Linkage/Eukaryotic Gene Mapping (Chapter 7)

Recall:
Segregation
Independent Assortment

Segregation: Each diploid organism has two alleles at each locus that separate (one allele per gamete) during meiosis.

Independent Assortment: Two alleles at one locus will act/sort independently of alleles at another locus.

Linked Genes

This principle allows allelic recombination to occur.

Genes location on close together on the same chromosome, and belong to the same linkage group. Linked Genes do not assort independently, and travel together in meiosis ending up in the same gamete.

Mendel's traits were not linked, so they sorted independently.

Recombination

The sorting of alleles into combinations
distinct from the parental generation.

9:3:3:1 Ratio?

F2 progeny do not appear in the 9:3:3:1 ratio expected with independent assortment, since they lie close together on the same chromosome.

Linked Genes Can Still Recombine- But Only When Crossing Over Occurs

-Genes may switch from a chromosome to its homolog by crossing over in Meiosis 1.
-In Meiosis II, genes that are normally linked will then assort independently, and end up in different gametes.

Crossing over takes place in meiosis and is responsible for recombination

Notation for Crosses with Linked Genes

For a cross between 2 linked loci (AA BB X aa bb) --> A, A are on top of each other and B,B are on top of each other X a,a on top of each other and b,b are on top of each other DOUBLE BARS

Simplified: One line
A,a on top of each other and B,b on top of each other

Simplified: Use of Slash
AB/ab


REFER TO PPT

If a heterozygous parent (AB/ab) is test-crossed to a homozygous recessive (ab/ab), what will the progeny phenotypic ratio be if genes A and B are completely linked?

a. 1AB:1Ab:1aB:1ab
b. 9AB:3Ab:3aB:1ab
c. 1AB:1ab
d. none of the above

Answer: C

C. 1AB:1ab

A Testcross Will Reveal the Effects of Linkage

-Start with a heterozygous individual (mate a homozygous dominant for both traits with a homozygous recessive for both traits)

-Cross the het/het with double homozygous recessive

Complete Linkage (no crossing over)

Only Nonrecombinant progeny are produced.

Independent Assortment (crossing over)

Half the progeny are recombinant and half the progeny are not

Crossing Over Creates 2 Recombinant Chromatids

In incomplete linkage, sometimes crossing over occurs.

No crossing over
1-Homologous chromosomes pair in prophase 1.
2-If no crossing over takes place, each gamete receives a non recombinant chromosome with an original combination of alleles.

Crossing over
1-A crossover may take place in prophase 1.
2-In this case, half the resulting gametes will have an unchanged chromosome (nonrecombinant) and half will be recombinant chromosomes.

Calculating Recombination Frequency for Incompletely Linked Genes

Must deserve the phenotype of the progeny. Progeny that have the same phenotype as one of the parents are non-recombinant progeny. Those that display a phenotype distinct from one of the parents are recombinant progeny.

From the cross AB/ab X ab/ab, what is the recombination frequency if the progeny numbers are 72 AB/ab, 68 ab/ab, 17 Ab/ab, and 21 aB/ab?

a. 0.213
b. 0.271
c. 0.500
d. 787
e. cannot be determined

A. 0.213

Cis Configuration

If both wild type allele are on the same chromosome, they are said to be coupled, or in the cis configuration

p+ b+
---------
p b

Trans Configuration

If the wild type alleles are on opposite chromosomes, they are said to be repulsed, or in the trans configuration

p+ b
---------
p b+

Alles in Coupling Configuration VS Allele in Repulsion Configuration

Conclusion:
The phenotypes of the offspring are the same, but their number differ, depending on whether alleles are in coupling or in repulsion.

RECOMB: LOWER #

REFER BACK TO THE PPT

Results of a Testcross (Aa Bb X aa bb) with Independent Assortment

Aa Bb (nonrecombinant) 25%
aa bb (nonrecombinant) 25%
Aa bb (recombinant) 25%
aa Bb (recombinant) 25%

Results of a Testcross (Aa Bb X aa bb) with Complete Linkage (genes in coupling)

Aa Bb (nonrecombinant) 50%

aa bb (nonrecombinant) 50%

Results of a Testcross (Aa Bb X aa bb) with Linkage with some Crossing Over (genes in coupling)

*Aa Bb (nonrecombinant)
*aa bb (nonrecombinant)
* = more than 50%

^Aa bb (recombinant)
^aa Bb (recombinant)
^ = less than 50%

Predicting the Results of Crosses Given the Recombination Frequency

Geneticists have determined that the recombination frequency between two genes in cucumbers is 16%. How can we use this information to predict the results of this cross?

Because the recombination frequency is 16%, the total proportion of recombinant gametes is 0.16.

The predicted frequency of progeny is obtained by multiplying the frequencies of the gametes.

Refer back to PPT

Test for Independent Assortment Vs. Linkage: The Chi-Square Test of Independence

-Sometimes it is not immediately obvious if two genes are incompletely linked, far enough apart on the same chromosome for crossing over to occur 100% of the time, or on different chromosomes and sorting independently.

Remember, we always have to account for the confounding variable of chance.

