Fall2012week 9 Turorial Crossing Over Linkage

Teaching Genetic Linkage and Recombination through Mapping with Molecular Markers

Lisa McDonnell and Jennifer Klenz

Note for instructors: this can be assigned as homework or for students to work on during a TA-led tutorial. The last question involves student’s analyzing banding patterns. The gels are in a separate file (Handout D, Supplemental Material S5b).

Problem Set - Mapping genes with molecular markers – Student version

Relevant Learning Objectives:

• From analysis of data from a cross (phenotypes and/or genotypes) recognize crosses that must involve linked genes. Be able to justify your analysis by describing the information in the data that allows you to determine genes are linked.
• Explain how crossing over (or recombination frequency) between two linked loci affects the genotype frequencies of the products of meiosis compared to loci that are unlinked.
• Infer the genotypes of individuals involved in a cross and explain how recombinant allele configurations were created from the parental arrangements.
• Describe what a molecular marker is, and how it can be used to map the location of genes/loci.
• Describe the crosses you would do, and why, to map the location of a gene with molecular markers. Be able to predict the outcome of the crosses if the gene(s) is linked or unlinked to the marker.
• Use molecular markers to map the location of genes (and be able to explain how this is dependent on linkage analysis)

Dutch elm are a kind of tree. Some are very susceptible to developing Dutch elm disease, which has destroyed trees throughout North America. Trees affected by the disease exhibit a leaf wilting phenotype. There is great interest in identifying genes involved in resistance to the disease. A variety of elm trees called “Jefferson” has been identified to be resistant to the disease. The resistance phenotype (measured by wilting of the leaves when a tree is infected with the disease) is recessive to wildtype (wilting phenotype). Most other elm varieties, such as one called “Pioneer”, are susceptible to the disease. A number of Simple Sequence Length Polymorphisms (SSLPs) molecular markers (a kind of microsatellite) that differ between Jefferson and Pioneer have been identified and mapped on every chromosome of the elm genome. See examples of 3 markers in the table and gel below (there are many more because elm has 28 chromosomes). In the gel below we are looking at the banding pattern for 3 markers which are described in this table.

Note that Jefferson is referred to as “Jeff” or “J” here, and the Pioneer variety of elm are called “Pio” or “P” here.

A true breeding Jeff elm was crossed with a true-breeding Pio elm to generate a heterozygous F1 (J x P).

DNA was extracted from the two parent plants and the F1, and then the three different SSLP markers were amplified using PCR.

a) How could a variety of elm such as Jefferson be resistant to Dutch Elm Disease, but another variety, such as Pioneer, not be resistant?

b) Explain what the following statement means in your own words:

“A number of SSLP molecular markers that differ between Jefferson and Pioneer have been identified and mapped on every chromosome “

c) Explain what you know about the molecular differences between the Jefferson and Pioneer alleles for each of the 3 sets of markers (chr1.1, chr 1.2 and chr 4.1) based on the bands in the gel above.

d) Diagram a genetic crossing scheme that would allow you to map the location of the Jefferson locus involved in resistance to Dutch elm disease to a specific chromosomal location. Because you will need to isolate DNA, perform PCRs and run the results on an electrophoretic gels, explain which individuals in your crossing scheme you will use for your marker analysis. Predict what the data will look like if there is genetic linkage between the resistance locus and a specific SSLP, and predict what the data will look like if the resistance locus is not genetically linked to any of the SSLPs.

When you have completed your predictions, and they have been checked by your TA, the TA will give you some banding patterns to analyze and determine linkage between the resistant locus and various molecular markers.

ANSWERS – for Instructors

Note for instructors: this can be assigned as homework or for students to work on during a TA-led tutorial

Problem Set - Mapping genes with molecular markers – Student version

Relevant Learning Objectives:

• From analysis of data from a cross (phenotypes and/or genotypes) recognize crosses that must involve linked genes. Be able to justify your analysis by describing the information in the data that allows you to determine genes are linked.
• Explain how crossing over (or recombination frequency) between two linked loci affects the genotype frequencies of the products of meiosis compared to loci that are unlinked.
• Infer the genotypes of individuals involved in a cross and explain how recombinant allele configurations were created from the parental arrangements.
• Describe what a molecular marker is, and how it can be used to map the location of genes/loci.
• Describe the crosses you would do, and why, to map the location of a gene with molecular markers. Be able to predict the outcome of the crosses if the gene(s) is linked or unlinked to the marker.
• Use molecular markers to map the location of genes (and be able to explain how this is dependent on linkage analysis)

Dutch elm are a kind of tree. Some are very susceptible to developing Dutch elm disease, which has destroyed trees throughout North America. Trees affected by the disease exhibit a leaf wilting phenotype. There is great interest in identifying genes involved in resistance to the disease. A variety of elm trees called “Jefferson” has been identified to be resistant to the disease. The resistance phenotype (measured by wilting of the leaves when a tree is infected with the disease) is recessive to wildtype (wilting phenotype). Most other elm varieties, such as one called “Pioneer”, are susceptible to the disease. A number of Simple Sequence Length Polymorphisms (SSLPs) molecular markers (a kind of microsatellite) that differ between Jefferson and Pioneer have been identified and mapped on every chromosome of the elm genome. See examples of 3 markers in the table and gel below (there are many more because elm has 28 chromosomes). In the gel below we are looking at the banding pattern for 3 markers which are described in this table.

