AP Biology Linked Genes and Non-Mendelian Modes of Inheritance

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In the previous lecture, we talked about Mendelian modes of inheritance.0002

And today, we are going to go on to discuss the topics of linked genes and some modes of inheritance that are non-Mendelian.0006

We will start out with the review of one of Mendel's law, the law of independent assortment,0014

so we can go on to see how linked genes do not follow this law.0020

Relating the law of independent assortment to meiosis, recall that the law of independent assortment states0025

that alleles for a particular trait assort independently of the alleles for other traits.0032

For example, let's look at the traits of eye color and height.0040

And here, we have an individual who is undergoing gametogenesis, the formation of gametes either sperm or egg.0048

Let's say that this is a male, so it is sperm formation.0057

This male is diploid, and in any human male, there would be 46 chromosomes, but for simplicity, we are just going to look at two chromosomes.0061

And we are going to call the large chromosome, chromosome 1 and the small chromosome, chromosome 2, and this individual has two of each.0069

This male inherited a chromosome 1 from his mother.0078

And we are going to say that the purple chromosomes were maternally inherited, and the green are of paternal origin.0084

This male inherited a chromosome 1 from mom, a chromosome 2 from mom, a chromosome 1 from dad, a chromosome 2 from dad.0101

If, let's say, eye color is on chromosome 1, just for example, and let's say that height is on chromosome 2 and that big B is dominant,0113

and it is brown eyes for eye color, and little b is recessive; and it is blue, so one allele, brown, the other could be blue.0131

If an individual, recall from Mendelian genetics, had big B-big G, they would have brown eyes, little b-little b, blue eyes.0146

And because brown is dominant, the heterozygote would have brown eyes.0155

Height, let's say, is on chromosome 2, and that we have big T is tall. Little T is short, and tall is dominant, so the heterozygote would be tall.0160

Let's just say that this individual inherited the brown-eyed allele from the mom.0176

And on the chromosome 1 from dad where eye color is, this individual inherited blue-eyed allele.0184

Chromosome 2, say mom was tall, and dad donated the short allele, so this individual is heterozygous for both eye color and height.0190

Now, when his germ line cells form sperm, what is going to happen is segregation of homologous chromosomes.0204

One cell will end up with one chromosome 1 and a chromosome 2, instead of two copies of each.0218

The other cell will end up with a chromosome 1, the large chromosome and chromosome 2.0224

And then, remember, during meiosis 2, the sister chromatids will separate.0230

For right now though, we are just focusing on this first step where segregation occurs.0236

So, here, we are getting one migrating to this pole of the cell. This two is migrating to this pole of the cell.0243

And the paternally derived one is going to this pole of the cell, and the maternally derived two is going to the other pole, so it ended up splitting.0254

These two are together, and these two are together.0267

We got the segregation where only one allele for each trait goes to one gamete.0270

Now, what does independent assortment tells us? It tells us that it does not matter what the blue-eyed alleles are doing.0277

The tall alleles are going to assort independently that this big B from mom is going to on to one chromosome.0285

And maybe dad's chromosome 2 will end up in there. Maybe the maternal chromosome 2 will end up in there.0293

So, we end up with big B, in this case and little t. It could have just as easily been big B ended up with big T.0299

Here, we ended up little b and big T, so the alleles for eye color and height are assorting independently.0311

They are not affecting...these two are not affecting what these two do.0324

This could go this side. This could go this side, and then, these assort on their own.0330

Now, that is simple Mendelian genetics. However, this only holds up if the traits are on separate genes, or they are very far apart on the same gene.0336

If traits are on the same gene, we say that they are linked.0348

What we are going to talk about now is linkage and how that is going to affect the offspring's genotype and phenotype.0357

Linked genes are those that are located on the same chromosome, and they, therefore, do not assort independently from one another.0367

What we just talked about were traits that are located on different chromosomes, like I said, eye color and height.0375

Eye color is on chromosome 1, and height is on chromosome 2.0386

Now, let's say chromosome 1 also contains genes for hair type, and that could be curly dominant; and let's say straight is recessive.0392

If that is the case, then, the genes for eye color and hair type are said to be linked.0405

Let's say the low side for eye color is here and also on the paternally derived chromosome.0413

And then, height is here, actually hair type, and then, height is over here on chromosome 2.0420

Therefore, height is going to assort independently, whereas, hair type and eye color may not.0434

So, if the maternally derived chromosome has the allele for brown, and the hair type curly,0444

and the paternally derived has the allele for blue eye color and straight hair, it is more likely that these two are going to stay together,0452

that the gamete is going to end up with this combination of brown and curly or blue and straight.0465

