genetics Flashcards
Chromosomes
A chromosome is a DNA sequence containing coding and non-coding regions, associated with proteins that are histone and non-histone proteins.
It is visible only duringthe metaphase. Before that, the
DNA is not packaged enough to be seen.
Each chromosome is constituted by a long DNA molecule associated with proteins. The total
length of DNA is 2 meters, but each chromosome is constituted by a segment of the 2 meters long DNA molecule.
Due to the phosphate groups, negative charges are present on the double strand of DNA. These negative charges should be neutralized to get the chromatin, because charges
with the same sign try to seperate.
Chromatin is formed by DNA, histones, non-histone proteins and also RNA, because DNA is not a static structure but a dynamic structure. Therefore transcription of some genes can be initiated, and in that case, it is possible to find RNA associated with DNA in the chromatin.
Chromatin
Chromatin is formed by DNA, histones, non-histone proteins and also RNA, because DNA is not a static structure but a dynamic structure. Therefore transcription of some genes can be initiated, and in that case, it is possible to find RNA associated with DNA in the chromatin.
What are non histone proteins that are associated with chromatin ?
Non histone proteins are regulatory proteins that are involved in transcription,replication, recombination and DNA repair.
To interact with DNA, these proteins should show some specific domains.
Let’s consider three different motivfs that are able to interact with DNA:
1. The helix-loop-helix motif:
When the double strand is wrapped around a hypothetical longitudinal axis, 2 grooves are formed, major and minor grooves, that give access to the information.
Because the backbone of the 2 strands is formed by phosphate and sugar, they are identical, but inside we have the only variable part of the molecule which is the sequence of
nitrogen bases.
So information in the molecule of DNA is represented by the nitrogen bases which are of 4: adenine, guanine, thymine and
cytosine.
The grooves are important because it’s only at the level of these grooves that proteins can interact with DNA inducing the opening of the double helix and giving access to the information contained in the sequence of the nitrogen bases.
So proteins that show this helix-loop-helix can interact with the grooves of the DNA.
This protein is organised showing 2 alpha helix as a secondary structure joined together by a short segment of amino acids.
One helix recognizes the major groove, the other binds DNA in a specific manner. In this case only
the major groove is involved in the recognition.
- Zinc fingers:
Another motif can make zinc fingers, that can be considered as loops that are stabilized by a zinc atom that connects two cysteines and two histidines.
Usually, in the protein, there are several zinc fingers that can interact with different major grooves. - Leucine zipper:
This motif is represented by 2 strings of leucine amino acids that
can recognize the major grooves on DNA.
These 3 examples are motifs that should be present in proteins in order for them to interact with the grooves of DNA. These are non histone proteins.
Why DNA packaging ?
Let’s take DNA from one cell and eliminate all the proteins that interact with DNA, so we obtain a naked linear DNA molecule, with a length of about 2 meters. To fit this lenght o DNA inside the nucleous, we have the process of DNA packaging. Histones play a crucial role in this process.
What are histones ?
Histones are basic proteins containing a large amount of amino acids that show positive charges ( arginine and lysine). Due to these protein charges, histones are able to neutralize the negative charges that are present on the double strand of DNA. Histones are divided into two classes: ● class 1 - contain H1 histones. ● class 2 - contain H2A- H2B- H3- H4 histones. ➝ they differ by their arginine/lysine ratio
The packaging and de-packaging of DNA is due to modifications at histone levels.
Structure of the nucleosome: an histone octamer, DNA that is
wrapped around and the tail of H2A, H2B, H3 and H4. These tails can be modified and this modification can induce packaging or de-packaging.
To package nucleosomes, histone tails should be methylated. So the modification that induces the packaging is the methylation of
tails: this forms chromosomes.
To uncondensed/unpackaged DNA, these tails should be
acetylated.
Package: methylation
Unpackage: acetylated
The packaging of DNA
- lowest level of chromosome packaging. It involves histone proteins.
In 1974, Roger Kornberg proposed an entirely new structure for chromatin:
-DNA and histones are organized into repeating subunits, called nucleosomes that contains a nucleosome core
- nucleosome core: supercoiled DNA wrapped almost twice around a disk-shaped complex of 8 histone molecules.
- histone core : 2 copies of the histones H2A,H2B, H3, and H4 assembled forming an octamer.
Nucleosomes are joined by a naked DNA; DNA linker.
The DNA linker is accessible for non histone proteins to be opened and eventually replicated and transcribed.
The DNA that is wrapped around is not accessible to the non histone proteins.
This structure is not static. On the contrary, it is highly dynamic : DNA is wrapped around and unwrapped immediately.
For this reason all the DNA will be accessible in the time to be replicated or transcribed.
