Homework - Genetics linkage problem

Homework - Genetics linkage problem

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A farmer purchases f1 hybrid seeds. These seeds contain four heterozygous loci that give a significant increase in crop yield when compared to either homozygote. Two of these loci are on the same chromosome and show partial linkage. If the farmer allows the f1 plants to self-fertilize, what is the probability that a seed from the resulting f2 generation has all four heterozygous loci?

a) at least 1/2

b) between 1/2 and 1/4

c) between 1/4 and 1/8

d) between 1/8 and 1/16

e) less than 1/16

I think it's either d or e. For the first two genes (unlinked), there is 1/2 chance to get heterozygote. So, 1/2 * 1/2 = 1/4. For the next two linked genes, if they were unlinked, it would be 1/2*1/2 = 1/4 for a total of 1/16. But since they are linked, I'm thinking that the chance for heterozygote might be reduced, so maybe the answer is e) less than 1/16?? Not sure where to end, though…

I agree with your reasoning! If the two genes are perfectly linked, then the answer is $frac{1}{8}$. If they are perfectly unlinked, then the answer is $frac{1}{16}$. The answer therefore must be somewhere between $frac{1}{8}$ and $frac{1}{16}$. Good job!

Of course the answer e) is pretty silly as any other answer ('a','b','c' and 'd') would yield to 'e' being necessarily true. So, if you have to choose only one, I would just go with 'd' and highlight that the answer 'e' is silly.

I still don't understand how if the two are completely linked, the answer would be 1/8.

Let's consider that at both loci, we have two allelesAandB. Here is a representation of the first F1 individual


Here is a representation of the second F1 individual


Of course, they are the same as the parental lines are pure breed. Now, if the two loci are perfectly linked, then the offspring can have the following possible genotypes






, but the offspring cannot be


or an other combination requiring recombination between the two loci. Therefore the probability to be heterozygote for the F2 at two perfectly linked loci is the same as the probability to be heterozygote at one of these two loci (if we assume the mutation rate is negligible).

Why can't parents be: parent1: -------B----A----- -------A----B----- parent2: -------A----A----- -------B----B-----[?]

Because parental lines are pure lines. When we talk about an F1 generation, we talk about the hybrid between two pure line. By pure line, we mean that we got rid of all genetic variance. So, one parental line is fixed for theAalleles and the other line is fixed for theBalleles. In other words here are the genotypes of the pure parental lines.




One parent can therefore only give


while the other one can only give


so, the F1 generation is therefore necessarily heterozygote at all loci.

Examples of Genetic Linkage in Humans (With Applications)

In this article we will discuss about the examples of genetic linkage in humans. Also learn its clinical applications.

1. Nail-Patella Syndrome and ABO Blood Groups:

This genetic linkage has been demonstra­ted by Reneisck and Lawler (1955). The nail- patella syndrome or hereditary onychosteodysplasia is characterised by a constellation of abnormalities. The nail dystrophy is vari­able, ranging from a triangular lunule, through moderate abnormality with discoloration, a longitudinal crack and reduced size, and, finally, to severe dystrophy with absence of a large part of the nail. Thumb and index finger and first and second toes are especially affec­ted.

The patellae are usually rudimentary or may be absent altogether. Some of the affec­ted persons exhibit curious conical bony pro­jections from the middle of the iliac bone, known as iliac horns. The elbows also often show abnormalities.

Like this, the abnormali­ties are numerous but they are not very severe and length of the life and fertility do not appear to be affected. This nail-patella syn­drome is due to a dominant gene showing per­fect regularity of transmission. The perfect regularity of transmission means affected per­sons always have an affected parent and the mating affected X normal gives affected and normal offspring in equal proportion.

An analysis of the pedigree of nine family groups indicated that the locus for the nail-patella gene is situated on the same chromosome as the ABO locus and their distance apart is about 10 unit of crossing over. It should be remembered here that the nail-patella gene is not associated with any particular ABO gene.

In some family it is found to be on a chromo­some containing B gene, in some other on a chromosome containing the O gene and in some exceptional cases it has been to the A gene. Taking the children of affected parents in different pedigrees, the distribution of blood groups amongst nail-patella syndrome and normal are as follows:

2. Linkage of the HLA Loci:

The region on the chromosome 6 that carries the major histocompatibility complex is now known as HLA. The different gene loci are designated as A, B, C, D and the specifi­cities or alleles at each locus are identified by numbers 1, 2, 3 etc. The actual order of the loci is thought to be D, B, C, A and are situ­ated close to each other and, therefore, the alleles at each of these four loci will nearly always be inherited together.

The particular combination of specificities on the same chro­mosome is called the haplotype and the haplotype of one chromosome is not of course necessarily the same as the other chromo­some. In one study of 1,362 children whose parents had been typed, it was found that only 11 children had recombinant haplotypes. The crossing over occurs only between locus A and B.

The following is the schematic representa­tion of the gene map of part of the short arm of the chromosome 6 showing the arrange­ment of HLA loci:

3. Linkage between Lutheran and Secretor Types:

This linkage group has already been detec­ted but the chromosome pairs remain unidenti­fied. These are at least two alleles at the Lutheran locus (Lu) which denote antigenic differences in the blood.

The dominant secretor allele allows the detection of the blood group antigen A and /or B in the saliva. It has been demonstrated by studying several pedigree that there is a fairly close linkage between Lu and Se loci. However, according to Green- wait (1955) Lu and Se are 16 unit apart i.e. 16% recombi­nation occurs between Lu and Se.

4. Linkage between Duffy Blood Type and Lamellar Cataract:

There is a special type of blood antigen known as Duffy antigen due to the presence of a gene Fy in the population of Great Britain and Negroes. This antigen is harmless. Lamellar cataract is a particular form of eye defect due to presence of a dominant gene. According to Renwick and Lawler there is a linkage between Duffy antigen gene and cataract gene.

Clinical Applications of Genetic Linkage:

(i) Helps to find marker genes that are linked to serious dominant conditions.

(ii) Helps in the prenatal diagnosis of severe dominant condition where the pre­sence or absence of marker gene could be detected in amniotic fluid.

(iii) Genetic marker would generally be a series of alleles at a locus that is closely linked to the particular disease gene, and the closer it is the better it is as a marker, because uncer­tainty due to possible crossing over between the marker and disease genes is reduced to a minimum.

(iv) Genetic linkage does not imply that the disease in general is associated with any particular allele.

(v) Genetic linkage helps in predictions on the transmission of the disease in small family pedigrees.

(vi) Of course, the prediction cannot be made with 100% accuracy of the possibility of a crossing over.

Genetics Mapping

In the tomato, three genes are linked on the same chromosome. Tall is dominant to dwarf, skin that is smooth is dominant to skin that is peachy, and fruit with anormal tomato shape is dominant to oblate shape. A plant that is true-breeding for the dominant traits was crossed to a dwarf plant with peachy skin and oblate fruit.The first generation plants were then test crossed to dwarf plants with peachy skin and oblate fruit. The following results were obtained.

151 tall, smooth, normal
33 tall, smooth, oblate
11 tall, peach, oblate
2 tall, peach, normal
155 dwarf, peach, oblate
29 dwarf, peach, normal
12 dwarf, smooth, normal
0 dwarf, smooth, oblate

Construct a genetic map that describes the order of these three genes and the distances between them.

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A chromosome possesses many genes & all genes present in the chromosome are inherited together.

Linkage: Study of inheritance of all genes present in a chromosome together.

  • All genes in a chromosome are together referred as linked genes & they form a linkage group.
  • The total number of linkage group in an organism is equal to its haploid number of chromosomes.
Bateson & Punnet:
  • While working on sweet pea (Lathyrus odoratus) observed that genes for flower color & pollen shape remain together and do not assort independently according to Mendel’s law of Independent assortment.
  • Test cross failed to produce 1:1:1:1 ratio & instead produced 7:1:1:7.
  • They gave the Coupling & Repulsion theory:
  • Coupling: when genes come from the same parent they enter the same gamete & are inherited together.
  • Repulsion: genes are inherited separately when genes come from different parents & they enter different gametes

T H Morgan’s Linkage:

He worked on Drosophila.

  • linked genes are genes present on the same chromosome.
  • Linked genes are inherited together.
  • Genes are linearly arranged in a chromosome.
  • Strength of linkage: Genes which are closely located show strong linkage & genes which are located far show weak linkage.
  • He stated that Coupling & Repulsion are two aspects of Linkage.
Linked genes show 2 types of arrangement:
  1. Cis arrangement: dominant alleles of 2 or more genes are present in one chromosome & its recessive alleles in its homologue. AB/ab. This is Coupling.
  2. Trans arrangement: the dominant allele of one pair & recessive of the other pair together lie in a chromosome.Ab/aB. This is Repulsion.
Linked genes are of 2 types :
  1. Complete linkage : genes for 2 or more characters appear together for two or more generations in their parental combination.They are closely located in the chromosome.
  2. Incomplete linkage: the parental combination of 2 or more characters are not retained in the next generations.They are not closely located in the chromosome.
  • Incomplete linkage occurs due to crossing over.

