I.    Early evidence for Linkage (the association of alleles of different genes on the same chromosome)

    A.  Bateson and Punnett early 20th century were working with sweet peas with flower color and shape of pollen grains.

Using P = purple, p=red and L=long, and l=round for pollen shape.

Crossed PPLL x ppll to get F1 dihybrid. Selfed the dihybrid:

	Number of progeny in F2
Pheno and geno type	Observed	Expected
Purple long (P-L-)	4831	3911
Purple round (P-ll)	390	1303
Red long (ppL-)	393	1303
Red round (ppll)	1338	435
total	6952	6952

Found that a dihybrid cross did not give expected ratios of 9:3:3:1

Nor did it look like any of the gene interactions described in previous chapter.

What was unique was that the traits of the parental generation appeared to influence the out come. PL and pl seemed somehow coupled together.

Could see that a single gene followed Mendelian ratios.

 

    B.    It took T. H. Morgan to come up with explanation:

1.Using Drosophila genes:

Eye color: pr+ = red or pr = purple
Wing morphology: vg+ = normal wings or vg = vestigial wings

pr+ pr+ vg+vg+   x   pr pr vg vg

F1 pr+ pr vg+ vg

F2:

phenotype	Expected ratios	Observed ratios
Red eyed normal wings (pr+_ vg+_)	9	>9
Purple eyed normal wings (pr_  vg+_)	3	<3
Red eyed vestigial wings (pr+_  vg_)	3	<3
Purple eyed vestigial wings (pr_  vg_)	1	>1

 

2.Morgan test crossed the F1 female pr+ pr vg+ vg with pr pr vg vg male.  This allowed him to "see" how the genes from the heterozygote female segregated because the male tester only contributes recessive alleles

Male gametes	pr vg
Female gametes (down)
pr+ vg+	pr+ pr vg+ vg
pr vg+	pr pr vg+ vg
pr+ vg	pr+ pr vg vg
pr vg	pr pr vg vg

 

Progeny were:

phenotype and genotype	expected	observed
Red eyed normal wings (pr+pr vg+vg)	0.25	0.47	parental
Purple eyed normal wings (prpr vg+vg)	0.25	0.054	nonparental
Red eyed vestigial wings (pr+pr  vgvg)	0.25	0.053	nonparental
Purple eyed vestigial wings (prpr  vgvg)	0.25	0.42	parental

 

 

3.  Morgan's observations
- 2 larger than expected classes carry the parental genotypes
- 2 smaller than expected classes carry the nonparental genotypes
- 1:1 ratio between the parental genotypes
- 1:1 ratio between the nonparental  genotypes

Proposed the coupling of the dominant alleles PL and the coupling of the recessive alleles pl

 

4.  Next, he tried a cross for each of the parents homozygous for one of the recessive alleles.

pr+ pr+ vg vg X pr pr vg+ vg+ ----> F1 pr+ pr vg+ vg
test crossed F1 female pr+ pr vg+ vg X male pr pr vg vg
This allowed him to "see" how the genes from the heterozygote female segregated because the male tester only contributes recessive alleles

Male gametes	pr vg
Female gametes (down)
pr+ vg+	pr+ pr vg+ vg
pr vg+	pr pr vg+ vg
pr+ vg	pr+ pr vg vg
pr vg	pr pr vg vg

 

Progeny were:

phenotype and genotype	expected	observed ratio
Red eyed normal wings (pr+pr vg+vg)	0.25	0.067	nonparental
Purple eyed normal wings (prpr vg+vg)	0.25	0.46	parental
Red eyed vestigial wings (pr+pr vgvg)	0.25	0.41	parental
Purple eyed vestigial wings (prpr vgvg)	0.25	0.063	nonparental

 

Again, the ratios were not 1:1:1:1. In this case the non allelic dominant genes appear to be repelling each other. Bateson and Punnett called this situation repulsion.

5.  Explanation of the larger than expected classes that had the parental phenotypes.......T. H. Morgan's explanation was that the two genes were on the same chromosome. This would explain their being together for most of the progeny.

6.  Explanation of the smaller than expected classes that had the nonparental phenotypes....  To explain the two other smaller classes, Morgan, using cytological evidence seen in the chiasmata seen at diplotene of prophase I of meiosis, suggested that the chiasmata represent a breaking and reunion of chromatids and a physical exchange of chromosome parts.  These "recombined" or recombinant chromosomes were responsible for the nonparental types.


(from An Introduction to Genetic Analysis, 6^th ed. By Griffiths et al. W. H. Freeman and Company)

7.  Linkage can be in the coupling (or cis) conformation where both recessive alleles are on the same chromosome of the homologous pair or the repulsion (or trans) conformation where a dominant allele of one gene is linked to a recessive allele of another gene.

