Lecture Script for “Teaching Genetic Linkage and Recombination
through Mapping with Molecular Markers (McDonnell and Klenz)
Part 1: Mapping Genes using Molecular Markers in a Test Cross (Dihybrid X
homozygous recessive strains)
Slides 1 – 5: Context of lesson and brainstorm activity
Slide 6, Setting up the example
“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 rice9-10 and corn11), 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 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.”
Slide 7, we discover a drought tolerant plant
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.”
Slide 8, the goals of our experiment
1) Is the drought tolerant phenotype heritable?
2) Is it caused by a single gene?
3) Mapping – where in the plant’s genome is the mutation/locus involved in drought tolerance?
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Slide 9-10, is drought tolerance heritable and caused by one gene?
“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 wildtype/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 truebreeding 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.”
Slide 10, what do we not know?
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
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tolerance! And for that matter, we do not know how the drought tolerance allele differs from the wildtype 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.
Slide 11, mapping goal and introducing genotypes
“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”.
Slide 12-13, microsatellite markers for mapping genes
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).”
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.”
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Slide 14, how do we use microsatellites to map genes?
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.”
Slide 15-16, Summarize the experimental plant and the parental plants we are using
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.”
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.
Slide 17, review the experimental plan and which crosses we will do next
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).
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Slide 18-19, generating a heterozygote and predicting the F1 phenotypes
“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. 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.”
Slide 20-22, predicting outcomes of the testcross
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 and slide 22). 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).
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
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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.
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.”
Slide 23, How to calculate the frequency of recombinants
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.”
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Slide 24-28, analyze banding patterns to determine linkage
“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. After a few minutes, pose the clicker question on slide 27 to determine if
students are on track. 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.
Slide 28, Summary of the Lesson
Lesson Part 2: Mapping Genes using Molecular Markers in a Dihybrid Self Cross
Slide 30-32: Introducing the F1 x F1 cross and experimental process
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. 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. 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.
Slide 33- 35, Group Activity - Predicting Outcomes of the F1 x F1 cross
“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.”
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Slide 33-35: Scenario and group activity instructions
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 10-15 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 35-37 of the Lesson Slides (Supplemental Material S1a).
Once the mini-groups have completed their predictions, ask them to re-assemble into their original
team of six and tape their posters up on the wall. Distribute a copy of Handout B (Supplemental Material
S3) to each team of six. They should now take the next 5-10 minutes to teach each other about their
predictions by working through the questions on Handout B. As an incentive to complete the questions,
we tell each group that they will need to hand-in a completed Handout B (with all six team member
names and student numbers on the worksheet) at the end of class. Once groups are done discussing, ask
students to take their seats. The students’ answers to the questions on Handout A and B will be
reviewed by going through the clicker questions on slide 36-38, with feedback provided in response to
clicker question results.
Slide 35, Punnett Square to predict outcome of cross – for instructors
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.
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Slide 36-38, feedback on F2 predictions using clicker questions
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).
Recall that drought sensitive is the dominant phenotype. 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.
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). 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).
Slide 39-40, how do we calculate map distance using F2 from the F1 x F1 cross?
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. 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.
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Explanation: “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.” 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.”
Slide 41, when two loci are so tightly linked, there is no recombination between them
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.
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Slide 42-45, additional practice assessing linkage by analyzing results of an F1 x F1 cross
This portion of the lesson involves students assessing genetic linkage between molecular markers and a
new trait of interest. Here, we present a fictitious 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.”
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.
Slide 46, additional information about what we do next
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.”
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