It is more likely than ending up with curly-blue, brown-straight, so we say that genes that are on the same chromosome are linked.0470

In 1906, Bateson and Punnett noticed these phenomena, and they called it coupling, not linkage, but that certain alleles or traits seemed to be coupled.0481

And they were studying pea plants, and what they did is they did some crosses where they studied flower color.0497

And there are two possibilities that they were studying red versus purple, so big P, purple is dominant, and little p, red, is recessive.0506

They also studied the shape of pollen grains- pollen grain shape.0518

Elongated shape is dominant, so for short, we will just call that long big L, so long or elongated grain shape is dominant.0526

And round pollen grains are recessive, so they went on, and they did their crosses.0535

They started out with the parental generation, and they got true breeding plants that were purple-flowered with elongated pollen grains.0543

And they crossed those with true breeding plants that had red flowers and round pollen grains.0554

They ended up with the F1 generation, dihybrids, that had the big P from one parent, little p from another, big L from one parent, little l from another.0561

These plants would have purple flowers and the long pollen grains.0574

And then, they went ahead and did the dihybrid cross self-pollinating the F1 generation, and they got an F2 generation.0581

And here is where things deviated from what was expected according to Mendelian genetics.0588

What you will expect when you do this F1 cross is...when we do the dihybrid cross, what we expect is 9:3:3:1 phenotypic ratio.0595

And make sure that you review or watch the Mendelian genetics lecture so that you have this down0616

because we are going on to concepts that build on what you learned from Mendelian genetics such as dihybrid crosses.0624

So, this 9:3:3:1 ratio would be dominant-dominant phenotype, so dominant would be purple with long pollen grain.0630

9 of those to every 3 that would have 1 dominant phenotype purple.0641

But, the other recessive, round to 3 that would have the other phenotype being recessive.0646

So, that would be red but then, dominant for that second trait, which would be long, and then, recessive phenotype for both traits would be red round.0654

That is expected in F2. That is not what they saw.0668

What they actually found is that purple and long and red and round were overrepresented.0672

These are what we call the parental phenotypes, and you look at the F2; and you might say "well, the F1 is their parents".0681

That is not what we are actually talking about. We are talking about the parents of the F1 where things originated.0687

And what they found is that purple-round and red-long were less common, so the parental generation, the phenotypes were purple-long, red-round.0695

And then, purple-long and red-round were overrepresented.0711

Purple seemed to stay with long, and red stayed with round.0727

There were some purple-round and red-long recombinant phenotypes, so these two are called the parental phenotypes.0732

Red-long and purple-round are the recombinant phenotypes. They are different than what it started out with.0749

And Bateson and Punnett were not really sure of the mechanism for this, but they did know that somehow, certain traits were coupled.0753

Let's look at gamete formation at this F1 generation and see if we can explain this.0767

Here, I have the F1 generation, and when the gametes form, we could get big P with big L, one gamete.0772

We could get big P with little l. We could get little p-big L, and we could get little p-little l.0784

And what we expect if these are in two different chromosomes, if the traits that we are studying0793

- flower color and pollen grain - were on two different chromosomes, we would expect mixing and matching0799

that the trait on this chromosome could be purple, and then, this one over here is long.0806

And this is red, and this one is round; and these will mix and match.0811

But if they are on the same chromosome, they are not as likely to mix and match, and that is what happened.0815

So, if I happened to have purple and long together inherited from the mother, and then,0820

let's say round and red together inherited from the other parent, then, these may stay together.0831

This was inherited from one parent, purple and long, and red and round were inherited from the other parent; and these tended to stay together.0841

These two tended to stay together, and these two tended to stay together because they were on the same chromosome.0852

If they were not on the same chromosome, they could mix and match much more easily.0862

And in the early 20th century, Thomas Hunt Morgan used a fruit fly to study genetics and to do various crosses.0866

And one of the things he was able to study was this type of linkage, and he tried to explain why some genes do not assort independently.0877

Now, I just mentioned that more often, if two genes are on the same chromosome,0887

those particular alleles will stay together in the gametes and be passed on together.0896

However, that does not always occur. This combination of purple and long can be split up, and you can end up with purple and round together.0901

How would that occur if they are on the same chromosome?0915

Crossing over is the mechanism by which parental alleles can be separated.0919

Recall from the lecture on meiosis that crossing over occurs during prophase 1.0923

And in prophase 1, the homologous chromosomes pair up, and corresponding segments of DNA are exchanged.0930

If we have that flower color allele and the pollen shape allele on the same chromosome,0938

crossing over is an opportunity for those alleles to be swapped with ones on the corresponding chromosome.0948

