Genetics: Linkage and Mapping

Chapter 5. Linkage and Genetic Mapping

• During the early 1900’s when scientists realized that chromosomes contained the genetic material they suspected a conflict may occur between Mendel’s law of independent assortment and chromosomes during meiosis.

• Probably thousands of genes yet only a handful of chromosomes.

• Some genes may reside on the same chromosome.

• This chapter:

– Inheritance of genes on the same chromosome.

– How cross data is used to describe the order of genes on a chromosome.

– Discussion of unique sexual reproduction of fungi (distinctive mapping approaches).

• Linkage and Crossing Over.

– Chromosomes carry a few hundred to a few thousand genes.

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– Chromosomes are sometimes called linkage groups, because each contains a group of genes linked to each other.

• Humans have 22 autosomal linkage groups, an X-chromosome linkage group and a Y-chromosome linkage group.

• Geneticists are often interested in doing dihybrid or trihybrid crosses and the inheritance pattern will depend on whether or not the genes are linked.

– Recombinant chromosomes and phenotypes result from crossing over.

• During prophase I of meiosis crossing over may occur.

– Occurs after tetrad formation (two pairs of sister chromatids), where a chromatid from one pair will cross over with a chromatid from the homologous pair.

– Bateson and Punnett discovered two traits that did not assort properly.

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• Crossed a true-breeding purple flower (PP) color long pollen grain (LL) producing plant with a red flower (pp), round pollen (ll) plant.

• The F₁ generation were all purple flower and long pollen producing offspring (PpLl).

• Even though there were four phenotypic categories in the F₂ generation they did not observe a 9:3:3:1 ratio.

• Found that there was a larger proportion of the parental phenotypes purple long and red round.

• The suggested that these phenotypes may be coupled somehow and not easily assorted in an independent manner.

– Did not know that this was due to linkage on the same chromosome.

– Morgan observed linkage of X-linked genes and proposed crossing over at the X-chromosome.

• His earlier studies set the framework for demonstrating the linkage of X-linked traits to the X-chromosome.

• Morgan studied many X-linked traits, for this experiment he crossed normal males to females that had yellow bodies (yy), white eyes (ww) and miniature wings (mm).

• Wild-type alleles = y+ (gray body), w⁺ (red eyes) and (normal wings) m⁺.

• F₁ generation contained normal females and males with yellow bodies, white eyes and miniature wings.

• linkage was only revealed once the F₂ generation was analyzed.

• Instead of equal proportions of the eight phenotypes he observed a larger percentage of the parental phenotypes.

– 758 flies with gray bodies, red eyes and normal wings (males of the parental gen.).

– 700 flies with yellow bodies, white eyes and miniature wings (females of the parental gen.).

• Morgan’s explanation was that all three genes are located on the X-chromosome and are therefore transmitted as a unit.

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– There were six other combinations of phenotypes that were not found in the parental generation.

• Also needed to explain the quantifiable difference between nonparental combinations of body and eye color vs eye color and wing length.

– Can be visualized by looking at data from pairs of genes.

• Morgan realized that his data was consistent with crossing over which was proposed by Janssens (1909).

• Morgan’s hypotheses:

– 1) Genes for body color, eye color and wing length are all located on the same chromosome.

» Likely they are inherited together.

– 2) Due to crossing over the X-chromosome (females only) can exchange pieces of DNA allowing for nonparental combinations of alleles.

– 3) Likelihood of crossing over depends on distance between the genes.

» The further apart two genes on the same chromosome are the more likely that crossing over will occur between them.

• If you examine the females X-chromosomes before and after oogenesis you can better understand the effect of crossing over on the F₂ phenotypes.

– No crossing over gives the parental phenotypes.

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• He proposed that the genes for eye color and wing length occasionally crossed over to produce nonparental offspring.

– Gray body, red eyes and miniature wings or yellow body, white eyes and normal wings.

– Fairly likely because the genes are far apart.

• Proposed body color and eye color are close together and crossing over between them would occur infrequently.

– Gray body, white eyes and miniature wings or yellow body red eyes and normal wings.

• Something very unlikely that did occur was a double cross over- gray body, white eyes and normal wings.

– Chi square can be used to distinguish between linkage and independent assortment.

• How do we as geneticists decide if something is linked or assorting independently?

• The Chi square can be used not only to test the goodness of fit of between a hypothesis and the observed data.

– Can be used to decide if a dihybrid cross is consistent with linkage or independent assortment.

• Chi square steps to show linkage:

– 1) Propose a hypothesis: For most dihybrid crosses it is that the genes are not linked.