Morgan’s experiment:

Crossing over:

  • It is the exchange of segments between non-sister chromatids of homologous chromosomes.
  • It occurs during pachytene stage of prophase I in meiosis.
  • Crossing over always occurs between linked genes.
  • It produces recombination of linked genes which play very important role in evolution.
  • Recombination frequency helps in finding out the distance between genes. Given by Sturtevant.
  • Recombination frequency helps in the construction of genetic maps of the chromosomes.

In a dihybrid cross:

  • When the test cross result is 1:1, genes are linked and there is no crossing over.
  • If the test cross result is 1:1:1:1 means the genes are independently assorting (present on separate chromosomes).
  • If in the test cross result parental combination is more than 50% & recombination is less than 50%, the genes are linked and crossing over has occurred.

Ramneet Kaur

Ramneet Kaur is professor of Biology.She helps MBBS aspirants to achieve their dream to get into a Medical College.Her ability to make difficult concepts simple and understandable makes her one of the most loved teachers. She has over 30 years of experience of teaching the students. She has mentored around 1 lakh students till date.Being extremely passionate about her students, she wants to benefit the students by helping them giving easy review and weekly MCQs through this blog

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Nice notes
Helpful in self studies

its very relevant and exact informations thats what i want to know than other articles…
please mam help me to make the punnet square to obtain 7:1:1:7 ration…

mam i need your notes with full tricks so from where i can get it please please

Lecture 14: Genetics 3—Linkage, Crossing Over

In his third lecture on genetics, Professor Martin picks up from the last lecture on eye color in fruit flies, and then continues with Mendelian inheritance. He then talks about linkage, crossing over, and gene mapping.

Instructor: Adam Martin

Lecture 1: Welcome Introdu.

Lecture 2: Chemical Bonding.

Lecture 3: Structures of Am.

Lecture 4: Enzymes and Meta.

Lecture 5: Carbohydrates an.

Lecture 9: Chromatin Remode.

Lecture 11:Cells, The Simpl.

Lecture 16: Recombinant DNA.

Lecture 17: Genomes and DNA.

Lecture 18: SNPs and Human .

Lecture 19: Cell Traffickin.

Lecture 20: Cell Signaling .

Lecture 21: Cell Signaling .

Lecture 22: Neurons, Action.

Lecture 23: Cell Cycle and .

Lecture 24: Stem Cells, Apo.

Lecture 27: Visualizing Lif.

Lecture 28: Visualizing Lif.

Lecture 29: Cell Imaging Te.

Lecture 32: Infectious Dise.

Lecture 33: Bacteria and An.

Lecture 34: Viruses and Ant.

Lecture 35: Reproductive Cl.

ADAM MARTIN: And so I wanted to start today's lecture by continuing what we were talking about in the last lecture. So I'm just going to hide this real quick. And so we're talking about the fruit fly and the white gene and the white mutant, which results in white-eyed flies. And we talked about how if you take females that have red eyes and cross them to males, the white-eyed male, then 100% of the progeny has red eyes in the F1 generation.

And so I asked you guys, would you get the same results if you did the reciprocal cross? So what if we took white-eyed females and mated them to red-eyed males? So what about this? Actually, I'm going to move this over to over here so that maybe it's more visible. So what if we have white-eyed females and crossed this to red-eyed males?

So let's unpack this sort of a little bit at a time. So what's the genotype of these white-eyed females here? Miles?

AUDIENCE: So if you designate the eye gene as the letter A, a female would be X lowercase a, X lowercase a.

ADAM MARTIN: Yes. So Miles is exactly right. So the dominant phenotype is red eyes, because the gene encodes for an enzyme that's important for the production of the red pigment. And so X lowercase a here would be a recessive mutant that lacks the pigment.

And because it's a recessive allele-- because you need only one copy of this gene to produce the pigment. So the recessive allele results in the white phenotype. Therefore, this has to be homozygous recessive. How about this red-eyed male? Yeah, Ory?

AUDIENCE: Wouldn't you have a Y and then an X capital A?

ADAM MARTIN: Yes. So this would be this phenotype, right, where capital A is the gene that produces-- is a normal functioning gene that produces the pigment. So then in your F1 here, are you going to see something similar to this or something different?

AUDIENCE: Something different.

ADAM MARTIN: Different, great. Who said different? Javier? Do you want to propose what you might see?

AUDIENCE: Yeah. For the males, they're going to inherit the Y gene from the father and the [INAUDIBLE].

ADAM MARTIN: Exactly. So the males are going to get the Y from the father, and they're going to get one X from their mother. So all the males are going to be of this genotype here, which means they're going to have what color eye? Javier is exactly right. That means they're going to have white eyes. So all the males will have white eyes. And what about the females?

ADAM MARTIN: Yeah. So Ory is saying the males are going to get red eyes, right, because they're-- or the females are going to have red eyes, because they're going to get the X chromosome from their father, which has the dominant gene that produces the red pigment. So all the females are going to be heterozygous, but have a functional copy of this gene. So all of the females will have red eyes.

OK, does everyone see how-- now, how would this compare with Mendel's crosses and pea color? Would there be a difference in Mendel's crosses if you switch the male versus the female if these were autosomal traits? Ory is shaking his head no, and he's right, right? In that case, it doesn't matter. You can do the reciprocal crosses, you get the same result. But because this is sex linked, which one is the male and which is the female is relevant.

And this actually relates to something that we just saw on the MIT news. I just got this email this morning, but it came out I think yesterday, which is that biology-related research in the mechanical engineering department-- specifically, the CAM lab-- they've been able to design a 3D sort of model for ALS disease, which is also known as Lou Gehrig's disease. And so what they've done in the CAM lab is to take cells from either patients that have ALS or from normal individuals, and they coax these cells to become neurons.

Here, you're seeing a neuron in blue and green here. And you see the neurites extend from this neuron. And they have a model where this neuron can then synapse with a muscle. And so they're using this 3D sort of tissue model to model ALS and to look for drugs that might affect ALS, potentially curing ALS.

And so last night, I started reading about ALS and was pleased to find that there's actually a very rare X-linked, dominant form of the disease that can be passed on from generation to generation. And the inheritance pattern of this X-linked, dominant version of ALS would have an inheritance pattern that's similar to what we observe for the white mutant in the fruit fly, right?

Whereas, if you have an affected father, and this is a dominant mutant on the X chromosome, then all of his daughters will get that X chromosome and be affected, whereas the sons will all be unaffected. However, if you have the reciprocal situation, where you have an affected mother and an unaffected father, then the sons and daughters get the disease randomly. So this is a sort of form of inheritance, which is relevant if you're considering human disease and some forms of it.

Most versions of ALS, well, are sporadic, but inherited forms are usually autosomal dominant. So this is a rare case here. But I thought it was interesting in that it's relevant to what we've been talking about.

So now, just to recap-- here, I'll throw this up so everything's up. So in the last lecture, we talked about Mendelian inheritance. And we talked about when you take two parents that differ in two traits and you perform a cross, you get a hybrid individual that is heterozygous for both genes. And now this is the F1 individual.

Let's say we want to know what types of gametes this F1 individual produces. We can perform a type of a cross known as a test cross, where we cross this individual to another individual that is homozygous recessive for both these genes, which means that you know exactly which alleles are coming from this parent. And they're both recessive, so you can see whether or not the gamete produced by this individual has either the dominant or the recessive allele.

Let me see. I'll boost this up. So now we can consider the different types of progeny that result from this test cross. And some will have the two dominant alleles from this parent and will, therefore, be heterozygous for the A and B gene. And it would exhibit the dominant A and B phenotype. I think that is what I'm showing here.

So if the chromosomes, during meiosis I, align like this, then the two dominant alleles segregate together, and you get AB gametes. And you also, reciprocally, get these lowercase a and b gametes, as well. So that's the other class here. So you can get these two classes of progeny.

And the phenotypes of these two classes will resemble the parents, right? So these are known as parental gametes. So these are the parentals. But you know because Mendel showed that if you have genes and their alleles on separate chromosomes, they can assert independently of each other. So an alternatively likely scenario is that the chromosomes align like this, where now the dominant allele of B is on the other side of the spindle. And therefore, these chromosomes are going to segregate like this during the first meiotic division.

And that gives rise to gametes that have a different combination of alleles than the parents. So you have some that look like this. So each of these would be different classes of progeny. And you have one last class that would look like this. And so neither of these look like the original parents, and so they're known as non-parental.

And so if these two genes are behaving according to Mendel's second law, where there's independent assortment-- if you have independent assortment, what's going to be the ratio of parental to non-parental? Rachel?

ADAM MARTIN: Yeah, Rachel says one to one, and I think a number of others also said one to one. So you have 50% parental, 50% non-parental, right? Because it's equally likely to get either of those alignments of the homologous chromosomes during meiosis I.

So now I'm going to basically break the rules I just explained to you in the last lecture and tell you about an exception, which is known as linkage. Gesundheit. And in the abstract sense, linkage is simply when you have two traits that tend to be inherited together. So just considering probability. So you have traits inherited together. They're exhibiting what is known as linkage.

But that's an abstract way to think about it. It's just based on probability, right? So a physical model for what linkage is, is that you have chromosomes. The genes are on the chromosomes. And for two genes to be linked, those genes are physically near each other on the chromosome. So the physical model is that two genes are near each other on the chromosome.