Here, we have purple and long with red and round.0956

If crossing over occurs between these two, what we could end up with is this chromosome having0961

the purple allele with the round and the other chromosome still having the red allele but now with the long.0967

Instead of the parental combination of purple-long and red-round, we end up with purple-round, red-long.0976

So, these were the parental types, and these are the recombinant types.0989

Now, the closer together two loci are in a chromosome, the less likely they are to be separated,0998

the less likely recombination is going to occur between them.1006

And with this knowledge, we can develop linkage maps, and linkage maps show the relative positions of genes on a chromosome.1011

So, linkage map shows the relative positions of genes on a chromosome.1024

And Thomas Hunt Morgan's lab studying the fruit fly Drosophila was able to develop this idea of the linkage map.1041

One map unit is equal to a 1% frequency of recombination.1049

The farther apart two loci are, the higher the recombination frequency will be.1063

Once we get to 50%, recombination frequency are 50 map units, these genes are functionally unlinked.1071

The chances of them staying together are just as good as the chances of them separating during meiosis, so at this point, they are functionally unlinked.1081

Even though they are physically linked, they are located on the same chromosome, they behave as though they are not linked.1091

Let's look at an example of a linkage map. Let's say that we have 3 loci: X, Y and Z.1098

And the recombination frequency between X and Y is 20%1108

Then, we do some studies and see that between X and Z, the recombination frequency is 5%.1116

And finally, between Y and C, the recombination frequency is 15%, and then, we want to make a linkage map.1125

I know that X and Y have the greatest recombination frequency.1138

So, the chances of this crossing over occurring between them are much higher than between the X and Z.1142

These two must be the farthest apart, so I am going to put them in the ends; so I have got that.1147

Now, I know that X and Z, they are pretty close together. They are closer together than Y and Z.1156

So, Z is somewhere between these two, and looking at the recombination frequency between X and Z is only 5% versus 15%.1164

I am going to put Z over here. That is closer to X.1175

There is less of chance of recombination occurring between these two. These two are more likely to stay together, so that is 5.1180

From here to the end, I have got 20 between these two, and that leaves my recombination frequency between Y and Z at 15; and that all works out.1187

X and Y are 20 map units apart. X and Z are 5, and Y and Z are 15.1203

So, this is a recombination, a linkage map that is based on recombination frequencies.1208

And I could have easily put Y on this end and X on this end and a Z over here.1213

If I have the X over here, Z next to it and then, Y over there, it is just the relative order. It could have gone right to left.1219

Another type of map that you will hear about is the physical map.1228

A physical map would actually get the distance between loci and nucleotides like X and Z, R.1232

But literally, 500 base pairs are nucleotides of R something, so that is a different type of map. This is a linkage map.1241

So far, we have been talking about genes that are located on chromosomes in the nucleus, and they show certain inheritance patterns.1252

However, you will see a very different type of inheritance pattern for genes that are located outside the nucleus on organelles.1261

And these are called extra nuclear genes.1270

Recall that when we talked about cell structure and organelles,1273

we said that certain organelles like mitochondria and chloroplast actually carry their own DNA.1277

So, there are some genes located outside the nucleus, and they have a maternal pattern of inheritance.1283

Another name for these is cytoplasmic genes because they are located in organelles in the cytoplasm.1292

And it turns out that a zygote derives its cytoplasm from the egg.1300

Fertilization occurs with a sperm uniting with an egg, and the sperm donates chromosomes. It donates a haploid set of nuclear chromosomes.1309

In humans, that would be 23 chromosomes. The egg also donates 23 chromosomes, a haploid set, so the result is zygote that is diploid.1321

In addition, the egg also donates its cytoplasm in organelles, so this cytoplasm in the zygote is derived from the egg.1332

Cytoplasm in the zygote is derived from the egg. Therefore, extranuclear genes such as mitochondrial genes are inherited via mother to offspring.1341

And it does not matter if these offspring are male or female.1363

If a woman has two sons and three daughters, all five of those children will receive her mitochondrial DNA.1367

And there is actually a set of disorders called mitochondrial disorders.1376

Recall that energy production occurs in mitochondria, and so mitochondrial disorders result in a decrease in the amount of ATP produced in the cell.1382

And the result is less energy available for cell functions.1391

There are set of disorders, and they show this inheritance from mother to offspring. You cannot inherit a mitochondrial disorder from your father.1396

Mitochondrial DNA has also been useful in matching up family members if there is a generation missing or something.1405

They can use a grandmother's DNA, and then, look at her mitochondrial DNA, and compare that with her possible grandchildren.1413