Chromatin is a dynamic cell component in which histones, regulatory proteins and a variety of enzymes move in and out of the nucleoprotein complex to facilitate the complex tasks of DNA transcription, packaging, replication, recombination and repair.
this level of packaging isnot enough. - other histone involved,
histone H1.: outside of the nucleosome. It’s a linker histone because it binds to a part of the linker DNA that connects one nucleosome core particle to the next. H1 molecules continuously dissociate and re-associate with chromatin. They bind near the sites where DNA enters and leaves the nucleosome.
It shows a H2N head and COOH tail.
Its tasks are to block the DNA on one nucleosome and to keep close the adjacent nucleosomes.
So we have an attraction between the tail of histone 1 of the first nucleosome and the head of histone 1 of the adjacent nucleosome.
Therefore, there is an attraction between H1 proteins, and this
attraction keeps the nucleosomes very close to one another.
- formation of solenoid. in the solenoid the DNA linker is not visible because all the nucleosomes are packaged very close to one another.
Therefore information is not accessible.. But it’s still not enough.
4.This structure will now form loops on
a protein called scaffold. At this stage, we reach 300 nm width. This still not enough
- The structure finally forms superloops to
reach the maximum packaging of DNA that
is visible at metaphase. Each segment of
DNA has a width of 700 nm.
Why are there 2 segments of DNA that are linked at the centromere level ?
➜ These chromosomes are only visible at metaphase which is one phase of mitosis and meiosis. Before these 2 processes, we observe DNA duplication. So before starting mitosis, each cell duplicates its DNA content.
After, during prophase, DNA is packaged, and reaches the maximum packaging at metaphase. we see
2 segments. They are called sister chromatids. They are identical because one is the copy of the other.
Chromatin
2 types of chromatin
1. euchromatin
2. heterochromatin
The difference between these 2 is morphological (structural) but all the DNA is duplicated, but the difference between euchromatin and heterochromatin is at transcription level.
Euchromatin : not packaged and accessible for transcription. when a DNA should be transcribed it is present as euchromatin.
Heterochromatin: packaged and not accessible for
transcription so it’s always at transcription level that we find a difference between euchromatin and heterochromatin.
2 forms of heterochromatin:
- facultative: chromatine that can be transcribed during particular moments of the cell life. So there is a possibility to have the expression of this DNA.
- constitutive: chromatin that does not contain information (consists primarily of repeated sequence and contains relatively few genes) and is never transcribed. An example is the DNA at centromere level.
centromere: sequence of DNA transcriptionally inactive, also called primary constriction. It is fundamental for meiosis and mitosis because at the primary constriction, during these 2 processes, is added a protein structure named kinetochore necessary for the binding of spindle fibers.
The localization of the primary constriction lets us
see a short arm p and a long arm q .
In human chromosomes, only 5 pairs of chromosomes contain a secondary constriction which is the site of rRNA synthesis.
In human cells, there are 46 chromosomes, that
come in pairs: 22 pairs of autosomes and 1 pair of sex chromosomes.
It is possible to identify these chromosomes looking at the position of the centromere and also looking at the possible different bandages that are formed after coloring chromosomes with different dyes.
The territory model
Chromosomes occupies a specific area in the nucleus and this localization should be respected.
These chromosome territories change according to
the cell type, the development phase and in pathological situations because this have influence on gene expression.
DNA sequences that participate in a common biological response but reside on different chromosomes can come together within the nucleus where they can influence gene transcription. So DNA sequences move, into the nucleus, to reach the region where an intense transcriptional activity is present.
The genes involved in the same process co-localize in the
same area to influence transcription activity.
e.g if there are 2 different chromosomes containing
information for proteins that are needed, these chromosomes,
keep in touch, move, to be transcribed together forming
transcription factories.
eg of inter-chromosome interactions comes
from a study in which human cultured breast cells (normal and
malignant) were treated with estrogens. Before the treatment, genes that are a target for estrogens are located far from each other, one on chromosome 2, the other on chromosome
21.
After the incubation of these cells with estrogens, chromosome territories have moved into close physical proximity to one another and the 2 loci are co-localized.
Inside the nucleus, genes involved in the same response tend to become co-localized in the same site called transcription factories where transcription machinery is concentrated.
They physically move inside the nucleus thanks to the nuclear skeleton/nuclear matrix which is a structure similar to the cytoskeleton.
In human cells, we find pairs of homologs.
what are homologs
Homologs: are chromosomes that contain the same information, one is from paternal origin, the other of maternal origin. They contain genes codifying the same traits but they may contain alternative forms of the same gene that are different alleles which arelocated on corresponding loci of homologs.
e.g on the homolog that comes from dad, is present a gene that is responsible for the expression of the shape of the nose, on the
corresponded locus of the homolog that comes from mom, there is going to be the gene that is responsible for the expression of the shape of the nose. But the nose can show 2 alternative forms, so the dad chromosome can carry the information to get the potato shape and mom can carry the sequence that is responsible for the expression of another nose shape. Depending on the relation between these 2 different sequences, one being dominant, the other recessive, we will get the expression of the nose shape.