OK, so let's consider again these generic genes, A and B. If A and B resulting from this cross-- if these two happen to be on the same chromosome, now they're physically coupled to each other. Then they're going to tend to be inherited together. No matter how these align, they're always going to go together during the first meiotic division. And that's just going to only give you the parental gametes.

So if there's linkage, you're going to have-- let's consider the case where you have complete linkage. If you have complete linkage, 100% of the gametes are going to be parentals, and you're going to have 0% non-parental. That's if the genes are really, really, really close to each other, and maybe you don't count so many progeny. You won't see any mixing between the two.

But there is a phenomenon that can separate these genes, and it's known as crossing over. And another term to describe it is recombination. So the alleles are getting recombined between the chromosomes. Recombination. And what crossing over or recombination is, is it's a mixing of the chromosomes, if you will. Or it's an exchange of DNA.

So there's a physical exchange of DNA from one of the homologous chromosomes to the other, OK? So you can think of this as an exchange of DNA between the homologous chromosomes, OK? And that's important. It's not an exchange between the sister chromatids, but between the homologous chromosomes that have the different alleles.

And what's shown here is a micrograph showing you a picture of the process of crossing over. You can see the centromeres are the dark structures there. And you can see how the homologous chromosomes intertwine. And there are regions where it looks like there's a cross. Those are the homologous chromosomes crossing over and exchanging DNA such that one part of that chromosome gets attached to the other centromere.

So I'll just show you in sort of my silly cartoon form how this is, just to make it clear. So let's say, again, you have these A and B genes, and they're physically linked on the chromosome. During crossing over, you can get an exchange of these alleles, such as a bit of one chromosome goes to the other homologous chromosome and vice versa, OK? So now you have the dominant A allele with the recessive b allele and vice versa.

So now during meiosis I-- after meiosis II, this will give rise to two types of gametes, one of which is non-parental. And the same for the one down here. You get two types of gametes. One is non-parental-- the lowercase a, uppercase B gamete.

So this happens if there's incomplete linkage. That means there can be a recombination event that separates the two genes. And I'm going to give you an example of a case where data was collected with what fraction of each class there is. So now we're considering an example where you have a linkage. So A and B are on the same chromosome. And so we'll consider a case where, in this class, there are 165 members. For this one, there is 191.

So I'm kind of-- line down like this. And then for the first recombinant class, 23 individuals. And for the last, there are 21 individuals. So you can see there are many more of the parental class than the recombinant class, but we can calculate a frequency, or recombination frequency, between these two genes. And in this case, the recombination frequency is 44/400, which is equal to 11%, OK?

So 11% of the progeny from this cross had some sort of crossing over between the A and B alleles. It would've been up here. Now, this frequency is interesting, because it is proportional to the distance that separates these two genes. So this recombination frequency is proportional to the linear distance along the chromosome between the genes.

Now, it also depends on the recombination frequency in a given organism or in a given part of the chromosome. So when you're comparing recombination frequencies between different organisms, there's actually differences in the real different-- they're not equivalent. You can't compare them, basically. And also, there are regions of the chromosome where recombination happens less frequently than others. And so, again, you can't compare distances along those.

But overall, you can use this as a distance in order to map genes along the linear axis of a chromosome. And maps are useful, because you can see where stuff is, right? So in this example here, I'll highlight a couple of places. Here's Rivendell. Here's Lonely Mountain. Here's Beorn's house.

So let's say we are able to determine the distance between Rivendell and Lonely Mountain, and the distance between Lonely Mountain and Beorn's house, and the distance between Rivendell and Beorn's house. You'd be able to get a relative picture of where all of these places are in relation to each other. So this is a two-dimensional map I'm showing here. It's not one dimensional, but chromosomes are one dimensional, so it's a bit more accurate, OK?

So this idea that recombination frequency can be used to measure distances between genes and that this could be used to generate a map is an idea that an undergraduate had while working in Thomas Hunt Morgan's lab back in 1911. And what I find fascinating about the story is this guy basically blew off his homework to produce the first genetic map in any organism. So the person who did it was Alfred Sturtevant, and he was an undergraduate at Columbia working for Thomas Hunt Morgan.

And I'll just paraphrase this quote here. In 1911, he was talking with his advisor, Morgan, and he realized that the variations in the strength of linkage attributed by Morgan to differences in the separation of genes-- so Morgan had already made this connection, that the recombination frequency reflects the distance between the genes. But then Sturtevant realized that this offered the possibility of determining sequences in the linear dimension of the chromosome between the genes, OK?

So then-- this is my favorite part-- "I went home and spent most of the night, to the neglect of my undergraduate homework, in producing the first chromosome map." And this is it. So the first chromosome map was of the Drosophila X chromosome, which we've been talking about. There's the white gene, which we've been talking about in the context of eye color. There's a yellow body gene here. There's vermilion, miniature, rudimentary, right?

These are all visible phenotypes that you can see in the fly. And you can measure recombination between various alleles of these different genes.

All right, so now I want to go through with you an example of how you can make one of these genetic maps. And it's essentially the same conceptually as to what Sturtevant did. And it involves what is known as a three-point cross. So a three-point cross. So there are going to be three genes, all of which are going to be hybrid, and I'll start with the parental generation that is little a, capital B, capital D. And we'll cross this fly or organism to an organism that is capital A, lower case b, lowercase d. Yes, Carmen?

AUDIENCE: So when you write the gametes up there, does that imply that they were analogous parents?

ADAM MARTIN: So what I'm writing here is the phenotype, basically. And so these are homozygous for each of these, yes. I could also write this as-- but I'm not going to draw the chromosomes, because it kind of gets more confusing. I'll draw the chromosomes here in F1, because we have now, basically, a tri-hybrid with one chromosome that looks like this, right? They got that chromosome from this individual here. And another chromosome will look like this.

See? So this F1 fly is heterozygous for these three genes, and it has these two parental chromosomes. So now we can look at the gametes that result from this fly by doing a test cross, just like we did before. And so we want to cross this to a fly that's homozygous recessive for each of these genes. And now we can look at the progeny.

And just by looking at the phenotype, we're going to know the genotype, because we know all of the flies from this cross have a chromosome from this individual that has recessive alleles for each gene. So we can consider now this first one here. That's one potential class of progeny. Another class would be this one. And these two, you can see, resemble the parents, right? So these are the parental classes of the progeny. So this is parental.

All right, now you can consider all other combinations of alleles. And so I'll quickly write them down. You could have something-- progeny that look like this and this. These are just kind of reciprocal from each other. You could have progeny that look like this and this. And the last class would be this and this.

So all of these progeny that I drew down here are recombinant, because they don't resemble the parents, right? Because there are three genes, now there's many more ways to get recombinant progeny, as opposed to having just two genes, right? So you can have many different combinations of these different alleles. And so now I'm going to give you data from a cross with three such genes.

So you might get 580 individuals that look like this, 592 like this, 45 and 40, 89, 94, 3, and 5. So this is data that is, I believe, from fly genes. I've just ignored the fly nomenclature, because it's confusing, and just given them lettered names, OK? But this reflects data from some cross somewhere.

So now we want to know-- let's go back to our map. We want to make a map, OK? And so to make our map, we're going to want to consider all pairwise distances between different genes. So we'll start with the A and B gene. I'll write over here. So let's consider the A/B distance. And remember, to get a distance, we're looking at the number of the frequency which there is recombination between these two genes, OK?

So we now have to look through all of these recombinant class of progeny and figure out the ones that have had a recombination between A and B, right? So on the parent chromosome, you see little lowercase a started out with capital B and vice versa. So any case where we don't have lowercase a paired with capital B, there's been some type of exchange.

So here, lowercase a's with lowercase b. So that's a recombinant. Here, uppercase A's with uppercase B. That's a recombinant, too. So we have to add all these up. So 45 plus 40. How about here? Recombination here or no? Yes. I'm hearing yes. That's correct. Here, recombination, yes or no? Carmen's shaking her head no. She's exactly right.

So we just have to-- these are all the recombinants between A and B. So it's 45 plus 40 plus 89 plus 94, which equals 268 over a total progeny of 1,448. And that gives you a map distance of 18.5%. Because this method was developed in Morgan's lab, this measurement is also known as a centimorgan. It was named in honor of Morgan. So that's what I refer to when I have lowercase c capital M. That's a centimorgan. So you can also use centimorgan here.

All right, so that's A and B, but now we have to consider other distances. So how about the A/D distance? And again, we have to go through and figure out where the alleles for A and D have been recombined, OK? So little a is with capital D, and upper case A is with lowercase d. So we have to find all the cases where that's not the case.

Here, this is lowercase a with capital D and capital A with lowercase d. This is parental from the respect of just the A and D genes, but all the rest of these guys are recombinants, OK? So this is 89 plus 94 plus 3 plus 5, which comes out to be 191 over 1,448. And this is 13.2 centimorgans. So that's the distance between A and D. Distance between A-B, distance between A-D.

So the last combination, then, is just B and D. So if we consider the B/D distance, again, we have to look for all cases in which lowercase b and d become separated and uppercase B and D become separated. Here, they're separated. Here, they're separated. Wait, no, not here, sorry. Here, they're not separated. Here, they are separated.