So, there are uses for looking at this type of inheritance.1422

A second topic that deviates from what we have been talking about so far as far as1427

how inheritance and expression of phenotype works is genomic imprinting.1433

When we talk about inheriting eye color, we said "OK, you could get a brown allele from your mom or you could get it from your dad”.1440

Either way, if you get that brown-eyed allele, you will have brown eyes. You will express it.1446

Your phenotype will be brown eyes. It did not matter which parent you got it from.1450

However, for some genes, it does matter.1454

For some genes, expression only occurs for the allele inherited from the father or the allele inherited from the mother.1459

This type of differential expression based on the parental origin of the allele is due genomic imprinting.1469

And the way that genomic imprinting works is a functional group is added to a gene that causes the allele to be expressed or not expressed.1476

Remember when we talked about chromosomes and cell expression of genes and control of gene expression,1484

one mechanism for control of gene expression was methylation.1490

Methylation can be used to turn off, for example, an allele turned off a gene.1495

To give you an example of this type of imprinting would be IGF-2., so insulin-like growth factor 2.1510

Even though, you would inherit an allele from your mother for this gene and an allele from your father, only the paternal allele is expressed.1526

The allele from the mother that you inherit from the mother is turned off.1540

And what happens is you inherit genes from your mother and genes from your father, and the ones from your father have this paternal imprint.1544

The alleles like IGF-2 that should be turned on or tuned on, certain ones in your mother maybe turned and others turned off.1552

This is just for a small number of genes, so you inherit this with the paternal and the maternal imprint.1562

And then, when gamete formation occurs in a male or female, that imprinting pattern is erased so that they can imprint it with the correct pattern.1568

Let's say a female, when gametogenesis occurs, and she is making eggs, the imprinting pattern that she received from her parents is erased.1581

And then, the genes are reimprinted with the maternal pattern, and her offspring will receive that.1592

If both of the alleles end up expressed in a gene that is only supposed to have one of the alleles expressed, cancer can actually result.1600

This is an interesting area of study for cancer as well.1610

OK, today, we covered some topics that fall outside very simple direct Mendelian inheritance1614

such as linkage, extranuclear genes and genomic imprinting, so let's go ahead and look at some examples.1622

Example one: the recombination frequencies between three genes - E, F and G - have been established.1629

Use this information to develop a linkage map of the three loci.1638

Remember, a linkage map is a map that gives you the relative positions and spacing and order1644

of genes on a chromosome based on the recombination frequencies between them.1651

Let's go ahead and just draw up a line to represent a chromosome.1656

I would start with the two that are farthest apart, and here, that is F and G.1659

So, I am going to put F on one end and G on another, and there are 19 map units between these two.1665

Now, E and G are 13 map units apart, so E is somewhere in here; and it is 13 from G, but it is only 6 from F, so I am going to put it much closer to F.1676

There is less of a chance of recombination between those two.1694

And that leaves me with...so I have E and F are 6 apart, and then, I have E and G are 13 apart.1700

And this all adds up because I have 13 and 6, 19, with E being closer to F than it is to G.1711

A scientist is studying characteristics of a species of birds.1720

She notes that there are two types of wing color: green and red, so wing color, green and red, and the allele for green is big G.1724

And that is green, and the allele for red is little g, so the allele for green is G for dominant. The allele for red- little g.1737

Now, I just need to add in something here, and that is about the second characteristic- tail feathers.1749

Tail feathers can come in two types in this experiment: long, which is dominant; and short, which is recessive.1758

This scientist is studying wing color and tail feathers.1769

And the scientist performs a cross between true breeding birds that are green with long tail feathers and those that are red with short tail feathers.1773

OK, true breeding, so we have green with long tail feathers, and that would be homozygous dominant for both traits.1786

It is a parental generation crossed with red with short tail feathers.1802

So, red is going to be little g, and short tail feathers are going to be little l.1811

She gets her F1 generation, which are going to be heterozygotes- big G-little g, big L-little l.1818

And they are going to be green with long tail feathers, but they are heterozygous.1826

She, then, performs a test cross with this F1 generation.1832

Now, she knows what the genotypes are, but she is performing this test cross, anyways, and we are going to see what the outcome is.1837

Let's go ahead and write down the test cross here or talk about what the test cross is.1846

Recall that test cross is when you cross an individual who has dominant traits, dominant phenotype.1853

And we are going to go ahead and cross that individual with an individual who has the recessive phenotype.1866

In this case, the recessive phenotype would be red wing feathers and short tail feathers, so dominant phenotype.1875

And we do know the genotype, which is big G-little g, big L-little l, times recessive phenotype, little g-little g, little l-little l- homozygous recessive.1887

