Chapter 8

Linkage, Crossing Over, and Chromosome Mapping

Morgan first mapped genes that were on the X-chromosome of the fruit fly. The alleles where followed from females to males to females, through several generations.
The next level of mapping determined the relative positions of various genes on the individual chromosomes (Figure 7.1).
It was discovered that genes are arranged in a linear order along the length of chromosomes with no branches or discontinuities.
Sturtevant, an undergraduate who worked in Morgan's lab, was the first to construct a chromosome map.
- He did it, start to finish, in just one evening.
- He had some results from dihybrid crosses that failed to conform to Mendel's rule of independent assortment.
- He brilliantly deduced that the deviation from Mendel's expected outcome was directly proportionate to the closeness of the genes on a chromosome.
The basic idea was that if genes are on the same chromosomes, they should all be inherited together.
The genes are linked together if on the same chromosome.
- The opposite of linked is independent assortment found when genes are on different chromosomes.
- It was discovered that the linkage of genes was not absolute. Some genes are "more linked" than some others.
- It turned out that recombination occurs within a pre-gamete cell between the homologous chromosomes.
- In fact, chiasmata (singular chiasma) were observed. Homologous chromosomes appeared to be exchanging linearly corresponding segments.

Mendel's Flawed Principle

In fact not all genes independently assort!
An early example was found in sweet peas.
Flower color and pollen length did not yield the expected 9:3:3:1 ratio in the F₂. Instead a 23:1:1:9 was observed. See Figure 7.3.

Frequency of Recombination is a Measure of Linkage Intensity

If you have a long ribbon stretched across a room, which has two beads on it, and you ask a blind person to cut the ribbon, would you expect the ribbon to get cut between the beads or on one or the other side of the two beads?
- Wouldn't it depend on how close the beads are to each other?
- If the beads are right next to each other, only rarely would they be separated by the cut.
- If the beads are far apart on the ribbon, the cut would almost always separate them.
The slogan for mapping genes by crossover frequency is "Genes that are close together tend to stay together, genes that are far apart tend to separate".
The rule will apply even to bacteria and viruses.
Another important point is that genes, no matter how far apart from each other, will never have higher recombination frequencies than the value that would be calculated if they were independently assorting.
For a dihybrid, test crossed to a double-recessive, the maximum value would be 50%.
- For example, a hybrid tall pea plant with smooth seeds is crossed to a double recessive, short plant with wrinkled seeds.
- (Whenever mapping, a heterozygote is crossed to a recessive homozygote. The events under investigation are those happening within the heterozygous individuals' germ cells. By crossing to recessives, what happened within the heterozyotes' germ cells can be seen without interference. The recessive alleles of the homozygote cannot hide or mask the alleles contributed to progeny from the heterozygotes.)
- With independent assortment, we would expect a 1 tall, smooth seeded : 1 tall, wrinkled seeded : 1 short, smooth seeded : 1 short, wrinkled seeded; ratio.
A parental type of progeny in a mapping study is one that has the alleles arranged on the chromosome the same way they are found in the hybrids' mitotic cells.
- The way the alleles are arranged in the dihybrids' mitotic cells depends on how they were contributed by the parents of the dihybrid. Both sets of parents of the dihybrid must be homozygous for all alleles involved in the mapping effort.
- For example, if one side of the dihybrids' parents were purebred tall and wrinkled seeded, the other parents of the dihybrids would have to have been short and smooth seeded. (Remember that the dihybrid is a dihybrid.)
- Then, one parental type would be tall and wrinkled seeded, and the other parental type would be short and smooth seeded. (For genes located on different chromosomes, this discussion is almost meaningless, but for those located on the same chromosomes, it is an important concept to eukaryote chromosome mapping.)
- For a two-point cross such as the dihybrid cross above, the other two outcomes would be classified as the recombinant types. They are tall and smooth seeded; and, short and wrinkled seeded.
When genes are linked closely enough to have recombination frequencies of less than 50%, the two most frequent types of progeny are always the parental types. Another way to correctly state this is that the most frequent type of progeny and its reciprocal are the parental types.
Often, we will not be given information about what the parents of the dihybrid were, but the most frequent types of progeny generated from the dihybrids' cross to double-homozygotes informs us of this, anyway.
In the example given above of a 1:1:1:1 ratio outcome, because the number of progeny for all four different phenotypes are equal, we could call any one of the progeny types a parental type. For example, lets label the tall, wrinkled progeny a parental type.
- Then, the short, smooth progeny would also be a parental type because it is the reciprocal of the tall, wrinkled seeded progeny. The reciprocal is used because the diploid dihybrid had different alleles for each of the two genes.
- Because the parents of the dihybrid were given as tall, wrinkled seeded, and short, smooth seeded; these two would be the best choice as the parental types.
- Lets not forget though that a 1:1:1:1 ratio is not indicative of gene linkage. It is consistent with alleles that independently assort. This example is being discussed solely to demonstrate that a 50% recombination frequency is the value expected for independent assortment.
To calculate the mapping distance for these two genes you would total the progeny counted for all categories. This value would generate the denominator. The numerator would be the number of recombinants.
- Imagine that the 1:1:1:1 ratio was derived by finding 100 tall, wrinkled seeded; 100 tall, smooth seeded; 100 short, wrinkled seeded; and 100 short, smooth seeded; among 400 progeny of the dihybrid - double homozygote cross.
- To calculate the mapping distance between the two genes, place the number of recombinant types over the total number of progeny.
- Recombinant types over the total are 200 (for the tall and smooth seeded; plus the short and wrinkled seeded) / 400, or .5. The recombination frequency would be 50%.
Therefore, if the genes are independently assorting, their recombination frequency would be .5 or 50%.
- This means 50% is the greatest value you can observe, and is equal to the value you would expect if the genes were independently assorting.
- To map genes, we must have recombination values that are meaningful. The values must be less than 50% for the dihybids' recombinant types.
- Below is an image that links to a Flash animation (.swf) that will help you understand two-point crosses. Watch it several times and be sure to understand all its details.

