Introduction: Mendel's law of indepependent assortment only applies to gene pairs on separate chromosomes. Linkage analysis allows us to infer the relative position of genes on a chromosome. Today we will do a simulation of linkage, relying on our knowledge of crossing over and meiosis.

In order to understand the effects of linkage consider the following test cross involving Drosophila:

In fruit flies, a certain recessive mutant allele leads to purple eyes as opposed to the normal red eye color. A second allele at a separate locus leads to vestigial wings as opposed to normal length wings. A test cross was conducted for these two loci by the biologist J.B. Morgan in the early part of the 20th century. This test cross took F1 females from a standard dihybrid cross and crossed them with a male pure breeding for purple eyes and vestigial wings.

1. Assuming that the genes are autosomal, write the genotypes of the parents assuming independent assortment:

2. Find the expected phenotypic ratios for the offspring from this test cross. Show your work carefully:

Table 1 shows the phenotypic (eye color, wing length) results for his test cross:

You can use a chi square test as we have done before to test for linkage in which case you would calculate the expected number of each phenotype assuming independent assortment and calculate your observed chi square just as we have done before.

3. Complete the table and do a simple chi square test of the data against the hypothesis that the genes related to eye color assort independently of the genes for wing length.

Do you accept or reject the hypothesis of independent assortment? Explain your reasoning.

Usually in testing linkage an even simpler chi square test is used. The data from the test cross is condensed into a simpler table by pooling the offspring that have the same phenotype as the parents. These offspring are called the parentals. The offspring from the test cross that phenotypes with respect to the characteristics being studied are called non-parentals or recombinants. Nonparentals (non-recombinants) are observed when there is a cross over event.

They represent offspring formed from gametes in which there has been crossing over. For instance, in a standard dihybrid cross, the F1s are A B/a b if the gene loci are linked and there is no crossing over. But when there is a cross over event you get a total of for possible chromosomes after meiosis from just one cross over event. You should at this point review cross over in meiosis to convince yourself of this fact. Chromosome models are available for you or you might show what happens in prophase I of meiosis if you start out with a tetrad that looks like this:

4. Draw the recombinant chromatids resulting from a single cross over event:

This consideration allows you to test for linkage against the null hypothesis of no linkage by comparing the number of parentals observed and the number of non parentals observed against the number expected if there is no linkage-that is independent assortment.

5. Complete Table II for the fruit fly data given earlier:

Carry out a chi square test of the null hypothesis that there is no linkage using the data organized in this way. Note this chi square test only has one degree of freedom.

Note: This test is a better test to use than the first one because it controls for most simple viability effects. The viability effects of a gene or genotype refer to effects on the ability of an individual to survive as opposed to effects on the fertility of the individual. To understand this, suppose that flies with purple eyes are less viable meaning that there were less likely to survive than flies with wild type red eyes. Thus we would see fewer flies in the offspring from the test cross with purple eyes than we would expect to see. This would inflate our chi square value for our first approach. Comparing the parental versus non parental numbers cancels out this viability effect since non-parental flies with red eyes ought to have equal viability to parental flies with red eyes. So we will always do tests for linkage using this sort of chi square test.

When two genes are linked, geneticists often use a line to divide the genes on one homologous chromosome from the other. So the F1 from a standard dihybrid cross might be represented as A B/a b where A B represents a chromosome(or the chromatid of an unduplicated chromosome) with the genes A B while a b represents a chromatid from a homologous chromosome or the other unduplicated chromosome with genes a b. This is important because if you have an individual with genotype A B/a b, this individual is different in terms of its chromosome arrangement than one with genotype A b/a B !

6. Remember that two loci are said to be linked if they are on the same chromosome. For the fruit fly cross, write the genotype of the F1 females and the recessive males using fruit fly notation (use v+, v for the normal wing and vestigial wing alleles respectively and p+, p for the red and purple eye colors respectively.

7. For the female fly, write all possible chromosome combinations that can happen when she produces gametes. There will be two parental types and two non-parental types. What process in meiosis produces the non-parental types?

8. When do you think you will see the larger number of non-parental chromosome types: if the loci for the genes are far apart, say at opposite ends of the chromosome or adjacent to each other? Explain.

The amount of linkage between two loci is experimentally quantified by a quantity called r, which stands for the frequency of recombination. This is calculated by the following formula:

r = (number of recombinants / total offspring from the test cross) * 100%

9. What would be indicated if r is zero or close to it? What would the value of r be if the loci are unlinked?

The test cross procedure and the frequency of recombination can be used to map the relative locations of gene loci along a chromosome. This type of analysis has been done for many years and we have detailed genetic maps for some of the major eukaryotic organisms used in genetics.

The basic technique is the so-called three-point test cross. These crosses start put with parents that are homozygous for the dominant form of each of three loci, crossing these parents with another parent which is homozygous for the recessive form of each of three loci. The F1's are then test crossed with an individual who is homozygous recessive for all three loci, very much as we did in the previous situation.

The three point cross allows us to tell the relative order of the loci, in terms of which one is in between the other two As part of this analysis we calculate a quantity called the map distance, in terms of map units. One map unit is equal to one percent recombination. So if r between two loci is 23.8% then the loci are said to be 23.8 map units apart.