OK, we have got our cross here.1901

If the genes for wing color and tail feather type are linked, what will be the most common phenotypes in the test cross offspring?1903

Alright, we have to just take this very systematically.1913

Test cross offspring, if we make this blue, it will be easier for us to follow, so let's make this blue.1918

And let's say this is the mother. This is the father, just to keep track of things more easily- mom and dad.1928

And this is going to be the F2 generation. What are the possibilities?1938

One possibility is that gametes could form here. That would be big G with big L.1943

That is one gamete possibility, Big G-little l, little g-big L, little g-little l, OK?1956

Over here, what gametes are we going to get? Well, all these gametes are going to be the same.1970

They are all going to be little g-little l, so that makes things much simpler.1974

When we look at the F2, what are the possibilities? They could get, no. 1, big G-big L, and from dad, they are always going to get little g-little l.1978

So, we will just go ahead and write that in because they are always going to get these recessive alleles from dad.1991

No. 2: from mom, gametes that form could have been little g-little l, 3: big G-little l, 4: little g-little l.1999

OK, I have four possibilities based on the gametes that were formed.2019

And what this is asking me is if the genes for wing color and tail feather are linked,2025

what are going to be the most common phenotypes in these offspring, the offspring of the test cross?2034

Well, let's go ahead and look at what phenotypes are possible.2043

The first is green with long tail feathers, green-long. The second possibility is red with short tail feathers.2047

Third, because green is dominant, we are going to get green and actually three.2060

Let me see, 1, 2, actually, this should be big L, so that is going to be long.2068

OK, so no. 2: little g-big L, little g-big L, red-long, 3: green with short tail feathers, and then, finally, red with short tail feathers.2079

If these two are linked, then, we are going to expect an overrepresentation of the parental phenotypes,2105

parental phenotypes meaning the parent of the one who is forming the gametes.2118

So, let's look at this individual in the F1 generation.2122

Her parents were GG, big G-big G, big L-big L and little g-little g, little l-little l. Therefore, from one parent, she got the big G-big L.2127

From the other parent, she got the little g-little l.2139

If these are linked, I expect them to stay together. I expect the big G and the big L to stay together more frequently than normally.2143

Big G and big L are more likely to stay together. Little g and little l are more likely to stay together.2154

So, I expect these two gametes to be overrepresented,2162

and these two gametes only to show up when there has been a crossing over during meiosis allowing the parental types to be split.2166

I expect the most common phenotypes to be green and long and red and short, which is what we saw on the parental generation- green-long, red-short.2175

So, the answer is, the most common phenotypes I expect in these offspring are green with long tail feathers or red with short tail feathers,2187

whereas, the recombinant types, red with long tail feathers and green with short tail feathers, I expect to be less common.2195

Example three: match the following terms to the correct descriptions.2207

One: genomic imprinting, crossing over, linkage mapping and extranuclear genes.2213

Genomic imprinting, so let's see: exchange of DNA between homologous chromosomes that occurs during meiosis. No.2221

When we have an exchange of DNA between homologous chromosomes during prophase 1, recall that that is crossing over.2232

Let's go ahead and find that in the list and put it here and knock that out.2239

OK, genomic imprinting, linkage mapping and extranuclear genes are left.2245

Genes found in organelles such as mitochondria and chloroplast rather than on nuclear chromosomes, so this one, the name even tells you.2251

They are on nuclear chromosomes. They are extranuclear genes, and recall that these genes have a maternal inheritance pattern- extranuclear genes.2259

Method of determining the relative order of genes on a chromosome based on recombination frequencies.2274

OK, we talked about finding the order of genes based on recombination frequencies as being linkage mapping.2283

And the closer together two genes are, the more tightly linked they are, the less likely recombination is going to occur between them.2292

Expression of an allele based on which parent the allele is inherited from, that must be genomic imprinting, and we discussed that, as well.2301

And we said insulin-like growth factor 2 is an example of that.2309

The paternally derived allele is expressed, whereas, the maternally derived allele is turned off.2313

Example four: Leber's hereditary optic neuropathy is a disorder causing the loss of vision in both eyes during early adulthood.2321

It has been observed that inheritance of this disorder is only through the maternal line.2330

Where might the genes responsible for Leber's optic neuropathy be located?2337

Well, maternal inheritance occurs with extranuclear genes such as mitochondrial genes.2342

So, the genes responsible for this may be located on the mitochondria, and in fact, they are located on the mitochondria.2349

So, the inheritance pattern would be from mother to her offspring and then, from the female offspring of her to their offspring.2358

It could not be transmitted from the father to his offspring.2366

That concludes this section of genetics and inheritance for Educator.com2372