We use a slightly more complicated version of the Chi Square test to test for this (chi square of independence)

Chi-Square Test of Independence

1. Start with a testcross (you will be given the total counts of the various progeny).

2. Make a contingency table. Genotypes of one locus across the top and the other locus along the side. Fill in progeny numbers, add up the rows AND column, then get a grand total.

3. Make a table of observed vs. expected progeny. To calculate expected, multiply row total by column total then divide by grand total.

4. Solve the standard Chi-Square formula. You will need to sum 4 different values for the 4 different possible progeny. You now have your chi-square value.

5. Calculate the degree of freedom. In these problems, you can calculate this as (rows-1) X (columns-1).


Refer to the PPT

Degrees of Freedom

df= (number of rows-1) X (number of columns-1)

Simple Gene Mapping Using Recombination Frequencies

-There is a direct relationship between the distance separating two genes and the frequency of recombination.

-Genes that are near each other will not recombine as often, so they are closer

-Genes that are far away from each other are more likely to have crossing over occur, so they are farther apart.

-Draw a genetic map to present the order of genes and the relative space between them.

-Measured in Map Units (m.u.)

-1 m.u. = 1% recombination

Gene Mapping Using Recombination Frequencies

Example:

A and G 20
W and E 3
A and W 5
E and G 12

CAN ALSO FLIP THE MAP

Refer to PPT for more examples

Gene Mapping Using Recombination Frequencies CONSIDERATIONS

1. If the recombination frequency is 50%, then the genes are in different linkage groups (you must draw a second line to plot these genes on)

2. The further away 2 genes are, there is a greater likelihood of a double crossover. When this happens, the recombination frequency will miss some recombinants, so the frequency is often an underestimate.

Gene Mapping and Recombination Frequencies

1-A single crossover will switch the alleles on the homologous chromosomes
2-But a second crossover will reverse the effects of the first, restoring the original parental combination of alleles
3-And producing only recombinant genotypes in the gametes, although parts of the chromosomes have recombined.

Simple Gene Mapping Using Recombination frequencies- Two Point Testcross Practice

D to B = 10
B to C = 20
D to C = 28

How can it be 28? It is not, use the two numbers to get the more accurate measure; therefore, D to C is 30.

Refer back to PPT

A Three-Point Testcross is Used to Determine the Order of Three Linked Genes

Conclusion: Recombinant chromosomes resulting from the double crossover have only the middle gene altered.

Refer to PPT

Three-Point Testcross

-First, ID the non-recombinants (2 most) and the double crossovers (2 fewest)
-Mark the location of the crossover on the strand inherited from the het parent
-Next calculate recombination frequency between the middle gene and each end gene
-Then draw a genetic map

Coefficient of Coincidence and Interference

-Oftern the occurrence of a second crossover is influenced by a first crossover (usually prevents it from occurring)
-This results from fewer observed double-crossover progeny than would be expected.

Coefficient of Coincidence=
# of observed double crossovers/# of expected double crossovers

The degree to which one crossover inhibits the second is called interference:

interference= 1-coefficient of coincidence

In a three-point testcross the nonrecombinant progeny are A+B+C+ and abc. The double crossover progeny are A+B+c and abC+. Which locus (A,B, or C) is in the middle?

a. A
b. B
c. C
d. A or B

c. C

Multiple Crossovers

Sometimes the second of two crossovers will involved a third or fourth strand. However, the average is still 50% detectable recombinants. This means that the distance between genes that are far apart may be underestimated.

Other Forms of Gene Mapping

-Using molecular markers (Genomewide-Association Studies)
-Deletion Mapping
-Somatic-Cell Hybridization
-Physical Mapping
-Whole Genome Sequencing

Genomewide-Association Studies (GWAS)

1-A series of of different hapotypes occur in a population.
2-When a mutation first arises, it is associated with a special haplotype of allele at other loci.
3-With no recombination, the mutation remained associated with A^2 B^2 C^4 haplotype.
4-With crossing over, the mutation is now also associated with A^1 B^3 and C^5 haploype.

Studies that look for nonrandom associations between the presence of a trait and alleles at many different loci scattered across a genome- that is, for associations between traits and particular suites of allele in a population.

Single Nucleotide Polymorphisms

Determine the difference between A and a; there are DNA changes.

Single-base-pair differences in DNA sequence between individual members of a species

Deletion Mapping

Technique for determining the chromosomal location of a gene by studying the association of its phenotype or product with particular chromosome deletions

If the gene product is present in a cell line with an intact chromosome but missing from a line with a chromosome deletion, the gene for that product must be located in the deleted region.

Refer to PPT

Somatic-Cell Hybridization

-Human fibroblasts are fused with Mouse Cell lines

-The nuclei fuse and human chromosomes are randomly lost.

-Fusion of membranes create hybrid cells called heterokaryons.

-Identify which lines contain your gene of interest and which human chromosomes those lines contain. (Refer to the PPT)

Physical Mapping- IN City Hybridization

Fluorescently labeled DNA probe

Whole Genome Sequencing

Is the process of determining the complete DNA sequence of an organism's genome at a single time.

Complementation Test

-Perform a cross between the mutant in question and a homozygous mutant for the other mutation.

-Helps to determine whether a mutation is at the same locus.

-If offsprings are normal, then mutations are complement of another

-wild type = different loci

-mutant = same loci

REFER BACK TO CHAPTER 5