Everyone see how I'm doing this? Are there any questions about it? You can just shout it out if you have a question? So this distance is 6.4 centimorgans. Everyone see how I'm considering every pairwise combination of genes and then just ignoring the other one and looking for where there's been a recombination in the progeny?

So now that we have our distances, we can make our map, right? So the two genes that are farthest apart are A and B, OK? So that's kind of like here, Rivendell and Lonely Mountain. Those are the two genes that are at the extremities. So I'll draw this out. It doesn't matter which way you put it. We're just mapping these genes relative to each other. But B and A are the farthest apart from each other.

Now if we consider the distance between B and D, that's 6.4 centimorgans. So it appears that D is closer to B than it is to A, because it's 13 centimorgans away from A, OK? So D is kind of like Beorn's house here. It's closer to Rivendell than it is from Lonely Mountain. So we'll put that in there. This distance is 6.4 centimorgans. And then the distance here is 13.2 centimorgans.

As far as I know, no one of the field of genetic mapping has been assaulted by large spiders, but the field is still young. So one thing that should be maybe bothering you right now is if you add up the distance between B and D and D and A, you don't, in fact, get 18.4 centimorgans. Instead, you get 19.6 centimorgans. This is 19.6 centimorgans. So it seems like, somehow, we are underestimating this distance here, OK? So we seem to be underestimating this.

So why is it that we are underestimating this distance? Well, to consider that, you have to sort of look at how all of these classes were generated. So now I'm going to go through each class, and we'll look at how it was generated. I'm also going to-- well, I'll just draw new chromosomes. So we have to draw this order now.

We have B, D, a. So the first chromosome is B, D, a. That's right, B, D, a. The other chromosome is b, d, A. So now we look and see how this recombinant class was generated. So this is lowercase a. We'll start with b-- lowercase b, lowercase a, but capital D. So it's capital D, lowercase a. So that recombinant results from this chromosome here, where there's crossing over, and little b gets hooked up to big D and little a. See how I did that?

And then if we consider this class, this is uppercase B, lowercase d, capital A. So these two classes of progeny result from a single crossover between B and D, OK? So this is a single crossover between B and D genes. And now we can go through and look at how this is generated.

So to get all recessive alleles on the same chromosome, there would be a crossover here. And so this is a single crossover between D and A. So a single crossover between D and A. Now, these last couple classes of progeny are interesting in that they're the least frequent class. And so when we consider how they're generated, we'll start with uppercase B.

Let's see if I can get rid of this. So uppercase B, lowercase d, lowercase a. And so what this last class is, is actually a double crossover. So this is a double crossover. And it's least frequent because there's a lower probability of getting two crossovers in this region. But now you see that even though it doesn't look like there was recombination between A and D, in fact, there was. There were two crossovers, and it just looks like there was no recombination, if you didn't see the behavior of gene D.

So if we take into account that there are actually double crossovers between B and A, then if we add that into our calculation here, where you add in 3 plus 5 multiplied by 2 because these are both double crossover events, then you get the 19.6 centimorgans that you would expect by adding up the other recombinations, OK?

How is that? Is that clear to everybody? You're going to have to make maps like this on the problem set and possibly the test. So make sure you can given-- yeah, Ory?

AUDIENCE: I realized that you immediately [INAUDIBLE] overestimated the difference between A and B and not overestimated B to D or D to A?

ADAM MARTIN: It's because when you have two genes that are very far apart, you can have multiple crossovers. And when you have sort of crossovers that are in pairs of two, then it's going to go from one strand back to the other, and so you're not going to see a recombination between the two alleles. So it's an underestimate, because if you have multiples of two in terms of crossovers, you're going to miss the recombination events. You see what I mean? You understand that you can miss the double crossover events?

ADAM MARTIN: Yeah, right? So then you're going to underestimate the number of crossovers that actually happened in that genetic region. All right, now I want to end with an experiment that, again, makes the point that genes are these entities that are on chromosomes. So just like you can have linkage between two genes on the same chromosome, you can also have linkage between genes and physical structures on chromosomes, like the centromere.

So you could have genes like A and B here that are present on the chromosomes and present very near the centromere of those chromosomes, OK? So they could be right on top of the centromere, OK? And to show you how this manifests itself, I have to tell you about another organism, which is a unicellular organism called yeast.

And yeast is special and that it can exist in both a haploid and a diploid form. So it has a lifecycle that involves it going both as a haploid and as a diploid. And so you can take yeast-- and we'll take two haploid yeast cells. And much like gametes, these can fuse to form a zygote. So, in this case, I'm taking-- again, we'll consider two generic genes, A and B. And we'll make a diploid yeast cell that is heterozygous, or hybrid, for A and B.

And what's great and special about yeast and why I'm telling you about this is because as opposed to flies and us and other organisms, the product of a single meiosis is packaged in this single package, if you will. So the yeast can undergo meiosis, and the product of a single meiosis is present in this case, where each of these would represent a haploid cell that then can divide and make many cells.

But this is the product of a single meiosis in a package, OK? So you can actually see the direct result of a meiotic division, a single meiotic division. So this is the product of a single meiotic division. And that's special because when we make gametes, we have individual cells. All the products of meiosis are split up, and then just one randomly finds an egg and fertilizes it. So you don't know which of the gametes are from the product of a single meiosis.

And so being able to see the product of a single meiosis allows us to see things like genes being linked to physical structures on the chromosome like the centromere. So if we consider this case, these two genes are both linked to the centromere. And during metaphor phase of meiosis I, they could align like this, in which case you would get spores that are parental for both dominant alleles or parental for both recessive alleles.

So each of these cells is known as a spore. So I'll label spore numbers here. So this is spore number. And in this case, you get two spores that are dominant for both alleles and two spores that are recessive for both alleles. Because there are two types, it's known as a ditype. And this is a parental ditype, because you have two types of spores . And they are both parental

An alternative scenario is that these chromosomes would align differently, right? So you get parental spores there. However, alternatively, you could have this configuration, where this is now flipped. And during meiosis I here, these chromosomes move together. And they, again, produce two types of spores, so it's a ditype. But in this case, all the spores are non-parental. So another scenario is you get this.

And because there are two types and they're non-parental, this is known as non-parental ditype. That's a non-parental ditype. And if these genes are linked to the centromere completely, then you can only get these two classes of packages, OK? So if these genes are unlinked-- so the two genes are unlinked, but both linked to the centromere, then you get parental ditype-- 50% parental ditype, type 50% non-parental ditype. So I'm abbreviating parental ditype PD and non-parental ditype NPD.

So what has to happen to get another type of spore? And another type of spore would be-- you could have spores that are all different genotypes from each other and that you have A cap dominant A/B dominant A, recessive b recessive a, dominant B and lowercase a and b. And this is known as tetratype, because there are four types. So how do you get this tetratype? Anyone have an idea? Yeah, Jeremy?

AUDIENCE: You're crossing over. So one of A and B would switch in one of the [INAUDIBLE].

ADAM MARTIN: Where would the crossing over happen?

AUDIENCE: Between the two [INAUDIBLE] one [INAUDIBLE].

ADAM MARTIN: Between the allele and what?

ADAM MARTIN: The crossing over would occur between the gene--

AUDIENCE: Oh, and the centromere.

ADAM MARTIN: And the centromere, exactly. Jeremy is exactly right. So Jeremy said that in order to get a tetratype, you have to have a recombination event, but this time, not between two genes, but between a gene and the centromere. So at least one of the genes has to be unlinked to the centromere. And in that case, now you get a meiotic event that gives rise to four spores.

And there are four different ways to get this. So if you have two genes unlinked and at least one is unlinked to the centromere, then you get a pattern where you have a 1 to 1 to 4 ratio between all these different events. So you have a 1 to 1 to 4 ratio between parental ditype, non-parental ditype, and tetratype.

And we can see this in yeast. If you have two genes that are linked to the centromere, you only get parental ditypes and non-parental ditypes, where virtually everything else gives rise to tetratypes, except if they're linked. What happens if the two genes are linked to each other, irregardless of the centromere? If you have two genes that are linked, what's going to be-- what are your progeny going to look like?

ADAM MARTIN: You're only going to get the parentals, or you're going to get a lot of parentals. Javier is exactly right, right? If the two genes are linked, the parental ditypes are going to be much greater than any of the other classes. Now, this might seem esoteric, but I like the idea that you can have linkage between a gene and something that's just the place on the chromosome that's getting physically pulled. It all makes it much more physical, which I think is nice to think about.

All right, we're almost done. I just have-- yes, Natalie?

AUDIENCE: Can you go over what the PD [INAUDIBLE]?

ADAM MARTIN: Yes. So PD is parental ditype. So this is the parental ditype. It's a class of product here where you get four spores that are each of these genotypes, OK? So each of these 1, 2, 3, and 4 would represent one of these cells from a single meiotic event. Does that make sense, Natalie? Does everyone see what I did there? So these 1 2 and 3 are the spores of the meiotic event right here.

Linkage of Genetics: Features, Examples, Types and Significance

When two or more characters of parents are transmitted to the offsprings of few generations such as F1, F2, F3 etc. without any recombination, they are called as the linked characters and the phenomenon is called as linkage.