In female cells, there is always a pair of X chromosomes, one coming from the mother, the other from the father. In male cells, there is always one X chromosome coming from the mother and one Y chromosome coming from the father.
Homologs carry genes that codify for the same traits. But they can carry alternative forms of the same genes.
Sister chromatids and genomic kit
Sister chromatids: are 2 chromosomes bound at
centromere level. They are identical (at early
meiosis and during mitosis) because one is the copy deriving from the DNA replication process.
Ploidy indicates the number of chromosome
sets available in a cell. Each set is indicated by
the letter “n”.
In human cells, there are 2 sets of chromosomes (pairs of homologs) so they are diploid,2n.
A cell with one set of chromosomes will be haploid “n”, in
the gametes, only one chromosome of each pair of homologs is
present (so it’s haploid) but after fertilization, the number 46 is restored (diploid).
A cell with more than 2 sets of chromosomes will be polyploid.
A zygote, in eukaryotes, is the fertilized egg cell, until the start of segmentation.
The movement of chromosomes at meiosis.
We can consider one cell containing a pair of homologs called 1’ and 1’’ to underline the different origin of this pair of homologs.
Before meiosis DNA replicates.
If on homolog 1’ there is allele A then both sister chromatids have it, and allele a on both sister chromatids of
homolog 1’’.
After duplication, DNA can start the packaging process, and lead to visible chromosomes at the metaphase.
The meiosis spindle can take this pair of homologs at the metaphase plate (a virtual plate that actually does not exist) aligning the homolog 1’ on one side and the homolog 1’’ on the other.
These pairs of homologs are bound together and reach the
metaphase plate.
During anaphase of the first meiosis, we observe the segregation of these homologs that reach the opposite poles and will form 2 cells, one containing the homolog carrying allele A and the other containing the chromosome carrying allele a.
These 2 cells are already haploid because they contain one representative of the pair we are considering.
However, these chromosomes are constituted by 2 sister chromatids. Therefore, it is necessary that they undergo the second meiosis process which leads to the separation of
sister chromatids. So at the end, there are going to be 2 cells with chromosome carrying allele A and 2 cells carrying allele a.
During first meiosis, homologs separate and the product of the first meiosis is represented by cells that are already haploid.
After the second meiosis which is identical to mitosis, there
is separation of sister chromatids.
So at the end of meiosis, starting from 1 cell, we have
4 cells.
When we consider one trait, 50% of the cells will carry the dominant allele, and 50% will carry the recessive allele.
Now let’s consider 2 pairs: pair 1 and pair 2. We hypothesize that, on pair 1, we find a dominant allele on one chromosome, and recessive one on the homolog for gene A.
And on pair 2, dominant allele on one, recessive on the homolog, for gene B.
This situation is different from the precedent as we are now considering 2 different traits.
Before meiosis,DNA duplicates.
Sister chromatids have the same alleles.
Now meiosis can start.
Chromosomes are visible due to packaging, and they migrate thanks to the fibers of the meiotic spindle then reach the
metaphase plate. Here chromosomes can be organized in 2 ways:
1. The first meiosis yields 2 cells, one that will contain a chromosome that contains allele A and a chromosome with allele B, and another cell with chromosome of allele a and a chromosome with allele b. These cells are haploid because they
contain 1 copy of chromosome 1 and 1 copy of chromosome 2, formed by 2 sister chromatids. The concept of ploidy is not
quantitative but qualitative.
In terms of the amount of DNA, it is doubled but in terms of
information, it comes from 1 copy of homolog.
- after the second meiosis, from the first cell we will get 2 cells that contain information for the dominant trait of A and recessive trait of B. From the second cell, we will get 2 cells carrying the information for recessive A and dominant B.
So 50% of cells from the first possible organization, from which we will get cells carrying both dominant and recessive alleles such that 25% of cells are AB genotype and 25% of cells are
ab genotype.
In the other 50% of organization of the homologs we will get 25% of cells that are Ab genotype and 25% of cells that will be aB genotype.
Increasing the number of homologs, increases the number of hypothetical metaphase plates and also the number of gamete
classes.
Let’s consider the presence of diploid cells and consider just one trait. The subject we are studying is dominant homozygous for trait A, so it contains information to express the dominant trait. In this case, these subjects will form only 1 class of gametes. These cells will contain all one copy of A and are haploid. If I have a subject that is recessive homozygous for trait A, also these ones will form just 1 class of gametes that will carry one copy of allele a.
If the subject is heterozygous, therefore containing allele A and
a, we can get 2 classes of gametes, 50% will carry the dominant allele and 50% the recessive one, because to form haploid cells, the mother cell should undergo meiosis.