The crossover recombination that can be measured occurs within the F₁s only. The two most frequent types elucidate the genotypes of the parents of the F₁s.

Click anywhere within the above image to view the animation that generated it.

Whenever mapping genes on the chromosomes of diploid organisms, half of the pair are hybrid, while the other are homozygous recessive for all the genes to be mapped. The only convenient exception is for genes on the X-chromosome of organisms where one of the sexes has just one X. Drosophila males have just one X-chromosome. The F₁ females are crossed to males regardless of there genotype or phenotype. The male progeny of this cross are used to measure recombination frequency, since they have only one X-chromosome, and therefore no masking of alleles, and the one X-chromosome they have, they got from their female parent.
Mapping genes on the X-chromosome is easy because the males show any recombination that occurred in the females. (Figure 7.15)
What if genes are so far apart on the same chromosome that the mapping distance is greater than 50% (50 map units (mu))?
- Gene A is 25 mu from B and B is 35 mu from C. How many map units are there between A and C? The answer is 60 mu. However, if you test just A and C, your result would be just 50 mu.
- Only by finding a marker between A and C, could you determine the true map distance of 60.
- You can't get more random than the random you get with independent assortment, which is also what you get when the genes are on different chromosomes.
With the same example used above, what would have happened if the genes where not only on the same chromosome, but so close they never separated? This is called complete linkage.
- If the parents of the dihydrid were as stated above, instead of getting a 1:1:1:1 ratio, you would get a 1:1 ratio of progeny.
- Approximately half would be tall and wrinkled; the other half would be short and smooth.
- The other two possibilities would not occur because of the arrangement of the alleles on the chromosomes in the heterozygotes and the lack of recombination ((0+0)/1000 = 0 mu).