» Choose this hypothesis even if the observed data suggest otherwise. Why?

» Why not use linked? What do we not know?

– 2) Do the chi square with the hypothesis that it fits independent assortment if the data do not fit then we can reject that hypothesis.

• Example: Chi square on the values for body and eye color.

– 1159 gray body; red eyes.

– 1017 yellow body; white eyes.

– 17 gray body; white eyes.

– 12 yellow body; red eyes.

– Independent assortment predicts a 1:1:1:1 ratio.

• 1) The predicted results do not seem to be consistent with the observed data however we propose the hypothesis:

– Although we know that eye color and body color are X-linked they assort independently.

» We can now use the chi square and calculate expected values.

• 2) Expected values: Probability of each genotype is ¼.

– The total number of offspring = 1159+1017+17+12 = 2205

– ¼ X 2205 = 551 = expected number for each phenotype.

• 3) Apply the chi square:

• Interpret the calculated chi square value. With a degrees of freedom of 3 (n-1 = 4-1)

• The chi square value is enormous and no where near an acceptable number.

– Reject the hypothesis that the two genes assort independently.

– Experiment 5A: Crossing over produced new combinations of alleles with the exchange of homologous chromosomes.

– Creighton and McClintock used corn to study two linked genes and recovered parental and recombinant offspring.

– Looked at the chromosomes of the offspring because the parental chromosomes had odd structures.

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• Allowed them to correlate recombinant offspring with observable changes in the structure of the chromosomes.

– Studied trait inheritance in corn.

– Previous cytological examination of the chromosomes, indicated that some strains had abnormal chromosome morphology (chromosome #9).

• Had a darkly staining knob on one end.

• Also has a translocation (interchange) from chromosome #8.

• These changes allowed it to be distinguished under the microscope.

– Creighton and McClintock realized that this abnormality could be used to visualize crossing over.

• They knew that a gene near the knobby end of chromosome #9 was responsible for kernal color.

– Existed in two alleles: The dominant C (colored) and the recessive c (colorless).

• Knew that at the other end of the chromosome was a gene involved in kernel endosperm texture.

– Dominant Wx (starchy endosperm) and recessive wx (waxy endosperm).

• Reasoned a cross over between a normal #9 and the abnormal #9 would result in a knob or a translocation but not both and would be visually distinctive.

– Hypothesis: Offspring with the nonparental phenotypes are the product of a cross over. This should create nonparental chromosomes via an exchange of chromosomal segments between homologous chromosomes.

• Not a standard cross because neither parent was homozygous recessive for both genes.

– Creates ambiguity when we analyze phenotypes and genetic recombination.

• In this cross we analyze whether crossing over occurs in parent A (pA), which is heterozygous for both.

• Parent A can produce four gametes B (pB) can only produce 2.

• Two phenotypes are ambiguous: Colored/starchy and colorless/starchy.

– Could result from recombination or not.

• Analyze first the colored/waxy and colorless/waxy.

– Colored/waxy can only occur if recombination did not occur and pA passed on the knobbed/translocated chromosome.

– Colorless/waxy can only happen if recombination occurs and pA passed on a translocated chromosome #9 that was knobless.

» The chromosomes were analyzed and found to be a knobless #9.

– Taken together this is consistent with genetic recombination of alleles and the cytological presence of a chromosome showing an exchange of genetic material.

• Supports the idea of crossing over during meiosis.

• Supports the hypothesis that genetic recombination is the swapping of chromosome segments.

– Authors stated: “pairing chromosomes, heteromorphic in two regions, have been shown to exchange parts at the same time they exchange genes assigned to these regions.”

– Crossing over occasionally occurs during mitosis.

• Much less likely than in meiosis but it does occur.

• When it occurs, mitotic crossing over may produce a pair of recombinant chromosomes with a new combination of alleles.

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» If this happens early in embryonic development the daughter cells may divide several times to produce a patch of tissue in the adult that may have characteristics different than those in the adult.

– In 1936 Curt Stern found odd patches of tissue on Drosophila and proposed that it was from mitotic recombination.

– He was working on X-linked alleles affecting body color and bristle morphology.

• Recessive y allele confers yellow body color.

• Recessive sn allele confers short body body bristles that look singed.

• The corresponding wild-type alleles were for gray body color (y⁺) and normal bristles (sn⁺).

• Females that are y⁺y sn⁺sn were expected to be gray/normal and most were.

• When Stern looked at the flies under a microscope he observed two adjacent regions that were different from the rest of the body, twin spot.

• He proposed that this was due to a single mitotic recombination event during embryonic development.