8. Suppose you have a fly with the following genotype c+ p+ v+/ c p v

List all the chromosomes that this fly can produce during meiosis and explain how each might arise in terms of crossing over between nonsister chromatids. (Hint some happen if there is no crossing over, some if there is a single cross over event and some if there are two cross over events)

Each group of four students will be given a mystery chromosome pair and generate data from a three point test cross. Then we will share the phenotype data with the whole class and do a linkage analysis to infer the map distances and order of each group's genes on their chromosome.

We will be simulating a linkage analysis as might be done with fruit flies.

A group will work with the one of the following groups of traits. I give the wild type and mutant phenotypes for each trait to be used by each group. Assume the wild type is dominant.

Group 1

Group 2

Group 3

Group 4

Group 5

Group 6

Working as a group:

For your group's traits how would you prepare a 3 point test cross? Remember you need a parent that you know is heterozygous for all three traits and has one chromosome with all dominant alleles and the other chromosome with all recessive alleles for the three traits.

9. First. describe the cross you would set up to get this heterozygous parent.

10. Using the map distance data I will give you, assign letters to your group's traits using Drosophila notation and write the genotypes the heterozygous F1 flies here:

All people in your group should agree on a the same notation to represent the genes!

In order to do the 3 point cross you cross, you must this parent with another parent which is homozygous recessive for your traits.

Write the genotype of the homozygous recessive fly here using Drosophila notation:

In fruit flies we use the male as the homozygous recessive parent because there is no cross over in male fruit flies.

11. For your mystery chromosome show the possible cross over events for the female parent in Table 1II. Make sure you clearly assign letters to the dominant and recessive alleles for each trait. For each cross over event you list the corresponding phenotypes you get when the gamete with each cross over event for the female parent is crossed with a gamete from the homozygous recessive parent. Note that there will be 8 chromosome types and eight phenotype classes!

Once you understand this, you will generate your own cross over data for the female gametes using the random number tables I provide you. Each group will know their physical map locations, which we will assume for now, are between 0 and 1. Note that physical map distance is not the same as map distance as defined here!

Each individual within your group will generate 100 gametes from the female fly in the following way. To get the first gamete, select the first random number in your random number table.

Suppose your first random number happens to be:

0. 389

A cross over between two loci will be said to happen if this number is between the physical distances given between the two loci.

For your chromosome, is there a cross over event involving any of the loci? If not then write the original chromosomes here else write the resulting cross over chromosomes.

Then take these resulting chromosomes and the second random number and do another crossover event at that physical position on the chromosome. Suppose the second number is .780. Write your results here.

Use the procedure we have outlined to generate 100 gametes. Notice it only requires 50 pairs of random numbers to do this.

For your data proceed the same way tallying up all the possible chromosomes to complete table II. Be sure to tally all the chromosomes that result from each cross over event, not just the non parentals and remember that is there is no cross over event between the two loci you still tally the parentals!

12. Now get together with your group and pool your group's data in Table IV. This is the data we will share as a class!

Table IV

Crossover events and phenotype classes for the 3 point test cross. Pooled data for all group members.

After collecting this data, transfer it to a separate data sheet and bring your data on Table V with you to the next class where you will swap data with one of the other group's. Your goal will then be to try to infer the arrangement of another group's loci and map distances between the loci of another group's chromosomes.

Simulated Crossover phenotype classes for the 3-point test cross. Pooled data for all group members. Group ________________

Phenotypes of the F1 females used for the three-point test cross and the homozygous recessive males. Do not give any hints as to the actually order of the genes on the chromosome or actual physical distances between the loci!

Offspring from the test cross:

Analyze the data you collected in table V in the following manner.

13. Test the hypothesis that each pair of loci is unlinked. First due locus 1 against locus 2 then locus 2 against locus 3.

pgd created 02/11/03 revised 09/22/03

14 Calculating the map distances between each pair of loci.

Suppose you have group I's data: You would need to calculate the map distance between each pair of loci: Eye color and Bristle length, Eye color and Body color, Bristle length and Body Color.

You do this by using phenotype data from the group's simulation and getting the number parental and non parental phenotype progeny from the test cross and applying the formula for the frequency of recombination:

r = (number of recombinants / total offspring from the test cross) * 100%

Where the number of recombinants is the same thing as the number of non parental phenotypes from the test cross.

The term parental phenotype, remember, refers to the offspring of the test cross that have the same phenotype is the heterozygous parent or the homozygous parent in the test cross. These are produced by gametes having no detectable crossing over between the two loci being analyzed. The non parental phenotypes arise when there is a detectable cross over between the two loci being analyzed.

Locus pair 1: Traits: ______________ and ________________

Number of parental offspring

Number of non parental offspring

Total offspring:

r =

estimated map units

Locus pair 2: Traits: ______________ and ________________

Number of parental offspring

Number of non parental offspring

Total offspring:

r =

estimated map units

Locus pair 3: Traits: ______________ and ________________

Number of parental offspring

Number of non parental offspring

Total offspring:

r =

estimated map units

15. Finding the arrangement of the loci

Using you map unit results, find the loci with the biggest map distance between them. Assume the other locus is in the middle. Draw a chromosome showing the loci and the estimated map distance as best you can below:

16. Does the relative order of the loci suggested by the data match the physical distance on your chromosomes?

17. What can you say about the physical distances vs the map distances. The physical distances add up properly but do the map distances? If not, what might be going on?

18. Repeat the analysis with the mystery chromosome data you have been given from the other from the other group and be prepared to present your analysis and what you infer about the arrangement of genes along the mystery chromosome.