This is a deviation from the Mendelian principle of independent assortment.

Mendel’s law of independent assortment is applicable to the genes that are situated in separate chromosomes. When genes for different characters are located in the same chromosome, they are tied to one another and are said to be linked.

They are inherited together by the offspring and will not be assorted independently. Thus, the tendency of two or more genes of the same chromosome to remain together in the process of inheritance is called linkage. Bateson and Punnet (1906), while working with sweet pea (Lathyrus odoratus) observed that flower colour and pollen shape tend to remain together and do not assort independently as per Mendel’s law of independent assortment.

When two different varieties of sweet pea—one having red flowers and round pollen grain and other having blue flower and long pollen grain were crossed, the F1 plants were blue flowered with long pollen (blue long characters were respectively dominant over red and round characters). When these blue long (heterozygous) hybrids were crossed with double recessive red and round (homozygous) individuals (test cross), they failed to produce expected 1:1:1:1 ratio in F2 generation. These actually produced following four combinations in the ratio of 7 : 1 : 1 : 7 (7 blue long : 1 blue round : 1 red long : 7 red round) (Fig. 5.6).

The above result of the test cross clearly indicates that the parental combinations (blue, long and red, round) are seven times more numerous than the non-parental combinations. Bateson and Punnet suggested that the genes (such as B and L) coming from the same parent (BBLL × bbll) tend to enter the same gamete and to be inherited together (coupling). Similarly, the genes (B and 1) coming from two different parents (such as BBLL x bbll), tend to enter different gametes and to be inherited separately and independently (repulsion).

Morgan’s View of Linkage:

Morgan (1910), while working on Drosophila stated that coupling and repulsion are two aspects of linkage. He defined linkage as the tendency of genes, present in the same chromosome, to remain in their original combination and to enter together in the same gamete.’

The genes located on the same chromosome and are being inherited together are known as linked genes, and the characters controlled by these are known as linked characters. Their recombination frequency is always less than 50%. All those genes which are located in the single chromosome form one linkage group. The total number of linkage group in an organism corresponds to the number of chromosome pairs. For example, there are 23 linkage groups in man, 7 in sweet pea and 4 in Drosophila melanogaster.

Features of Theory of Linkage:

Morgan and Castle formulated ‘The Chromosome Theory of Linkage’.

It has the following salient features:

1. Genes that show linkage are situated in the same chromosome.

2. Genes are arranged in a linear fashion in the chromosome i.e., linkage of genes is linear.

3. The distance between the linked genes is inversely proportional to the strength of linkage. The genes which are closely located show strong linkage, whereas those, which are widely separated, have more chance to get separated by crossing over (weak linkage).

4. Linked genes remain in their original combination during course of inheritance.

5. The linked genes show two types of arrangement on the chromosome. If the dominant alleles of two or more pairs of linked genes are present on one chromosome and their recessive alleles of all of them on the other homologue (AB/ab), this arrangement is known as cis-arrangement. However, if the dominant allele of one pair and recessive allele of second pair are present on one chromosome and recessive and dominant alleles on the other chromosome of a homologous pair (Ab/aB), this arrangement is called trans arrangement (Fig. 5.7).

Examples of Linkage:

Maize provides a good example of linkage. Hutchinson crossed a variety of maize having coloured and full seed (CCSS) with a variety having colourless and shrunken seeds (ccss). The gene C for colour is dominant over its colourless allele c and the gene S for full seed is dominant over its shrunken allele s. All the F1 plants produced coloured and full seed. But in a test cross, when such F1 females (heterozygous) are cross pollinated with the pollen from a plant having colourless and shrunken seeds (double recessive), four types of seeds are produced (Fig. 5.8).

From the above stated result it is clear that the parental combinations are more numerous (96.4%) than the new combination (3.6%). This clearly indicates that the parental characters are linked together. Their genes are located in the same chromosome and only in 3.6% individuals these genes are separated by crossing over. This is an example of incomplete linkage.

Morgan (1911) crossed an ordinary wild type Drosophila with grey body and long wings (BB VV) with another Drosophila (mutant type) with black body and vestigial wings (bbvv). All the hybrids in F1 generation are with grey bodies and long wings (BbVv) i.e., phenotypically like the wild type of parents. If now a male of F, generation (Bb Vv) is back crossed with a double recessive female (test cross) having black body and vestigial wings (bbvv) only parental combinations are formed in F2 generation without the appearance of any new combinations. The results indicate that grey body character is inherited together with long wings.

It implies that these genes are linked together. Similarly, black body character is associated with vestigial wing. Since only parental combinations of character appear in the offspring of F2 generation and no new or non-parental combinations appear, this shows complete linkage. Complete linkage is seen in Drosophila males.

Types of Linkage:

Depending upon the presence or absence of new combinations or non-parental combinations, linkage can be of two types:

If two or more characters are inherited together and consistently appear in two or more generations in their original or parental combinations, it is called complete linkage. These genes do not produce non-parental combinations.

Genes showing complete linkage are closely located in the same chromosome. Genes for grey body and long wings in male Drosophila show complete linkage.

(ii) Incomplete Linkage:

Incomplete linkage is exhibited by those genes which produce some percentage of non-parental combinations. Such genes are located distantly on the chromosome. It is due to accidental or occasional breakage of chromosomal segments during crossing over.

Significance of Linkage:

(i) Linkage plays an important role in determining the nature of scope of hybridization and selection programmes.

(ii) Linkage reduces the chance of recombination of genes and thus helps to hold parental characteristics together. It thus helps organism to maintain its parental, racial and other characters. For this reason plant and animal breeders find it difficult to combine various characters.

You'll note here that all offspring are not pink. Your genotypic ratio is 25% (RR), 50% (Rr), and 25% (rr). The phenotypic ratio is also the same in this case, with 25% red (RR), 50% pink (Rr), and 25% white (rr).

You know that the possession of claws (WW or Ww) is dominant to lack of claws (ww). You also know that the presence of smelly feet (FF or Ff) is dominant to non-smelly feet (ff). You cross a male who is clawed and has smelly feet with a female who is clawed and has non-smelly feet. All 18 offspring produced have smelly feet, and 14 have claws and 4 are un-clawed. What are the genotypes of the parents?

Answer: Start with what you know early in the story: Dad is clawed, so he has at least one big W. You don't know whether his second allele is big W or little w at this point. He also has smelly feet, so again you know he has one big F but you cannot decipher the second allele at this time. Mom is clawed so she has at least one big W, but the other allele remains unknown. She has non-smelly feet, so she has the recessive characters and can only be "ff." So, based on the above, we know this much: Dad is (W ? F ?) and Mom is (W ? f f). OK, lets look at the offspring. All children had smelly feet. If Dad had a hidden little f, then it would match up with Mom's little f's and about about one-half of the children would have ended up with non-smelly feet (ff). That didn't happen, so Dad must be FF (homozygous dominant). Now then, look at any recessive individuals that may be un-clawed. There are four, and all must be ww. Each child got a little w from Dad and the other little w from Mom. So, both parents must be heterozygous (Ww). Note that just like the monohybrid crosses, how important the recessive offspring are in these types of problems. You automatically know that each parent had that hidden recessive allele based solely on the phenotype of the offspring. So, you figured out the problem without any Punnett squares and the parents are as follows: Dad is "WwFF" and Mom is "Wwff"

You have an individual who is totally heterozygous for 2 genes that are not linked (i.e., not on the same chromosome). One gene is for ear size (AA or Aa being big ears whereas aa is for small ears) and the other gene is for buggy eyes (BB and Bb for buggy eyes whereas bb represents normal eyes). If you testcross this individual, what are the resulting genotypes and phenotypes?

Answer: Remember that a testcross represents a cross with a totally recessive individual. These types of crosses are useful in weeding out hidden recessive alleles from your unknown. Remember the information on recessives if you don't remember anything else. By knowing the recessive, you automatically know both the phenotype and genotype. In the monohybrid cross, a testcross of a heterozygous individual resulted in a 1:1 ratio. With the dihybrid cross, you should expect a 1:1:1:1 ratio!

Homework - Genetics linkage problem - Biology

In the end of nineteenth century as a result of technological progress was be significantly increased the optical characteristics of microscopes, and was also significantly improved cytological research methods. This allowed scientists to make a series of important discoveries. Has been proven a leading role of cell's nucleus in the transmit of hereditary traits. They drew attention to the striking similarity between the behavior of chromosomes during the formation of gametes and fertilization, and scheme of inheritance of genetic factors, that Mendel described. On the basis of these data has been formulated the chromosome theory of inheritance. According to this theory a pair of factors localized in a pair of homologous chromosomes, and each of these chromosomes is carrier of one factor. Later, the term factor, which mean the basic unit of heredity, has been replaced by the term - gene. Thus we can say that genes, that located in the chromosomes, is the physical unit, through which the hereditary traits transmitted from parents to offspring. Each gene is represented in homologous chromosomes as a pair of alleles, which located in one locus, which means in the same place in these chromosomes. Now it was possible to explain the basic laws of inheritance in terms of the chromosome theory, as the characteristics of chromosomes motions during meiosis. Segregation of homologous chromosomes that occurs during anaphase 1 of meiosis and random distribution of alleles between the gametes is the basis for explanation of first law - Law of Segregation. And the independence of the segregation nonhomologous chromosomes during anaphase 1 of meiosis is the basis of the second law - the Law of Independent Assortment.