8.  Notation - when doing linkage problems it is advisable to use the "Drosophila notation" where all the alleles from one of the homologous chromosomes are written first followed by a slash and then the alleles from the other member of the homologous pair are written.  For example, for two genes on chromosome #1 (A and B):
AB/AB x ab/ab ----> AB/ab

 

II.   Recombination
    A.  Meiotic recombination is any process that generates a haploid product with a genotype that differs from both haploid genotypes that constituted the meiotic diploid cell.  Recombination is detected by comparing the output genotypes with the input genotypes.

(from An Introduction to Genetic Analysis, 6^th ed. By Griffiths et al. W. H. Freeman and Company)

    B.     Interchromosomal recombination
           1.  Mendelian independent assortment
           2.  A test cross yields 1 parental #1: 1 parental #2: 1 recombinant #1: 1 recombinant #2
           3.  Occurs for genes on separate chromosomes and genes that are very far apart on the same chromosome.

 

(from An Introduction to Genetic Analysis, 6^th ed. By Griffiths et al. W. H. Freeman and Company)

 

 

C.    Intrachromosomal recombination
            1.  Recombination generated by crossing over among nonsister chromatids of two homologous chromosomes.
            2.  Test cross yields >25% parental #1: >25% parental #2: <25% recombinant #1: <25% recombinant #2
            3.  Thus, a total recombination frequency between two genes of <50% suggests that the genes are linked on the same chromosome.
            4.  Cases where recombinants are not recovered suggest that the genes are so tightly linked that they never segregate.

(from An Introduction to Genetic Analysis, 6^th ed. By Griffiths et al. W. H. Freeman and Company)

III.  Special cases of linkage on the X chromosome

For crosses that involve genes on the X chromosome, a homozygous recessive male can be considered a tester strain because he only has one set of alleles to contribute.

 

Example of two linked genes on the X chromosome in mice:

eye color gene: a⁺ for wild type and a for apricot colored
coat color gene: g⁺ for for wildtype and g for grey colored

parental cross     female wildtype (a⁺g⁺/a⁺g⁺) X  male apricot eyes and grey body (ag/Y)

F1 were wildtype females (a⁺g⁺/ag) and wildtype males (a⁺g⁺/Y)

Selfed the F1 generation   female a⁺g⁺/ag    X male a⁺g⁺/Y

female gametes		male gametes
a+g+	parental	a+g+
a+g	recombinant
ag+	recombinant	Y
ag	parental

F2 progeny (Notice that since the male progeny only inherit the Y chromosome from the father, the female gametic contribution to the F2 males have no chance of being masked in the male progeny.  Thus, like a test cross, you can look at the frequency of each phenotype in the male progeny to determine the frequency of the female gametes in the F1 generation)

female	male	freq.
a+g+/a+g+	a+g+/Y	>0.25	parental
a+g/a+g+	a+g/Y	<0.25	recombinant
ag+/a+g+	ag+/Y	<0.25	recombinant
ag/a+g+	ag/Y	>0.25	parental

 

 

IV.  Linkage maps

    A.  Researchers had noticed that different recombination frequencies were obtained between different linked genes.  Morgan proposed that differences in recombination frequencies were due to differences in the distances between genes.  Specifically,  
the greater distance between two genes =  greater the chance for recombination = greater the recombination frequencies = less linkage

   
(from An Introduction to Genetic Analysis, 6^th ed. By Griffiths et al. W. H. Freeman and Company)

    B.  Morgan's student (Alfred Sturtevant) realized that "variations in strength of linkage, already attributed ... to differences in spatial separation of genes, offered the possibility of determining the sequences in the linear dimension of a chromosome"

    recombination frequency of 1% = 1 genetic map unit (m.u.) 

(100%)(# of recombinant progeny)/total # of progeny = # of genetic map units

1.  From the progeny of a dihybrid heterozygote test cross, we can determine the distance between the two genes.
For example, the cross AB/AB x ab/ab is made ----> F1 = AB/ab
The F1 is backcrossed to the aabb parent (AB/ab x ab/ab)
The F2 are  2/6 AB/ab  1/6 Ab/ab 1/6 aB/ab  2/6 ab/ab

How far apart are A and B?

Parental gametes = AB and ab    
Recombinant gametes = Ab and aB = 1/6 +1/6 = 2/6 = 0.33 or 33% or 33 map units

 

2.  From a map distance between 2 genes, we can calculate the expected progeny
For example the distance between A and B is 10 m.u.

What percent of the progeny will be AaBb in the cross AB/ab x ab/ab?

Parental gametes = AB and ab
Recombinant gametes = Ab + aB = 0.10
Thus, the freq. of the parental gametes = 0.90 and the freq. of each parental = 0.45 (AB = 0.45 and ab =0.45)
Thus, the freq of the AB/ab progeny will be 0.45 * 1 = 0.45

V.  Three-point test crosses

The three point test cross is a test cross in which one parent is heterozygous for three genes.