• The X-chromosomes of the female fly were y⁺ sn and y sn+ (genotype:y+y sn+sn; phenotype:gray/normal).

– Rarely a crossover event occurred during mitosis to produce two adjacent daughter cells that are y⁺y⁺ snsn and yy sn⁺sn⁺.

– As the cells continue to divide they can produce a patch on the fly that is gray/singed next to a yellow/normal patch on a fly that is normally gray body normal bristles.

^–These twin spots provide evidence for mitotic recombination.

• Genetic mapping in the diploid eukaryote.

– Genetic mapping (gene mapping or chromosome mapping) is to determine the linear order of genes that are linked to each other along the same chromosome.

– A simplified genetic map follows and shows that each gene has its own locus within a particular chromosome.

• Ex: gene vg that affects wing length on chromosome II.

• Gene b is located a moderate distance away and is involved in body color.

– Preparing a genetic map is an extraordinary amount of work why do it?

• Leads to an understanding of complexity and organization of a specie’s genome.

• Useful for evolutionary studies, can make it easier to clone the gene and helps with the portrayal of the basis of heredity.

• Mapping is often the first step in identifying a human genetic disorder and can be used in diagnosis and genetic counseling.

– We will now discuss techniques for generating a genetic linkage map.

– Particularly useful with organisms that are easily crossed and produce many offspring.

• Ex: Drosophila; humans are more difficult but there are molecular genetic ways to generate a map in for humans.

– The frequency of recombination between two genes can be correlated with distance along a chromosome.

• Experimentally, the percentage of recombinant offspring is correlated with the distance between two genes.

• If two genes are close together few recombinant offspring will be observed; for genes that are far apart many recombinants offspring will be observed.

• A test cross must be done in order to know if the characteristics of an offspring are due to crossing over during gamete formation in a parent.

– Most test crosses are made between a heterozygote for two or more genes and an individual who is homozygous recessive for these same genes.

– The goal of the test cross is to determine if recombination occurred during gamete formation in the heterozygous parent.

» New combinations are impossible in the homozygous recessive parent. Right?

» Example a cross between ss ee (ebony/short) and a ss+ ee+ (gray/normal) fly.

• How can we assure ourselves that our heterozygous parent has the dominant alleles s+ and e+ linked to one chromosome and the s and e recessive alleles linked to the other chromosome?

• Their are four possible offspring that these parents could generate with recombination between the s and e genes occurring.

• Parental offspring = ebony/short and gray/normal; recombinant offspring = ebony/normal and gray/short.

• All offspring inherit one chromosome with an s and e allele from the homozygous parent.

• The heterozygous parent gives a s e, s+ e+ or a recombinant chromosome.

• Recombinant possibilities s+ e or s e+.

• The data from this test cross can be used to calculate the distance between the two genes.

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– This is expressed as map unit (mu) or centimorgans (cM) in honor of Thomas Morgan.

• The s and e alleles are located 12.3 map units away from one another along the same chromosome.

• Experiment 5B: Alfred Sturtevant used frequency of crossing over to build the first genetic map in 1911.

– Sturtevant considered the outcome of crosses he conducted involving six different mutant alleles that affected the phenotypes of normal flies all were recessive and X-linked.

– Looked at y (yellow body color), w (white eye color), w-e (eosin eye color), v (vermillion eye color), m (miniature wings) and r (rudimentary wings).

• w and w-e are the same gene so he really only looked at 5.

– y+ (gray body), w+ (red eyes), v+ (red eyes), m+ (normal wings) and r+ (normal wings)

– Hypothesis: When genes are located on the same chromosome the distance between them may be estimated from the proportion of recombinant offspring. This provides a way to order the genes along a chromosome.

• Sturtevant began the experiment with several strains that contained the alleles he was studying.

• Used heterozygous females and hemizygous recessive males.

• Some of the dihybrid crosses resulted in low percentages of recombinant offspring (Ex: y and w = 1%) indicating close proximity of the two genes.

• Some showed relatively high percentages of crossing over (v and m = 26.9%) indicating a larger distance between them.

• Sturtevant made two assumptions the closer genes are linked the more accurate the distance and that if the distance between w and r was 33.7 and between w and v was only 29.7 then v was between w and r (closer to r).

– The later assumption was confirmed by the crossover data between v and r (3%).

• Sturtevant proposed the following genetic map:

• Started at y and mapped from left to right.

– Ex: y and v are 30.7 mu apart while v and r are 3 mu apart.

• Some genes (y and m; 37.5 vs. 57.6) show inaccurate map distances, the reason is that the closer you get to 50% the greater the chance of double crossover events occurring.