Let’s consider 2 genes, A and B, and a subject that is double dominant homozygous with genotype AABB in diploid cells.
In the haploid cells formed by this subject, we can get only one
class that carries the genotype AB.
If we have a subject with genotype aabb, we also get only 1 class of gamete with genotype ab.
If we have a subject with
genotype AAbb, we get 1 class of genotype Ab.
For a subject with genotype AaBB, we get 2 classes of
gametes, 50% carrying AB and 50% carrying aB.
From a double heterozygous, we can calculate the number of gamete classes by 2^n where n is the number of genes in heterozygous.
In the case of AaBb genotype, the genes that are present in heterozygous are A and B so 2^2 =4 different classes of gametes: AB, ab, Ab, and aB. We have these classes of gametes because these 2 traits are located on different pairs of homologs and they assort independently. This means that these 2 pairs of homologs can be organized at the metaphase plate in 2 different ways (cf. the example detailed
previously).
If we consider a subject that is heterozygous for 3 genes: A, B and D.
In this case 2^3 =8 different classes of gametes can be formed and 4 different metaphase plate organization.
To identify the different classes of gametes, we need to write all the possible organizations at the metaphase plate.
Mendel and his experiment
Mendel has to be considered a competent scientist because he decided to use the experimental model and made a good choice in selecting pea plants for his experiments.
Pea plants were perfect because they enabled him to organize controlled breeding and also to self fertilize them to get two breeding populations.
Mendel decided to study the inheritance of one trait at a time, thanks to this choice he was able to postulate his laws.
He was also lucky and clever because he chooses only seven traits because in pea plants there are only seven pairs of homologous, and also the choice of the particular traits was singular. It is highly possible that he had also obtained results that didn’t fit with the previous one, but he transmitted only the good ones.
The science of genetics is very young, it started in 1953 with the discovery of the DNA by Watson and Crick, and in the last century, it continued with several results until the publication of the complete human genome sequence (April 14, 2003).
THE CHARACTERISTICS OF PEA PLANTS:
Why was the choice of the plea plant as an experimental model winning?
In pea plant flowers is possible to appreciate the presence of both the male and female reproductive structure:
Stamen = is the male reproductive structure Anthers = pollen producers, male gametes Ovary = female reproductive structure Stigma = a sticky organ that is on the top of pistils on which the pollen grains rest.
This kind of plant can be fertilized, eliminating the anthers from one and transferring the pollen from another flower to the stigma, but is also possible the self-fertilization in which the pollen of the same plant is used to fertilize the plant.
Mendel decided to study 7 different traits present in pea plants:
- flower color
- seed color
- seed texture
- pod color
- pod shape
- plant height
- flower position
It is important to take in mind that these seven traits show just two alternative forms, so we have only two possible alternative forms.
He started considering one trait at the time.
THE EXPERIMENTAL PHASE:
Mendel started studying tall and dwarf plants to understand how the trait was inherited.
- parental lines: tall and dwarf plants coming from several cycles of self-fertilization, doing so he was able to have a homogeneous population in which all the plants of a line showed the same characteristic.
He obtained what is called true-breeding plants that had a homozygote formed trait.
- he crossed the two true-breed populations with the different traits obtaining the first progeny F1, in which was present only the tall trait, while the dwarf trait was completely disappeared.
- he crossed two plants of F1 he obtained an F2 progeny in which was presented the trait dwarf in a ratio of 1 : 3 (1 dwarf, 3 tall).
From this, he was able to determine which trait was dominant (tall) and which one was recessive (dwarf).
The observations that he derived were that:
Phenotypically: In the first generation the recessive traits completely disappeared, then in the second generation the recessive trait reappeared always in the ratio of 1 out of 4 (1:3)
Genotypically: in the first filial generation he always got all heterogeneous traits, then in the second generation he obtained 1:2:1, 1 out of 4 dominant homozygote, 1 out of 4 recessive homozygote, and 2 out of 4 heterogeneous.
Mendel first law
the dominance law:
all members of F1 produced by a cross of two true-breeding P generations, which differ in one character, show only one of the two characters, called dominant.
The alternate form which disappears is called recessive.
Mendel second law
the second law self-fertilized F1 members produced two types of progeny, showing both characters. These progeny consistently appeared in a ratio of 3:1
Test cross
A test cross is a way to determine the genotype of an organism that shows a dominant phenotype.
Because, as seen before, it is possible for the subject to have a homogeneous or heterogeneous genotype.
To apply test cross, it is necessary to cross the unknown subject with a recessive one, because the latter one necessarily has a homogeneous recessive genotype.
If in the progeny all the subjects present a dominant phenotype (a) our parental subject is necessarily homogeneous dominant.
Instead, if in the progeny there are different phenotypes (b) in the ratio 1 out of 2 recessive our parental subject has to have a heterogeneous genotype
why does this happen?