However, it is absolutly clear that every organism has a large number of traits and this quantity can be considerably greater than the number of chromosomes in haploid set. This is especially noticeable for species with a small number of chromosomes. For example number of chromosomes in haploid set in pea equal to 7, in rye is also equal to 7, fruit fly 4, and in roundworm 1. Then it is obvious, that in each chromosome must be located genes that determine the development at least a few different traits. Such genes are called linked and the number of linkage groups is equal to the number of chromosomes in the haploid set. Accordingly, these genes don't have to obey to the principle of an independent assortment - they must be inherited together as one unit.

Genetic Linkage Calculator

In the genetic calculator, for designation of genetic linkage in the parental genotypes notation, the linked genes must be concluded in brackets. For dihybrid there are two possible localization of dominant and recessive alleles in the chromosomes. In the first case, the dominant alleles are localized in one of the pair of homologous chromosomes and recessive in the other - (AB)(ab). This variant of alleles localization is called cis-position. Genotype (AB) (ab) can be obtained from crosses of parents with genotypes (AB)(AB) - phenotype AB and (ab)(ab) - phenotype ab. In the second case the dominant and recessive alleles of a gene localized in different homologous chromosomes - (Ab)(aB). This variant of localization is called trans-position. Genotype (Ab) (aB), respectively, can be obtained from crosses of parents with genotypes (Ab)(Ab) - phenotype Ab and (aB)(aB) - phenotype aB. The difference in the ratio of phenotypes in the offspring for Mendelian inheritance and genetic linkage can demonstrated in the test crossing. In this crossing the number of types of gametes is equal to the number of phenotypic classes in the progeny. In the case of independent inheritance the genotype AaBb will give the four types of gametes AB, Ab, aB and ab with the ratio 1:1:1:1. In the case of a genetic linkage genotype (AB)(ab) can give only two types of gametes (AB) and (ab). Accordingly, by crossing the individuals with the genotype (AB)(ab) and (ab)(ab), we obtain two classes of phenotypes AB and ab with the ratio 1:1. Genotype (Ab)(aB) will also give two types of gametes (Ab) and (aB). And by crossing parents with genotypes (Ab) (aB) and (ab) (ab), we also obtain two classes of phenotypes Ab and aB with ratio 1:1. As you can see in both cases of genetic linkage. the phenotypic classes of offspring depends on the location of dominant and recessive alleles in the chromosomes and identical with the phenotypes of parents, by crossing which was obtained the original dihybrid.

But such results can be obtained only in the case of complete linkage. Typically, complete linkage is quite rare. The fact is that during meiosis, homologous chromosomes can exchange of regions with each other. This process is called crossing-over or genetic recombination. In the process of genetic recombination, the alleles, which located in linkage group in the parents, can segregate and give the new combinations in the gametes. Phenotypes, which are obtained from these gametes are called recombinants or crossovers. Thus, the progeny will be not two but four phenotype, as in the independent inheritance. But for linked inheritance the ratio will be different. Classes with the parental phenotypes will be form the bigger part of offspring, and the recombinant classes - smaller part. For example, for the genotype (AB)(ab) will be more offspring with phenotypes AB and ab and less with phenotypes Ab and aB, and for genotype (Ab) (aB) vice versa. The exact phenotype ratio will depend on the distance between genes. The farther away from each other located linked genes, then greater the probability, that crossing-over occurs between them. Thus, the frequency of crossing-over or recombination can be a measure for determination of the distance between genes. If the single crossing-over occur between genes and we known the amount of crossovers, then the distance between the genes can be calculated by the formula: RF = (c/t) * 100%, where the RF - the recombination frequency (the distance between genes), c - the amount of crossovers, t - the total amount of offspring, obtained from test cross. If the value of crossing-over exceeds 50%, then we can talk about an independent assortment of genes, that is without linkage inheritance. In the examples that you see below, we will use the Crossing Over Map Calculator to calculate the distance between genes and the Genetic Calculator for modeling genetic crosses with genetic linkage. It is important to note that Crossing Over Map Calculator can give correct results only for test crosses. First, let's look at some of the features of this calculator.

Crossing Over Map Calculator

Crossing Over Map Calculator have a pretty simple and clear interface. It can be divided into two parts - in the part "Genes and phenotypes" you can enter the required data, and in the right - "Results" is shown the results of the calculations. You can find the radio buttons "Three Genes / Two Genes" in the left side. Since we are going to consider examples for two linked genes, then you should switch it to "Two Genes". And the specifications of genetic problems solution for the three linked genes will be considered later. Algorithm for data entry will look like this:
1) Choose the number of genes (in this case - switch the Radio Button in "Two Genes").
2) Write the dominant and recessive alleles of genes in the first table.
3) Press the "Calculate Phenotypes" button. The program automatically fills the first column in the second table by all possible combinations of phenotypes.
4) In the second column you will need write the number of individuals of each phenotype.
5) Press the "Calculate Results" button.

For examples with the two linked genes you can get the following results on the right side:
1) In the first table you can see the recombination frequency between genes. 1% recombination corresponds to obtaining one recombinant on 100 individuals. (1% recombination frequency = 1 m.u. (genetic map unit) unit of the genetic map, or a centimorgan (cM)).
2) Map Distance ( m.u.) - the actual distance between genes. This distance takes into account the effect of interference and possible double crossovers. The recombination frequency and Map distance are linked with each other by this formula ( Haldane’s Mapping Function ) : RF = (1 - e ^ ( -2 * map distance )) / 2 or map distance = - ( ln(1 - 2RF) / 2 ), where RF - the recombination frequency.
3) Parents genotypes. Based on the parental genotypes you can judge about genes localization - is they in cis- or trans- position.

Genetic Linkage Examples

Two linked genes in dihybrid test cross of tomato

Now, let's move to the concrete examples. In tomato genes that determine the height of the plants - T (tall) and t (dwarf) and the shape of the fruit - S (round) and s (pear-shaped ), located in one chromosome, ie they are linked. If we cross the homozygous plants with genotypes TTSS and ttss, it is also, as in the case of independent assorment, all the offspring will have the same phenotype. In this case, all the plants were high, with rounded fruits and have genotype TtSs. As a result of test cross, when these plants are crossed with homozygous recessive plants (ttss), was be obtained in the offspring 40 tall plants with round fruits, 40 dwarf plants with pear-shaped fruits, 10 tall plants with pear-shaped fruits, and 10 dwarf plants with round fruits. If the genes linkage was be complete, then in the offspring would be only tall plants with round fruits and dwarf plants with pear-shaped fruits in equal proportions, and if the genes were not linked, then the ratio of phenotypes would be 1:1:1:1. Thus we can say that in this case between the linked genes occurs crossing over, which gives a new recombinant phenotypes. Now let's analyze the obtained data.

  • Toggle Radio Button in "Two Genes".
  • Write the dominant and recessive alleles of genes in the first table:
  • Get the combination of phenotypes in the first column of the second table and write the amount of plants for each phenotype:
  • Push the button "Calculate Results" and get results.
  • In the first place in the "Parentes genotypes" we see parental genotypes (non-crossover genotypes)
  • Accordingly, our crossover genotypes is:
  • Second in the first table of the results we see recombination frequency = 20%. This is the distance between genes. The reliability of results you can check by the formula, that described above: RF = (c/t) * 100%. RF = ((10 + 10)/(40 + 40 + 10 + 10)) * 100% = (20/100) * 100% = 0.2 * 100% = 20%. Thus we can say that genes are localized in the trans-position at a distance of 20 santimorgan (centimorgan) (cM). Using the recombination frequency the program calculates the actual distance between genes = 25.541281 m.u. (In the field "Map distance")

Now let's simulate this cross in genetic calculator. (TS)(ts) and (ts)(ts) - it's our parents genotypes. In these parents genotypes notations we must include the recombination frequency between genes - this value should be concluded in percent characters. There are three options for this entry:
1) (T%20%S)(ts)
2) (T%20%S)(t%20%s)
3) (TS)(t%20%s)

All three options are correct and you can use any of them. We will choose the first option and write the parents genotypes like this:
(T%20%S)(ts) x (t%20%s)(ts)

As a result of this crossing, we get phenotypes ratio 4 TS : 1 Ts : 1 tS : 4 ts or with probability 40% TS : 10% Ts : 10% tS : 40% ts. Since it is test cross, the ratio of gametes will be identical to the ratio of phenotypes.

Two linked genes in dihybrid test cross of maize

Consider another example of test cross. In maize genes that determine the color of seedlings - Green (dominant) and yellow (recessive) and brightness of the color of leaves - Opaque (dominant) and brigh(recessive), located in one chromosome. All plants from crosses of pure lines of maize have a Green seedlings and Opaque leaves. As a result of test cross, when these plants are crossed with homozygous recessive plants with yellow seedlings and brigh leaves, obtained in the progeny 240 plants with Green seedlings and Opaque leaves, 220 plants with yellow seedlings and brigh leaves, 36 plants with Green seedlings and brigh leaves, and 24 plants with yellow seedlings and Opaque leaves. Let's analyze the obtained data.

