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LINKAGE MAPPING

Linkage Mapping

Written by Brian White. copyright 1995

If you analyze a cross involving two genes (A and B), the analysis can become more complex. If the two genes are on different chromosomes analysis is straightforward:

AaBb X aabb


However, if A and B are very close together on one chromosome than the original combinations of A and B alleles in the heterozygous parent will remain together in the progeny.

What are these 'original combinations of A and B alleles'? There are two possibilities:



Note: the AaBb type of notation cannot distinguish between these different possibilities, while the other types of notation clearly show the combinations of A and B alleles on each chromosome.

These original combinations of A and B alleles are:

case 1 :	a b 	and 	+ +
case 2 :	a +	and	+ b
These allele combinations are the parental types. They are called parental types because they are the allele combinations that the individual in case 1 or 2 inherited from its parents.

If A and B are close together on the same chromosome, recombination will occur during meiosis and four types of gametes will be produced:

In CASE 1:

				Gamete	            Type   	   Frequency 
 will produce gametes:	             parental type	   p/2
                                          parental type		p/2
                                          recombinant type	r/2
                                          recombinant type	r/2
or CASE 2:
			       Gamete	            Type	   Frequency

 will produce gametes:	  	    parental type	   p/2
                               	         parental type		p/2
                               		 recombinant type	r/2
                                     	 recombinant type	r/2
IMPORTANT: The genotypes which are parental or recombinant depend upon the genetic constitution of the heterozygous parent.

Note that the two parental types are produced with equal frequency as are the recombinant types. Also, p+r=1 and p is less than or equal to r.

Test Cross

Usually mapping crosses are a cross of a heterozygote - where recombination takes place - with a recessive homozygote. This results in progeny with genotypes that can be directly inferred from their phenotypes.

If is crossed with , four types of progeny will be observed:

    Progeny Genotype	Progeny Phenotype	Type		Frequency 
			a	b	parental	     p/2

			+	+	parental	     p/2

			a	+	recombinant	     r/2

			+	b	recombinant	     r/2
Note that produces only one type of gamete: . And since that parent is homozygous recessive, the genotypes of the progeny are directly revealed in the phenotypes.

Recombination Frequency

What would happen if the homozygous parent was homozygous dominant ?

In CASE 2: the parental and recombinant types would be exchanged.

That is: parental types = and . Recombinant types = and .

Using crosses like this, you can measure recombination frequency.
Recombination Frequency is defined as:

or , and is usually expressed as a percent recombination or map units measured in centimorgans (cM) = .

Recombination frequency between genes increases with increasing distance between genes and can be used to create genetic maps of chromosomes.

Note: when two genes are unlinked (on different chromosomes or very far apart on the same chromosome) their recombination frequency is 50%.

Why isn't it 100%? Look back at the case where A and B are on different chromosomes - all 4 types of gametes are produced with equal frequency. No matter which are the parental or recombinant types, each type represents 50% of the total.

Therefore: 0% < recombination freq. < 50%

Where near 0% recombination frequency means the genes are very tightly linked, and 50% recombination frequency means the genes are entirely unlinked.
Also, the recombination frequency can never be exactly 0%, that would require an infinite number of progeny. If no recombinants are observed in n progeny, you can only conclude that the recombination frequency is < . In humans, 1 cM is roughly 1 million nucleotide pairs (bp) of DNA.

Solving Linkage Problems

Tip: The most important part is to determine which progeny resulted from parental type gametes, and which from recombinant types.

In a plant, leaf color and leaf shape are controlled by two linked genes. Leaves of the wild-type plant are red. A recessive mutation in this gene causes white leaves. Wild-type leaves are pointed, and a recessive mutation in this gene causes them to be smooth. The following crosses were performed:

pure breeding white, smooth X pure breeding wild type

gives F1: all red, pointed

Now, the next cross:

red, pointed X pure breeding white, smooth

gives F2:
40 white, curly
36 red, pointed
10 white, pointed
14 red, curly

What is the recombination frequency between the gene for color and for shape?

Solution:

First, assign genotype symbols. Since the mutations are recessive to wild-type, use + for the wt allele and lower case letters for the mutant alleles:
w = recessive color allele for white
s = recessive shape allele for smooth
The first cross is: X giving progeny: (red, pointed)

The second cross is: X giving progeny where the genotypes can be unambiguously interpreted from their phenotypes.
The parental gamete types will be: and , resulting in [white, smooth] and [red, pointy] progeny.
The recombinant gamete types will be: and , resulting in [white, pointy] and [red, smooth] progeny.

Therefore, the recombination frequency is:

The following example is very complex because it includes many interacting factors: three genes, one of which is dominant; the genes are sex-linked; and the cross is to wild-type, not homozygous recessive (In this case, we use the Y chromosome as 'homozygous recessive' and do the mapping using only the male progeny). However, it illustrates all the important concepts for solving linkage problems. Working through this problem will help you to get a thorough understanding of most of the important principles we will cover in genetics. This problem is much more difficult than you would find on an exam.

Determine the inheritance pattern and linkage (if any) of these three traits:

Bar-eye, singed-wing, forked-antenna

Cross 1: pure-breeding bar-eye, singed-wing, forked-antenna female X pure-breeding wt male

F1: many progeny-- all females bar-eyed; all males bar, singed, forked

Cross 2: One of these bar-eyed females X pure-breeding wt male gave the following progeny:

Solution:

First, we need to find out whether the three genes (bar, singed, forked) are dominant/recessive and sex-linked/autosomal. It is easiest to treat 'each trait separately.

* Bar:

This data is consistent with bar being sex-linked dominant or autosomal dominant to wild-type. (Why? You should be able to convince yourself). Hopefully, the remaining data will help sort this out.

* Singed:

This data is consistent with singed being sex-linked recessive.

* Forked: The data are the same as with singed, so forked is sex-linked recessive.

We have now determined that:
* singed (sn) is sex-linked recessive
* forked (fo) is sex-linked dominant
* bar (B) is dominant, and could be sex-linked or autosomal

If B were linked to sn or fo, it would have to be on the X chromosome and we could conclude that B is sex-linked dominant.

So, try assuming bar is dominant sex-linked and see if the predictions match the data. See if it is linked to singed (this time, ignore forked).

+ = recessive wild-type allele

B = dominant bar allele (Upper case, since it's dominant)

So the first cross was: giving progeny:

50%