Because if an allele is completely dominant, one copy will produce enough product to allow the normal function development, this is called an haplosufficient gene, when this does not happen it is said to be haploinsufficient.
All the Mendelian traits show this relationship, but this is not true for all the characteristics in nature, in fact Mendelian traits are a strange case not the normality.
Mendel third law
Mendel decided to analyze two traits at a time. He used the same strategy seen before, chose 2 traits and let them self-fertilize till a pure breed parental generation. Then he crossed these two true-breeding parental generations and obtained an F1 that showed both dominant traits.
However, the generation F2 obtained by F1 was able to produce 4 classes of gametes.
why 4 classes of gametes?
The parent is diploid and in its cells, there are 2 pairs of homologs, one pair carries the gene responsible for the expression of the color of the seed and the other gene carries the information for the type of surface of the seed.
The homologs are represented in the prophase of the first meiosis and all the chromosomes are formed by two sister chromatids that are identical because one is the perfect copy of the other.
During the first metaphase the pairs of the homologs are carried to the equatorial plate and they can be reorganized in the two-way displayed.
For each possible combination the probability is 25%.
This is why there are 4 classes of gametes.
During this experiment, Mendel observed that F2 progeny showed in the ratio of 9:3:3:1
This ratio is due to the presence of a double heterozygote genotype in both the parents of F2 progeny, this genotype forms 4 classes of gametes. Writing the possible combination in the square of Punnett it is possible to observe the appearance of 9 completely dominant phenotypes, 3 and 3 half dominant for one character and only one completely recessive out of 16.
phenotype:
9/16 G - W
3/16 gg - W
3/16 G - ww
1/16 gg - ww
This observation drove Mendel to postulate his third law,
the principle of independent segregation:
The alleles of different genes segregate, or assort, independently of each other.
This happens due to the peculiarity of Mendelian traits that are localized in different pairs of homologs. In fact, all the seven traits considered by Mendel were dislocated in seven different pairs of homologs. Even this has to be considered a peculiarity of these traits, not the normality.
Independent assortment is at the basis of variability.
Inconplete dominance
An allele is considered to be dominant when it produces the same phenotypic effect (=observable feature, like eye color, hair color, etc.) both in heterozygosis (Aa) and in homozygosis (AA).
Meaning that the two genotypes Aa and AA must yield the same phenotype.
BUT, in some cases, the heterozygous (Aa) genotype yields a different phenotype (observable feature) from the homozygous (AA) genotype.
This difference between phenotypes is an example of incomplete, intermediate, or partial dominance.
For example, if we cross two snapdragon flowers, one possessing homozygous recessive genotype, rr, (which yields a white flower phenotype) and the other possessing homozygous dominant genotype, RR, (which yields a red flower phenotype) we would expect that ALL the organisms that arise from this cross have a red color because R is the dominant allele. But that isn’t the case. What happens is that 100% of the progeny will show pink flowers, because the two alleles R & r are in incomplete dominance. The dominant allele R is not fully expressed.
In incomplete dominance, a new, intermediate trait will be formed as a result of combining the red and white colors.
If we were to cross the plants belonging to F1 (pink flowers), we would obtain the phenotypic ratio that exactly matches the genotypic ratio: ¼ white aa, ½ pink Aa, ¼ red AA.
If we were to make a test-cross between a pink flower and white flower (rr), to further demonstrate that the pink flower is given by a heterozygous genotype (Rr), we would obtain ½ pink flowers and ½ white flowers, confirming that the genotype of the pink flowers is obtained by an incomplete dominance in heterozygosis (Rr).
Exercise (from lecture). Cross two plants that are double heterozygous & show the phenotypic ratio:
AaBb x AaBb
BUT allele A (dominant A) shows complete dominance on a (recessive a)
AND allele B (dominant B) shows incomplete/intermediate dominance on b (recessive b)
Important result: “in case of intermediate dominance, the phenotypic ratio is different than what mendel said, because the relationship between the alleles are different.
Co-dominance
Co-dominance refers to the contemporary expression of the two alleles. So, instead of obtaining an intermediate phenotype (as a result of incomplete dominance) both alleles will be present and will be expressed.
Two examples of Co-dominance are:
Leaves of clover
Blood type AB0
In this first example of the clover leaves, we have three different phenotypes:
One phenotype shows a white frame on each leaf (VhVh)
Another phenotype shows a white shade in the leaves (VfVf)
The third phenotype shows both the white frame and shades on the leaves (VhVf)
Two alleles are responsible for these phenotypes: Vh & Vf. The reason why we can obtain three different phenotypes is that they represent all the possible genotypic combinations possible.
The reason why this is an example of Co-dominance is that, whenever their genotype is expressed, their phenotype WILL be observed.
So, in heterozygosis, both the white frame and the white shade will appear.