• A test cross will only “ever” give a value of 50% even if the genes are more than 50 mu apart this is also due to multiple crossovers.

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• Much the same as doing it with a dihybrid cross but follow three traits and then when you analyze the data regroup the data according to pairs of genes.

– Using the product rule we can calculate the likelihood of a double crossover.

• If we want to examine the likelihood that a double crossover will occur multiply the likelihood of one crossover event times another.

– If a crossover between v and r is 3% and r and m is 24% then the likelihood of a double cross over is 0.03 X 0.24=0.0072 = .72%

» Out of 409 offspring you would expect 409 X .0072 = ~3 offspring due to double crossover.

– Often if you carryout the experiment you will get a lower than expected number of offspring resulting from a double crossover this is due to positive interference.

» When one crossover event occurs in a chromosome it often decreases the probability of another crossover event.

• Genetic Mapping in haploid eukaryotes.

– Much of our early understanding of genetic recombination was due to genetic analysis in fungi.

– Fungi can be unicellular or multicellular and fungal cells are often haploid (1n) and reproduce asexually.

– Fungal cells can also reproduce sexually by fusing with another cell to form a diploid (2n) cell.

– This diploid zygote can then proceed through meiosis to form four haploid daughter cells or spores.

• These four spores are often called a

• Can also be followed by mitosis to form an

• These daughter cells are in a sac known as an

– The position of the spores in the ascus can be unordered or ordered.

• If an ascus is not tight the spores may move around leading to unordered spores (S. cerevisiae and A. nidulans).

• An extremely tight ascus leads to ordered or linear spores (N. crassa).

• The position of the spores in the ascus reflects their position when they were formed by meiosis and mitosis.

• The tightness of the ascus keeps the spores next to the cell they formed from.

– Linear tetrad analysis can be used to map the distance between a gene and a centromere.

• Since you can visualize the centromere of the chromosome provides a way to correlate the location of the gene with actual cytological characteristics of a chromosome.

– N. crassa the gene A leads to orange pigmentation while the recessive allele (a) leads to albino pigmentation.

• So you can follow crossing over by visualizing spore color.

• If you have a 4 (orange):4 (albino) or M1 arrangement of spores it is called first-division segregation .

• 2:2:2:2 or a 2:4:2 or M2 is second-division segregation.

• Since second-division segregation (SDS) is due to crossing over the percentage of M2 asci can be used to calculate map distance between the centromere and gene of interest.

• A crossover will only separate a gene from its original centromere if the chiasma forms between the gene and the centromere.

– The chances of getting a 2:2:2:2 or 2:4:2 pattern depends on the distance between the gene and the centromere.

– Unordered tetrad analysis can be used to map genes in dihybrid crosses.

• Unordered tetrads contain spores that are randomly arranged.

• Dihybrid analysis can still be done by growing the spores and checking their phenotypes.

– This can determine if two genes are linked, assort independently this can be used to compute map distance.

• Ex: meiosis of a ura+ura-2, arg+arg-3 (heterozygous) diploid cell.

• This diploid was from a fusion of two haploid cells that were ura+arg+ and ura-2arg-3 respectively.

• Ura+ and arg+ are the normal alleles for arginine biosynthesis, while ura-2 and arg-3 are defective alleles (require uracil and/or arginine added to growth medium).

• The possible asci could contain all parental offspring (parental ditype, PD), all recombinant offspring (nonparental ditype, NPD), or 2 parental : 2 nonparental (tetratype,T) mixture.

• With linked genes what is the relationship between crossing over and the ascus that results.

– No Crossing over = parental ditype.

– Single crossover = tetratype.

– Double crossover between all four chromatids = nonparental ditype.

– Double crossover between three chromatids = tetratype.

– Double crossover between two chromatids = parental ditype.

• Allows us to calculate map distance as a percentage of offspring that carry recombinant chromosomes.

• Multiply ½ by tetratypes because only ½ of the spores in a tetratype are recombinant.

• This is fairly reliable but does not take into account double crossovers that occur when genes are further apart.

– Since we can calculate double crossovers with tetrad analysis we can use an equation that takes this into account.

• The total number of crossovers equals the single crossovers plus 2X the number of double crossovers.

• Overall double crossovers contain 50% nonrecombinant chromosomes (parental).

• There fore we must multiply the ratio by 50% or 0.5.

– Since parental ditype and tetratype are ambiguous (can be due to no crossing over single crossing over or double crossing over).

• We need to relate the equation to the number of parental and nonparental ditypes and tetratypes.

• Below is an accurate measure of map distance since it takes into account both single and double crossovers.