Crossing Over

During mitosis, homologous chromosomes do not pair and exchange pieces. The chromosomes within each somatic cell of a diploid organism are each still wholly as they were contributed by each parent.
During Prophase I of meiosis "Mom and Dad's chromosomes kiss goodbye".
- Only in the cells that are developing into gametes and in Prophase of Meiosis I, do the homologous chromosomes pair and exchange pieces. (Figure 7.12)
- Due to this process, the DNA from both parents mix for every given chromosome, except for the X-chromosome in heterogametic males (or heterogametic females like female birds). (Figure 7.7)
In each germ cell, during the S phase just prior to meiosis, the DNA replicates, so that during G_2, Prophase I and Metaphase I, there are four copies of each chromosome.
The two chromatids of the homologous chromosomes and the homologous chromosomes pair to form a tetrad (four-strands).
Strands break and exchange sections by reattaching to corresponding regions of their homologs. (Figure 7.9)
- This was first studied in an organism from the class Ascomycetes. (Figure 7.8)
- Cytological evidence came from other studies, especially one in maize.
- Physically different chromosomes could be observed to have exchanged pieces, and phenotypic changes could be correlated to these exchanges. (Figure 7.10)

Relating Crossover Frequency to Distances on Chromosomes

The further apart genes are from each other the greater will be the recombination frequency: The blind person and the scissors.
The most frequent group will be the parental types, sometimes called parental di-types, when mapping genes on a certain chromosome because there will be 2 parental types, no matter how many genes are being mapped at a time. This is because the organism is diploid and has two chromosomes, one from each parent.
- In the example above, more tall, wrinkled; and, short, round individuals than any other types, would be expected, if the genes were linked and close enough to be mapped because this was the arrangement of the alleles within the dihybrids' somatic cells.
If 40 tall, wrinkled; 40 short, smooth; 10 tall, smooth; and, 10 short, wrinkled were counted in the progeny, what would be the distance between the height and seed morphology genes?
- The parental types would be the most frequent group and its reciprocal: tall, wrinkled; and, short, smooth.
- The recombinant types would be tall, round and short, wrinkled. The mapping distance would be (10+10)/(40+40+10+10) = 20% or 20 map units. (Figures 7.4, 7.5 and 7.6)
Using two-point crosses, how would you order three genes?
- Imagine you have mapped the distance between two genes, A and B, and it was 50 mu.
  - You have also mapped the distance between A and C and it is 35 map units; and the distance for B and C is 30 map units.
  - You can now order the genes, and provide a map distance estimate for all three genes.
  - First, the two closest genes are B and C; B must be next to C.
  - Next, the distance between A and B is 50 map units.
  - This is the largest distance possible and so might actually be any number equal to or larger than 50 map units.
  - C is closer to A than B.
  - Therefore, the order must be B-C-A, and B to C is 30, C to A is 35, and B to A is 35 plus 30, or 65 map units.
- How do you know the correct order is B C A and not A C B?
  - Just how many ways can three genes be ordered?
  - When ordering three genes, the rest of the chromosome is temporally ignored.
  - There is no difference between B C A and A C B.
  - Therefore there are only three orders possible.
  - The only difference in the three orders is which gene is in the middle. (Figures 7.13 and 7.14)

Solving Three Point Crosses

You can map three genes at a time. The trihybrid would be backcrossed to individuals that are recessive for all three genes. With independent assortment, we would expect a 1:1:1:1:1:1:1:1 ratio.

			1Aa	1aa
			1Bb	1bb
		1AaBb	1aaBb	1Aabb	1aabb
			Cc	cc
1AaBbCc	1aaBbCc	1AabbCc	1aabbCc	1AaBbcc	1aaBbcc	1Aabbcc	1aabbcc