The steps for solving a three point test cross are:
1.  assign parental genotypes (classes with the most progeny)
2.  assign double recombinants (classes with the least progeny)
3.  determine the gene order based on comparison of the parentals and the double recombinants (the gene in the middle "flips" or switches with respect to the other two flanking genes in the double crossover when compared back to the parental)
4.  rewrite the cross if needed
5.  determine the frequency of single crossovers between genes A and B and genes B and C (for the order ABC)
6.  Draw map based on recombination frequency of 1% = 1 genetic map unit (m.u.) 

For example, in Drosophila

eye color:  v⁺ = wildtype and v = vermilion eye color
wing veination: cv⁺ = wildtype and cv = crossveinless
wing morphology:  ct⁺ = wildtype and ct = cut wing edge

Parental cross:   crossveinless, cut winged (v+ cv ct /v+ cv ct)  X vermilion (v cv+ ct+/v cv+ ct+)

F1 were triple heterozygotes (v+ cv ct/v cv+ ct+) wildtype phenotype

Test cross:  F1 wildtype (v+ cv ct/v cv+ ct+) X    triple mutant (v cv ct/v cv ct)

F2 progeny

1.  Parental phenotypes are the most numerous classes = v+ cv ct and v cv+ ct+
2.  Double crossovers are the least numerous classes = v cv+ ct and v+ cv ct+

3.  Determination of the gene order by comparison of the parental and double cross over classes (the gene in the middle "flips" or switches with respect to the other two flanking genes in the double crossover when compared back to the parental): so in this case, the gene order is v ct cv

4.  Rewrite the crosses:
Parental cross:   crossveinless, cut winged (v+ ct cv /v+ ct cv)  X vermilion (v ct+ cv+/v ct+ cv+)
F1 were triple heterozygotes (v+ ct cv/v ct+ cv+) wildtype phenotype
as F1 wildtype (v+ ct cv/v ct+ cv+ ) X  triple mutant (v ct cv/v ct cv)

5a.  Determine the frequency of crossover between the v and ct loci:
Parentals are v+ ct and v ct+ and recombinants are v+ct+ and v ct
(89+94+3+5)/1448 = 0.132 = 13.2% = 13.2 m.u.

5b.  Determine the frequency of crossover between the ct and cv loci:
Parentals are ct cv and ct+cv+ and recombinants are ct cv+ and ct+ cv
(45+40+3+5)/1448 = 0.064 = 6.4% = 6.4 m.u.

6.  Thus the map is    

VI.  Interference

Interference is the measure is the independence of crossovers from each other.

Interference (I) = 1- the coefficient of coincidence (c.o.c)

The coefficient of coincidence (c.o.c) is the ratio of the observed number of double recombinants to the expected number of  double recombinants (observed # of double recombinants/expected  # of double recombinants  OR
observed frequency of double recombinants/expected  frequency of double recombinants).

Thus, I = 1 - (observed # of double recombinants/expected  # of double recombinants)

If  I = 1,  the double crossover event never occurs.  If I = 0, there is no interference.

Consider the above three point test cross as an example.
The expected number of double recombinants = the probability of a single crossover between the v and ct loci * the probability of a single crossover between the ct and cv loci*total # of progeny

Expected doubles = (0.132)(0.064)(1448) = 12

I = 1 - (5+3/12) = 0.33

VII.  Calculating recombination frequencies from selfed dihybrids:

The three point testcross is the best way to examine linkage; however, many times a double recessive homozygote tester is not available.  Recombination frequencies can be calculated from selfed heterozygous dihybrids (although the frequencies obtained are not as exact as a test cross because of the mathematical manipulations involved).

Consider two genes A and B

Parental cross AB/AB  X ab/ab

F1  =   AB/ab

Self the F1   AB/ab  X  AB/ab

In this cross, the only complete genotype that we know is aabb (ab/ab) which occurs at a frequency of 55/381 = 0.144
We know that ab/ab resulted from male ab gamete fusing with female ab gamete and that the frequency of these gametes being produced is equal in the male and the female since they are both F1s.
So let a = frequency (or probability) of the ab gamete forming.
P(ab from male)*P(ab from female) = 0.144
a*a = 0.144
a² = 0.144
square root of a = square root of 0.144
a = 0.3799

ab and AB are both parentals and will form at the same frequency which we just calculated to be 0.3799
Thus, the total frequency of the parentals = freq. of ab + freq. of AB = 0.3799 +0.3799 = 0.7598

If the freq. of the parentals = 0.7598, then the frequency of the recombinants = 1 - 0.7598 = 0.2402

Thus the recombination freq. = 24.02% = and the genes are 24.02 m.u. apart

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