  • In the previous example, we marked alleles with a single letter. You probably noticed that it's not very comfortable, especially when you write the amount of individuals for each phenotype. But there is a better way, you can use the natural notations of traits for marking alleles. To do this, you need to conclude allele in these characters <Allell name>. So, we write the dominant and recessive alleles of genes in the first table in this way (Do not forget to switch the Radio Button in "Two Genes"):
  • Get the combination of phenotypes in the first column of the second table and write the amount of plants for each phenotype:
  • Push the button "Calculate Results" and get results.
  • In the field "Parentes genotypes" we see parental genotypes (non-crossover genotypes)
  • Accordingly, our crossover genotypes are:
  • In the first table of results we can see the recombination frequency = 11.5385% ( RF = (c/t) * 100%. RF = ((36 + 24)/(240 + 220 + 36 + 24)) * 100% = (60/520) * 100% = 0.115358 * 100% = 11.5385% ). Thus we can say that genes are localized in the trans-position at a distance of 11.5385 centimorgan (cM). Using recombination frequency the program calculates the actual distance between the genes = 13,118213 m.u. ( in the field "Map distance" )

Now let's simulate the cross in genetic calculator. Write the parents genotypes:
(<Green>%11.5385%<Opaque>)(<yellow><bright>) and (<yellow>%11.5385%<bright>)(<yellow><bright>)

As a result of the cross we obtain such probabilities of phenotypes: 44.2308% <Green><Opaque> : 5.76925% <Green><bright> : 5.76925% <yellow><Opaque> : 44.2308% <yellow><bright>. Frequency of crossovers is approximately equal to 5.76925 + 5.76925 = 11.5385, and frequency of non-crossovers is approximately equal to 44.2308 + 44.2308 = 88.4616% ( or 100% - 11.5385% = 88.4615% ). If the total amount of individuals in the offspring is equal to 520, then amount of non-crossovers will be equal to 520 * 88.4616 / 100 = 460, and amount of crossovers 520 * 11.5385 / 100 = 60. In the experiment the amount of non-crossovers equal to 240 + 220 = 460, and amount of crossovers 36 + 24 = 60. Thus, the results is absolutely correct.

You can experiment with the possible variation in practical results of these crosses. "Random statistic" module of genetic calculator give you the opportunity to simulate the random distribution of progeny according to probabilities of phenotypes. Check the box "Generate statistic" and write the amount of the progeny (for the first cross- 100, for second - 520) in the field "Sample size". Each time you press the button "Calculate results" you'll get a new variant of assortment.

Two linked genes in dihybrid cross of Garden pea

To calculate the distance between the genes we must use only the results of test the crosses. But with Genetic calculator, we can simulate genetic linkage also for another crosses. Let's see this on example of a garden pea. Garden pea, rightly, we can call - the first object of genetic studies, as this plant the Gregor Mendel used in his experiments. Exactly on the basis of these experiments, he formulated the basic laws of genetics. As you know, in the pea the number of chromosomes in haploid set equal to 7, and of course, we can say that Gregor Mendel was lucky to choose to study the inheritance such pairs of traits, the genes of which was not linked, that's mean it was located in different chromosomes. In pea genes that determine the color of the flower - Purple (dominant) and red (recessive) and the shape of pollen grains - Long (dominant) and round (recessive), located in one chromosome at a distance of 12 centimorgan. All the offspring from crosses between plants with purple flowers and long pollen and plants with red flowers and round pollen had purple flowers and long pollen. As a result of self-pollination of these hybrids in the offspring was be obtained 69.5% plants with purple flowers and long pollen, 19.3% with red flowers and round pollen, and 5.6% plants with purple flowers and round pollen and 5.6%, with red flowers and long pollen. If the linkage was be complete, then the ratio of the progeny was be approximately equal to 3 : 1, like for monohybrid crosses, and if the genes was not be linked, then the phenotypes ratio in the case of independent inheritance of traits for dihybrid crossing was be 9 : 3 : 3 : 1. In this case, the ratio of non-recombinant phenotypes really approximately equal to 3 : 1, and we have a small amount of recombinant phenotypes.

Now let's simulate the cross in genetic calculator. Write the parents genotypes:
(<Purple>%12%<Long>)(<red><round>) and (<Purple>%12%<Long>)(<red><round>).

As a result, we obtain our ratio of phenotypes: 69.5% <Purple><Long> : : 5.6% <Purple><round> : 5.6% <red><Long> : 19.3% <red><round>.

Several Linkage Groups of Genes

Using Genetic calculator we can solve the problem also with several linkage groups. A good demonstration of this possibility will be solving the following example: In maize recessive genes, which determine to the curly leaves (cr) and dwarfism (d), are localized in the third chromosome at a distance of 18 map units (18%), and the dominant genes of rust resistance (Rp) (rust) and narrow leaves (Nl) - in the tenth chromosome at a distance of 24 map units (24%). We have a plant heterozygous by all genes, that are in the cis-position. Define: 1) What types of gametes, and with what probability can form this plant? 2) With what probability in the offspring we can expect the formation of homozygous dwarf plants with resistance to rust and with normal leaves?

We can write the parents genotypes of our plants as follows: (<Cr>%18%<D>)(<cr><d>)(<Rp>%24%<Nl>)(<rp><nl>). To answer on first question we need to choose the - - "Gametes genotype 1" for type results and calculate it. In the second question we need to find the probability of the genotype - <Cr><Cr><d><d><Rp><Rp><nl><nl>. Change the type of results to "Genotypes" and go to the tab "Find". Click on each cell in the column "Combinations" and from the drop down list select a value that you need. For the first cell is <Cr><Cr><d><d>, and for the second - <Rp><Rp><nl><nl>. Push the "Find" button and you'll get an answer to the second question (0.011664).

Crossing Over Mapping

Chromosomes - have a linear structure, and respectively the genes in the chromosomes are also located in the linear order. Thus, the frequency of crossing-over or genetic recombination between the genes can be used not only to determine the distance, but also to determine the relative position of these genes in the chromosome. However, you must know, that recombination have random character and between genes can also occur the double crossing-over. And often it's difficult to estimate the true distance between the genes, that located far enough from each other, because he is not always detected. As a result, the crossing-over frequency between the outermost genes is less than expected and not equal to the sum of the frequencies of single crossing-overs. Only the presence of the third gene between studied genes (called marker gene) allows you to accurately find the distance and positions of genes. As evidence we can examine the example of trihybrid crossing.

Three linked genes in trihybrid test cross of maize

In maize the genes, that determine the color of seedlings - Green (dominant) and yellow (recessive), the brightness of the leaves color - Opaque (dominant) and bright (recessive) and the shape of the leaves - cutting (recessive) and normal (dominant), are located in one chromosome. All plants from crosses of pure lines of maize have a green seedlings and opaque leaves with normal shape. As a result of test cross, when these plants are crossed with homozygous recessive plants with yellow seedlings and bright cutting leaves, was obtained the offspring:
270 plants with green seedlings and opaque leaves with the normal form
235 plants with yellow seedlings and bright cutting leaves
62 plants with green seedlings and bright cutting leaves
60 plants with yellow seedlings and opaque leaves with the normal form
48 plants with green seedlings and opaque cutting leaves
40 plants with yellow seedlings and bright leaves with the normal form
4 plants with green seedlings and bright leaves with the normal form
7 plants with yellow seedlings and opaque cutting leaves
Let's analyze the obtained data.

  • In Crossing Over Map Calculator switch the Radio Button to the "Three genes" and write the dominant and recessive alleles of genes in the first table in this way:
  • We get the combination of phenotypes in the first column of the second table and write the amount of plants for each phenotype:
  • Push the button "Calculate Results" and get results.
  • First of all this is Genes localization map (the order of genes in chromosome):
    <yellow>-<bright>-<cutting> - in that order, these genes are located in the chromosome. Recombination frequency between <yellow>-<bright> equal to 18.3196%(Crossing over frequency), between <bright>-<cutting> is 13.6364%(Crossing over frequency) and between <yellow>-<cutting> it's equal to 31.9559%(Crossing over frequency).
  • From the obtained eight classes of phenotypes - two is non-crossovers and they phenotypically identical to the parents phenotypes, and six is crossovers, two of which is double crossovers.
    1. In the field "Parental genotypes" we see parents genotypes (non-crossovers genotypes)
    2. Becouse in the localization map the gen of leaves coloration (<bright>/<Opaque>) located in the middle, then the Double Crossovers is ( the lowest phenotypes ) :
      7 <yellow><Opaque><cutting>
      4 <Green><bright><Normal>
    3. First pair of single crossovers is:
      48 <Green><Opaque><cutting>
      40 <yellow><bright><Normal>
    4. Second pair of single crossovers is:
      62 <Green><bright><cutting>
      60 <yellow><Opaque><Normal>