Barring epistatic relationships, in addition to generating eight distinct genotypes, the cross would generate 8 different phenotypes as well.
The use of a triple homozygous recessive makes it possible to calculate the recombination frequencies generated by the trihybrids during meiosis.
How many different genotypes and phenotypes would be expected among the progeny of trihybrids crossed to triple homozygous recessives, if all three genes were completely linked?
- If completely linked, you would have just two alternate arrangements of three alleles for the six different alleles in the heterozygotes.
- This is also the case in somatic cells when the genes are not completely linked.
- This is because one chromosome comes from each of the two parents.
- For example, if the parents of the heterozygote were purebred tall, wrinkled and purple; and, short, round and white, the only types of progeny you would find from the F₁ cross to the homozygous recessive strain would be tall, wrinkled, and purple; or short, round and white. The ratio would be 1:1.
If the genes were not completely linked, the most common two phenotypes would be the parental types. The most common two types in the example above are the tall, wrinkled and purple; and, the short, round and white.
- If detectable crossovers occurred between all three genes, six different phenotypes would be present in the progeny.
- If double-crossovers also occurred, as many as eight different phenotypes would be present among the progeny.
- Eight different phenotypes of progeny would be present in approximately equal number if all three genes independently assorted.
It is an essential first goal to determine the order of the genes to map three-point crosses, successfully. You will fail two-thirds of the time if you try to skip this step.
- The order of the genes given to you in a genetics problem is most likely wrong because there are three possibilities and the order given is just one of them.
- Solve your problem by first determining which of the three possible orders is correct.
There is only one simple way to determine the order of the genes:
To do this you must find one parental type and the double-crossover type that is most like it.
- 'Most like it' means that only one allele of one of the genes differs.
- The one gene that differs is the one in the middle.
- If the parental type is Abc and the double-crossover type most like it is abc then the gene that is in the middle is A (or a). The order would be bAc (or cAb).
- A double crossover is when two crossovers have occurred between the three genes. The only gene that will differ from the parental type it is most alike will be the one in the middle.
How do we determine which are parental and which are double-crossover types?
- The most common type and its reciprocal are the parental types.
- There are 8 potential outcomes as you can see from the table above.
- Of the eight, the two rarest types will be double-crossovers because the chance of a double crossover is the probability of one of the single-crossovers and the probability of the other single-crossover type, and therefor is the product of the two probabilities.
- If you play with it a bit you can easily see that the two double-crossover types are reciprocal of each other.
- From the example above, if one of the double-crossover types is abc, the other double-crossover types is ABC.
If you have eight types of progeny, it is easy. Find one parental type and the double-crossover type it's most like and the gene that is different is the middle one. You got your order.
What if the double-crossover types are too rare to find?
- If you find you have 7 types, the missing one would be a double-crossover type. You know what it is; it's the reciprocal of the one double-crossover type that you found.
- If you have only six different types, can you find out what the double-crossovers would have been if you had any? Yes, there would always be 8 potential types.
- If you get only six, discover one of the missing combinations; it and its reciprocal would have been the double-crossover types.
- How can you determine what the eight possible types of progeny are?
  - Use the matrix system such as the one shown in the table above.
  - Check off the ones you find, and the missing ones would be the double-crossover types.
What if only 5 types were found?
- Then, one would be missing a reciprocal.
- The reciprocal of that one would be a single-crossover type.
- The last missing two would be double-crossover types.
What if there were only 4 different types of progeny? Then, the problem is not solvable.
When doing a three- point cross, which involves three genes, each with a pair of alleles, in order of expected frequency, you expect two parental types, the most common types; a single-crossover type and its reciprocal; another single-crossover type and its reciprocal; and, least in frequency, two reciprocal double-crossover types.
- After you have found the two parental types, record them on the table one under the other.
- Then find the next most common type and its reciprocal, and record them one under the other on the table.
- This will be a single-crossover type.
- Find the next most common type and its reciprocal, these will be the other single-crossover types.
- The two remaining are the double-crossover types, one of which you will use to determine the correct gene order.
You are now prepared to solve a problem.
- In Drosophila, three genes are linked on the same chromosome. For one of the genes, the allele w⁺ is dominant to w; for another gene, the allele y⁺ is dominant to y; and, for another gene, the allele ec⁺ is dominant to ec. Homogeneous, heterozygous female flies are crossed to male flies that have the recessive phenotype for all three genes. The following list is of all progeny from the crosses classified by their phenotypic consistancy to the alleles used to describe them:
  - Map these three genes.
  - Below is a Flash animation that might be useful to you. Simply click the image to play the .swf file.

Click anywhere within the above image to view the animation that generated it.

The slide below is similar to the one above, except the genes are on the X-chromosome. This makes it even more interesting to study.
The values and gene order are the same.
Look at the slide below closely. Observe all details and make sure you understand every part of it. It will explain to you how to calculate distances and draw your map. (Also, see Figure 7.16)



Eukaryotic Gene Mapping Problems Set I

Eukaryotic Gene Mapping Problems Set II

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