterns and decide if linkage is occurring between the BR locus and the microsatellite markers. Susceptible plant F1s: Susceptible Resistant plant X BRS A400/BRS A400 BRR A200/BRS A400 BRR A200/BRR A200 Marker Allele in resistant parent plant Allele in susceptible parent plant A 200 400 B 350 100 C 90 550 D 50 75 E 675 600 F 220 410 F1s self-fertilized Which F2 plants will we select for our analysis? - Beetle resistant - Normal Notes: Remember we want to analyze the F2 that are homozygous recessive because we can infer their genotypes completely (and therefore determine genetic linkage). So in this case we will look at the beetle resistant F2 plants.

Calculate map distance
Use handouts with gel data from different markers to determine the location Determine linkage Calculate map distance First group of 3 with the correct answer wins bragging rights!

A B For your interest: What do you do next?
If the genome for this plant is sequenced: Look at the genes in this approximate location. At a specific map location there could be anywhere from genes depending on the genome. Depending on the phenotype of interest it could be difficult or easy to make educated guesses as to possible target genes. For species like Arabidopsis there are databases with additional sources of helpful information such as whether any gene is expressed under specific conditions, including attack by pests such as the beetle. Genes in this region could be checked to see what their wild-type expression pattern looks like under pest attack. You could compare the DNA sequence of your beetle resistant plant to wildtype for your most likely candidate genes to look for sequence differences causing the mutant phenotype. You could take a wild copy of a likely candidate gene and add it to the mutant to see if the wild type phenotype is restored. A B Lecture Script: Recall that these crosses and linkage analysis were part of a larger goal: identify the drought tolerance (or beetle resistance) locus. The information on this slide pertains to what one might do next, once the locus of interest has been mapped to a specific region of the genome. “Ultimately, we would want to identify the gene and the specific mutation that are involved in resistance to the beetle and, hopefully, elucidate the function of the gene, as well as the mechanism by which this gene and gene product contribute to drought tolerance.” BR, Beetle resistant locus

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