The second example of codominance, which also ties in with the idea that a single gene can possess multiple alleles (that will be mentioned later) is the expression of blood types on human RBCs (red blood cells).
There are 3 alleles responsible for the expression of the blood group in an individual:
Ia: (codifies for the enzyme A-glycosyl transferase A)
Ib: (codifies for the enzyme B-glycosyl transferase B)
i: (doesn’t codify for any enzyme, no glycosyl transferase)
It is important to notice that both Ia and Ib are dominant on i, and also that Ia and Ib are co-dominant between each other, meaning that if they are both present, both glycosyl transferases A & B will be present.
there are 6 possible genotypes and 4 possible phenotypes.
So, if an individual is blood type A, they will present antigen A.
If an individual is blood type B, they will present antigen B.
If an individual is blood type 0, they will present NEITHER of the two antigens.
If an individual is blood type AB, they will present both antigen A & B.
The antigens mentioned are present on the RBC surface and they derive from a membrane glycolipid modification due to the action of the various glycosyltransferases (enzymes that establish natural glycosidic linkages).
To be more precise, these modifications entail the addition of sugars to these glycolipids.
A specific gene called gene H, found on glycolipids, regulates the expression of the enzyme fucosyltransferase, which is responsible for the attachment of fucose (a sugar) to a short chain of sugars (note: the professor pointed out that it is not necessary to know which sugars compose this short chain).
Upon addition of fucose, we can identify the antigen 0.
Then, depending on what alleles we possess in our genome, further modifications occur:
IF we possess the Ib allele, we will also find a B-glycosyltransferase and its job will be to add another sugar, called galactose, to the antigen 0. Yielding what is then defined as antigen B.
IF we possess the Ia allele, we will also find the A-glycosyltransferase and its job will be to add N-acetyl-galactosamine, to the antigen 0. Yielding what is then defined as antigen A.
IF we possess both Ib & Ia alleles, 50% of the antigens will possess galactose and 50% will possess N-acetyl-galactosamine because BOTH A-glycosyltransferase and B-glycosyltransferases are present! (hence: co-dominance)
IF we DO NOT possess either of Ib and/or Ia, no further modification will occur, and the antigen 0 will be maintained.
(The bonds are called glycosidic bonds)
(Blue molecule is not important to remember)
Type A1 (still Ia allele) and B have different glycosyltransferases, determined by only 4 amino acids, which represents the active site, and thereby determines the specificity to their substrates (being, as previously mentioned, N-acetyl-galactosamine and galactose respectively). Type 0 is inactive mainly due to the presence of deletions, and appearance of a premature STOP codon (stop codon will be covered in a future lecture).
The presence of fucose is absolutely fundamental, without it, the addition of N-acetyl-galactosamine and galactose are NOT possible.
the previously mentioned H gene that codes for the fucosyltransferase, allowing for fucose to be attached to the short sugar chain, a fundamental step leading to the presentation of any of the three antigens on the RBC surface.
The H gene has two alleles, H (completely dominant) and h (recessive).
Dominant H codes for the presence of the fucosyltransferase while recessive h doesn’t.
An individual with genotype hh (homozygous recessive) has a (rare) disorder called Bombay phenotype, where, because of the lack of fucosyltransferase, the individual doesn’t express ANY of the blood groups.
We should also take into account the Rh system.
The Rh system is given by antigens, integral proteins on the RBC surface.
We can find Rh-negative (Rh-) and Rh-positive (Rh+), which are determined by a series of alleles.
The Rh system has massive medical relevance because, if a mother who is Rh- is carrying a child that is Rh+ and, during delivery, the child’s erythrocytes come in contact with the mother’s circulation, they will induce the production of anti-Rh antibodies (that will stay in the body post delivery).
This is important because, in future pregnancies of Rh+ fetuses, it is possible that these antibodies reach the fetal RBCs and cause hemolysis (rupturing of RBCs).
To prevent this, right after the birth of a Rh+ child, a mother is treated with anti-Rh antibodies to destroy the newborn’s RBCs present in a mother’s circulation to avoid antibody production (activated by the immune system) and possible damage to future fetuses.
In the case that the mother has blood group A Rh- and the child has blood group B Rh+, it will NOT be necessary to treat the mother with anti-Rh antibodies because the mother, having blood group A, has anti-B antibodies that will hemolyse any B group RBCs (along with the Rh+ factor) that will enter the mother’s circulation.
Another relevant circumstance would be when we have a mother who has A Rh- blood group and the child has 0 Rh+ blood group. In this case, it is necessary to treat the mother with anti-Rh antibodies because the child has 0 group and its RBCs won’t be recognized and hemolysed by the mother’s anti-B antibodies (due to its genotype like mentioned earlier).