Let's check this results by the our formula: RF = (c/t) * 100%. For first pair of single crossovers( Recombination frequency between gens - <bright>-<cutting> ): RF = ((48 + 40)/(235 + 270 + 7 + 4 + 48 + 40 + 62 + 60)) * 100% = (88/726) * 100% = 0.121212 * 100% = 12.1% For second pair of single crossovers( Recombination frequency between gens - <yellow>-<bright>): RF = ((62 + 60)/726) * 100% = 0.168044 * 100% = 16.8%. So we can expect the recombination frequency between gens <yellow>-<cutting> equal to 16.8% + 12.1% = 28.9%. However, from cross is obtained a different results. The frequency of single crossing-over between gens <yellow>-<cutting> = 28,9%, which is on 3% less than expected. The contradiction between the theoretically expected and practically obtained results is eliminated if we take into account the double crossing-over between genes <yellow>-<cutting>. Recombination frequency - RF = ((7 + 4)/726) * 100% = 0.015151 * 100% = 1.5%. So recombination frequency between gens - <bright>-<cutting> must be 12.1% + 1.5% = 13.6% and recombination frequency between gens - <yellow>-<bright> must be 16.8% + 1.5% = 18.3%. Therefore recombination frequency between gens <yellow>-<cutting> will be 13.6% + 18.3% = 31.9%. The distance between genes with the presence of double crossing-over is equal to the sum of percentage of the single crossing-overs and a doubled of percentage of double crossing-overs. In our example the distance between genes <yellow>-<cutting> is : 16.8% + 21.1% + 1.5% * 2 = 31.9%.

Now let's simulate the cross in genetic calculator. Write the parents genotypes:
(<Green>%18.3196%<Opaque>%13.6364%<Normal>)(<yellow><bright><cutting>) and (<yellow><bright><cutting>)(<yellow><bright><cutting>)

As a result of this crossing, we obtain such phenotypes probabilitys:

<yellow><bright><cutting> 35.2711%
<yellow><bright><Normal> 5.56913%
<yellow><Opaque><cutting> 1.24907%
<yellow><Opaque><Normal> 7.91073%
<Green><bright><cutting> 7.91073%
<Green><bright><Normal> 1.24907%
<Green><Opaque><cutting> 5.56913%
<Green><Opaque><Normal> 35.2711%

The frequency of double crossing-over is approximately equal to 1.24907 + 1.24907 = 2.49814%. The frequency of single crossovers is approximately equal to 5.76925 + 5.76925 = 11.5385 + 2.49814 = 13.6364% and 7.91073 + 7.91073 = 15.82146 + 2.49814 = 18.3196%. Accordingly, the total frequency of crossovers is 13.6364 + 18.3196 = 31.956%. Thus, the results correspond to the experimental and absolutely correct.


Active Learning

In Class: In the classroom, the entire module engages students in active learning. From start to finish, students are active participants in authentic science as they collaborate on hands-on lab work and problem-solve in small groups to conduct DNA analysis. Students also engage with primary literature and, like scientists, actively maintain an accurate, up-to-date lab notebook for the duration of the module. Students also create a research poster of their results (including images of gels, chromatograms, etc.). Furthermore, students communicate like scientists as they present for one hour in a format mimicking an interactive, scientific research poster session (see Supporting File S5. Dog Genetics – Lesson 4. Final Genotype: Collaborative Poster).

Outside of Class: From the beginning of this module, students are asked to be active participants in their learning. Specifically, students are given pre-lesson reading and questions to gauge prior knowledge and build a conceptual framework for future learning. Also, students are given the materials and expertise to collect data from dogs at their home or in their community. For many, this raises engagement as they become interested in knowing specific genetic characteristics of animals that are close to them and their families. As students work on pre-lesson assignments, they model good scientific procedure by contributing to their ongoing lab notebook. Finally, a good portion of the creation of the final research poster is completed out-of-class. This is an important part of the active learning process, as students must combine content knowledge with 21 st century skills to set up meetings, communicate, compromise, and ultimately develop a final product that meets the expectations of all group members.


Assessments are embedded throughout the module and include (a) pre-and post-lesson worksheets (Supporting Files S1. Dog Genetics – Module Introduction, S2. Dog Genetics – Lesson 1. DNA Extraction and PCR Sample Preparation, S3. Dog Genetics – Lesson 2. Gel Electrophoresis, S4. Dog Genetics – Lesson 3. DNA Sequence Analysis), (b) lab notebooks (Supporting File S6. Dog Genetics – Lab Notebook Grading Rubric), and (c) student evaluation of participation of group members (Supporting File S7. Dog Genetics – Group Evaluation). We also include two summative assessments at the end of the module in the form of a quiz (Supporting File S8. Dog Genetics – Example Quiz Questions) and a collaborative poster (Supporting File S9. Dog Genetics – Poster Template). Additionally, students and instructors intermittently discuss the Cadieu et al. (2009) Science paper (1) which was the inspiration for this module during the first three lessons. Discussions occur during "down-time," e.g., when the gels are running during Lesson 2 (Supporting File S3. Dog Genetics – Lesson 2. Gel Electrophoresis). Students engage in discussion of the paper and are guided by worksheets provided by the Instructor (Supporting File S10. Dog Genetics – Cadieu et al. Discussion). For instance, discussion of the paper starts with understanding the primary objective and general methodology of the research presented and then transitions to specific methodology and results for each gene. The discussions are complemented by scaffolded homework (pre- and post-lesson worksheets) that integrates analysis of the scientific paper with understanding methodology and students' experimental results from their analyzed dogs.

We emphasize the value of record keeping in our introductory biology laboratory course, since scientists, health professionals, and teachers will be required to keep records of their experiments, patients, or students. Our students keep records by maintaining lab notebooks. Lab notebooks are required to include the following elements: date, experiment title, group members' names, purpose of experiment, data, sample identification numbers, calculations, figures/tables, results, and conclusions. Students also answer questions that are embedded in some of the lessons in their lab notebooks. Lab notebooks are informally scanned by instructors during the lab times to verify students are recording relevant information, data, and activities. At the end of the module, lab notebooks are formally assessed by grading selected elements from each lesson. Each student lab notebook is evaluated on the same elements that are listed on a grading rubric (Supporting File S6. Dog Genetics – Lab Notebook Grading Rubric), and each element appears at least twice on the grading rubric. This method of evaluating selected criteria is an effective grading strategy because we can give specific, informative feedback without grading the entire lab notebook.

Students are given a summative assessment at the end of the module in the form of a quiz (Supporting File S8. Dog Genetics – Example Quiz Questions). The quiz evaluates students' understanding of the first three lessons and the Cadieu et al. (2009) paper (1). In our course, students are given a quiz at the beginning of Lesson 4 and then work in their groups to create a poster summarizing their research and results from their two dogs. For most of our students, this is the first scientific poster they have made, so we discuss common mistakes and discuss expectations described in the grading rubric (Supporting File S5. Dog Genetics – Lesson 4. Final Genotype: Collaborative Poster). Students also investigate and present an additional, assigned gene that regulates a different aspect of dogs' phenotypes in a small section of their posters. We provide a modifiable PowerPoint template from which students design their posters (Supporting File S9. Dog Genetics – Poster Template).

During the final lesson, students present their results in a poster conference that models a professional conference. Students take turns asking and answering questions about their research during the conference and fill-out a questionnaire from this discussion (Supporting File S5. Dog Genetics – Lesson 4. Final Genotype: Collaborative Poster). Additionally, all participants of the conference evaluate the posters and vote to determine the best poster. The winning poster receives a "best in show" certificate, ribbon, and prize. Lastly, students self-assess their own participation in the group and the participation levels of other group members using a standard form (Supporting File S7. Dog Genetics – Group Evaluation).

Inclusive Teaching

Helping students find identity and purpose in science is an important aspect of inclusion. In this module, students are invited to have a personal investment in the project as they bring samples from their personal pets for analysis. Although this may seem trivial to some, students often respond to the module by referencing how exciting it is to analyze their own pets (see Student Reaction section).

In addition, students work in small, collaborative groups throughout the module and share expertise amongst themselves. In these situations, students of diverse backgrounds often find comfort and develop camaraderie with others as they are able to use their strengths and bolster their weaknesses in a collaborative setting. Finally, the breadth of the activities in this module promote inclusive learning through stimulation of multiple modalities. For example, students isolate and analyze DNA, conduct primary literature research, discuss/debate results, and co-create a poster presentation (which is modeled after a professional scientific poster session).

Genetics Questions

This material may consist of step-by-step explanations on how to solve a problem or examples of proper writing, including the use of citations, references, bibliographies, and formatting. This material is made available for the sole purpose of studying and learning - misuse is strictly forbidden.

Recombination frequency is a measure of genetic linkage. It's the frequency of a single chromosomal crossover (sometimes called crossing over - two parts of genetic material on homologous chromosomes swap places during meiosis – literally „cross over“ to the other homologous chromosome the resulting chromosomes are called recombinant chromosomes). A cM – centiMorgan (in honour of geneticist Thomas Hunt Morgan) – is a unit that describes a recombination frequency of 1%. Its parent unit, the Morgan, is rarely used (1 Morgan = 100 cM). For instance.

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Watch the video: Ο ηλεκτρικός βραστήρας δεν ανάβει καθαρισμός του θερμοστάτη (June 2022).


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