Note that Jefferson is referred to as “Jeff” or “J” here, and the Pioneer variety of elm are called “Pio” or “P” here.

A true breeding Jeff elm was crossed with a true-breeding Pio elm to generate a heterozygous F1 (J x P).

DNA was extracted from the two parent plants and the F1, and then the three different SSLP markers were amplified using PCR.

a) How could a variety of elm such as Jefferson be resistant to Dutch Elm Disease, but another variety, such as Pioneer, not be resistant?

The resistance is genetically based (inherited) rather than due to environmental causes

b) Explain what the following statement means in your own words:

“A number of SSLP molecular markers that differ between Jefferson and Pioneer have been identified and mapped on every chromosome “

There are a number of polymorphisms that differ between these two types that can be exploited for linkage mapping. The polymorphisms are visible as different sized bands on gels after PCR reactions are performed.

c) Explain what you know about the molecular differences between the Jefferson and Pioneer alleles for each of the 3 sets of markers (chr1.1, chr 1.2 and chr 4.1) based on the bands in the gel above.

For chr1.1 Jefferson has more nucleotides in the region amplified by the primers compared to Pioneer. For chr1.2 this time Jefferson has fewer nucleotides than Pioneer does within the region amplified by the PCR primers. Overall the chr1.2 primers amplify a smaller region than 1.1. For chr4.1 Jefferson has more nucleotides in the amplified region than Pioneer does but the size difference between the two alleles is not as great as were seen in 1.1 and 1.2 (ie the two alleles are closer in banding size because there the number of nucleotides that differs between the two is not as great).

TA guide – things you might need to ask students as they work on this problem:

• Consider that the morphological phenotypes are also related to changes in DNA sequence.
• What might help you is you draw a little mini gel beside your labeled alleles on chromosomes and put different band sizes for the different alleles.
• What does linkage means in terms of parentals vs. recombinants?
• What are parental phenotypes for this experiment? What are recombinant phenotypes?
• What kind of proportions of phenotypes should they see for independent assortment….what would this mean in terms of bands on gels?
• What about if linkage is occurring (not complete linkage, but some recombination)?
• One thing you could do is ask them to choose a hypothetical linkage (e.g. 10 or 20 map units) and draw their gel based on that. Then they can swap drawings with another group and ask them to analyze their gel.

There is more than one way you could do your cross and obtain data. You could do a test cross or you could do a dihybrid. TAs: you should check student work and encourage them to consider both options, as they will need to be familiar with a testcross and an F1 self.

If you do a dihybrid:

then you only need to look at data from homozygotes for wilting resistance to see whether you get a 1:2:1 or not. This means ¼ homozygous high bands: ½ heterozygous to ¼ homozygous low bands (the band sizes will depend on which alleles the wilting parents and non-wilting parents had). The 1:2:1 ratio on the gel would suggest independent assortment. Some students might say that the gel should show an equal number of parental vs. recombinant bands to indicate independent assortment. Not everyone understood that this would like on the gel. It is ok to count homozygous high bands as 2 high bands, homozygous low bands as 2 low bands and a heterozygote has 1 high and 1 low. So if you count this way you will indeed see ½ high and ½ low.

If there is linkage it means you get fewer recombinants. This is where it is very important to know which band represents a recombinant chromosome. When examining the homozygous recessive plants from an F1 self you are counting the number of recombinant bands (and each band represents a chromosome), and you divide that by the total number of bands (chromosomes) NOT the number of recombinant plants over total plants.

If you do a testcross what do you expect as a result that would indicate linkage or independent assortment? You would expect a 1:1 ratio of parental types to recombinant types if you were examining just the phenotype of interest plants, and the same result if you were examining the other phenotype plants. In a testcross situation it is OK to count the number of recombinant plants divided by the total number of plants (because all plants will get one chromosome from the tester plant so we really only have one chromosome per plant that could represent a recombination)

Here is a result on a gel where the nonwilting (n.w.) phenotype from Jefferson is linked to marker chr1.1. It is not complete linkage (recombinant types are still produced, but at a lower frequency than independent assortment would predict). THIS IS FROM AN F1 SELF, LOOKING AT THE HOMOZYGOUS RECESSIVE (JEFFERSON) F2 PLANTS ONLY

The F2 nonwilting plants mostly look like the Jefferson parent. Parentals are homozygous high bands with the wilted phenotype. Recombinants are low bands. Note there is one homozygous low band (n.w.14) so that is counted as two recombinant bands. On this gel you have to consider that there are really 42 F2 bands. You have 6 low bands from the heterozygotes and the 2 low bands from the homozygous low band plant

Recombinants/total *100% =recombinant frequency = map distance

8/40*100% =20% or 20 map units between the nonwilting mutation and marker 1.1

Here is a result on a gel where the wilting phenotype from Jefferson is assorting independently from marker 4.1

You see ½ high bands and ½ low bands.

You could also show data from F2 individuals that wilt (the dominant phenotype), but some will be homozygous wildtype and some will be heterozygous which complicates things more. If you just look at F2 homozygous non-wilting (disease resistant) mutants anytime they get a Pioneer-sized band they know it had to result from crossing over.