Multiple alleles
It was first established by Mendel that genes don’t exist in more than two allelic states. With further research, it was discovered that this isn’t the case and that genes are capable of possessing three, four, or more alleles. So basically, a specific phenotypic trait is not only given by ONE specific allele, but it can be a combination of multiple alleles.
For example, in drosophila (fruit fly), the eye color is determined by several alleles. We have:
W+ type: red eyes
Wsat type: deep ruby in white satsuma eyes
W type: white eyes
Because there are multiple alleles, we are capable of getting multiple genotypic combinations.
The formula to calculate the number of genotypic combinations is: n(n+1)/2
Another example is the color of the coat of a rabbit is determined by a gene that has four alleles:
c (albino): recessive to all other alleles
ch (himalayan): partially dominant only to c
cch (chinchilla): partially dominant to ch and dominant to c
c+ (wild-type): completely dominant to all the other alleles
Because the alleles are so numerous, we have many possible genotypic combinations and phenotypes, and, because C+ is dominant over all the other alleles, whenever C+ is present the phenotype will show wild-type. The figure on the right shows all the possible phenotypes and their genotypic combinations. Albino is recessive to all the other alleles.
Pleitropy
Pleiotropy is observed when an allele exerts effects on multiple traits.
For example in chickens, the trait that affects the feathers can be defective. This one defect causes a cascade of events that leads to alteration in several parts of the body, for example, as a final effect of this feather defect.
Another example would be sickle cell anemia (in humans) or albinism (affects three different phenotypes: lack of pigment in hair, eyes and skin).
Mendelian trait
If a trait respects a series of requisites it can be deemed a “Mendelian trait”.
These requisites are:
- One gene: one trait/character
- Each trait assorts independently from the other
- In every couple of alleles, one is completely dominant and the other is recessive
When we studied the crossing of double heterozygous (AaBb x AaBb) we know we obtain a phenotypic ratio of 9:3:3:1.
Gene interaction & Epistasis
Epistasis: a trait that is controlled by two independent genes (not alleles!) and the lack of expression or the expression of one gene masks the expression of the other gene (covered more specifically in lecture on 10th of december).
Gene interaction: a trait is controlled by two independent genes and the interaction of two products causes a new phenotype to appear.
Baetson and Punnett obtained evidence that a trait can influence more than one gene through breeding experiments with chickens.
They crossed chickens with different comb traits. There are 4 phenotypes of chicken comb traits, as shown by the figure on the side.
By crossing a chicken with the Rose comb trait and another chicken with Single trait, they obtained in F1 only chickens with Rose comb traits. This experiment showed that the trait Rose was completely dominant on the trait Single. So gene R (for rose) is dominant on gene s (single).
Then, they crossed a chicken with Pea trait with another chicken with Single trait, and they obtained in F1 100% of chickens with Pea trait, again indicating that s (single) gene is recessive while the P (pea) gene is dominant.
Following this experiment, they hypothesized that 3 alleles are responsible for the expression of comb shape in chicken:
R (dominant Rose)
P (dominant Pea)
s (recessive Single).
To further investigate their hypothesis, Beatson and Punnett crossed a chicken showing the Rose trait with a chicken showing the Pea trait. From this, they obtained the appearance of the Walnut trait, which further confirmed the 3 allele hypothesis.
Their current hypothesis was that:
Genotype PP: phenotype will be Pea trait
Genotype RR: phenotype will be Rose trait
Genotype PR: phenotype will be Walnut trait
Genotype ss: phenotype will be Single trait.
Furthermore, they crossed two chickens showing Walnut trait BUT this crossing yielded a chicken with Single comb trait, it is fundamental to notice that this result didn’t match their previous hypothesis because, from what they believed, in order to have Walnut trait we required both PR alleles together, and that wouldn’t include the presence of a single “s” allele.
To explain this phenomenon , Bateson and Punnett hypothesized that the comb-type is determined by two independently assorting genes, R and P, and NOT two alleles!
This means that:
the expression of gene R codes for Rose
The expression of gene P codes for Pea
The expression of both genes R & P codes for a new phenotype: Walnut.
The reason why we’re capable of obtaining Single trait when we cross two walnuts is because to obtain single we need “rrpp” genotype, which is possible because Walnut means that the two genes are heterozygous (RrPp). Check the figure on the side.
So, the real genotype of the various phenotypes is: RRpp: Rose rrPP: Pea RrPp: Walnut rrpp: single
At a molecular level, we understand that the Chicken comb shape is influenced by two genes R and P and the expression of the proteins they code:
When both genes R & P are NOT expressed, we observe the appearance of the Single phenotype.
If only R is expressed, the protein coded by the gene R (also known as protein R) allows for the expression of the Rose phenotype.
If only P is expressed, the protein coded by the gene P (also known as protein P) allows for the expression of the Pea phenotype.
When both genes ARE expressed, the interaction between the proteins coded by gene R and gene P (proteins R and P respectively), give birth to a new phenotype, Walnut.
It is important to not confuse this phenomenon with Incomplete dominance, as incomplete dominance is specific to pairs of alleles and NOT pairs of genes!
Alleles
Alleles are alternative forms of the same genes. One form is dominant (induces the expression of the dominant trait) and the other one is recessive (induces the expression of the recessive trait). We will study about the relationships we can find between two alleles
Epistasis
Epistasis
It is different from gene interaction because when we consider gene interaction, the appearance of a new phenotype occurs.
Epistasis is defined as a phenomenon when a trait is controlled by two independent genes and the lack of expression of one gene masks the expression of the other gene.
The gene which can mask the other gene is called Epistatic gene. The other gene which is controlled by the epistatic gene is known as hypostatic gene.
(Winner is epistatic gene, looser is hypostatic gene)
We have three types of Epistasis:
Recessive Epistasis :lack of expression of one gene that masks the expression of another gene.
Dominant Epistasis: the expression of one gene masks the expression of another gene.
Double recessive Epistasis - an ??
Bateson and Punnett experiment: they crossed two plants showing purple and white flowers (True breeding) to obtain F1 hybrids, which all had purple flowers.
When these hybrids were intercrossed Bateson and Punnett obtained in the F2 a ratio of 9 purple: 7 white plants .
The classic phenotypic ratio of 9:3:3:1 is not respected in this case. The 9 boxes which are purple are purple because they have at least one expression of one C and one P(at least one dominant allele of each gene) i.e Both genes are expressed.
The 7 white boxes are white because the lack of expression of P masks the expression of C as well as lack of expression of C masks the expression of P so all these 7 white boxes have lack of expression of one gene or other or both and in all these cases we got white flowers. Thus, they explained the results by proposing that two independently assorting genes C and P, are involved in the anthocyanin synthesis and that each gene has a recessive allele that blocks pigment production.
Explanation at molecular level
CASE-1) When both C and P are expressed
We have precursor 1, the expression of C allows the conversion of precursor into intermediate form (the flower is still white) and the expression of P induces the appearance of the pigment anthocyanin and we got flower color purple.
CASE-2) When C is absent, but P is present
We have white color because the first reaction could not take place so the precursor could not change into the intermediate form and even P is expressed but then too anthocyanin pigment cannot be made since intermediate form is absent
.
CASE-3) When C is present, but P is absent
We have presence of intermediate substrate but since gene P is not expressed no formation of anthocyanin so again white color.
CASE-4) When both C and P are absent
When both are not expressed so it’s easy to understand it will be white color.
Single recessive epistasis
The lack of expression of one gene masks the expression of another gene .
The example we take here is the rodent coat color and two genes are involved in the expression of this coat color. The crossing over is explained as follows:
EXPECTATIONS:
We expect 9(A-C-) mice showing Aghouti phenotype because 9 out of16 genotypes will have expression of both A and C genes.
We expect 3(A-cc) mice to express brown coat color and the other 3(aaC-) to have black coat color.
We expect 1(aacc) mouse to show albino.
RESULTS:
But after this cross we got different ratios.
Because A and C have an epistatic relationship. 9 is aguti and 1 is albino but in the: CASE of (A-cc) since gene C is epistatic on A means that the lack of expression of C masks the expression of A so in this case gene A is present but its expression is masked therefore it is also albino.aaC as expected to have black color.
Dominant epistasis
Dominant epistasis
Expression of a gene masks the expressions of another non allelic gene in other words the product of one gene may inhibit the expression of another gene.
Definition: caused by dominant allele or one gene masking the action of either allele of the other gene
Ratio is 12:3:1 instead of 9:3:3:1
What’s the reason?
The first one is white, the other one is green. When the breeding plants are crossed we got F1 showing white flowers (double heterozygous) coming from the cross of two breeding plants so when we cross these two double heterozygotes plants we expect 9:3:3:1 ratio but we get another ratio which is 12:3:1 because gene B is epistatic on gene A.
Explanation for every case
Case-1)When B is expressed
We see 12 boxes in white. In all those boxes there is an expression of at least B so all the time B is expressed therefore summer squash is white (expression of B is responsible for white).
Case-2)When B is not expressed but A is expressed
We have expressions of yellow color.
Case-3)When both is not expressed
We have expressions of green color.
Duplicate dominant genes
If the dominant alleles of two genes produce the same phenotype and they don’t have cumulative effect, the ratio will be 15:1.
HOW?
We consider the species Capsella bursa-pastoris and the trait “seed capsule shape”.
Two independent genes A1 and A2 express the triangular shape of the capsule that is dominant over the ovoid one.
- if A1, A2 or both genes are expressed in the seed capsule the shape will be triangular.
- if neither genes are expressed the seed capsule shape will be ovoid.
HINTS FOR FIGURING OUT GENE INTERACTION (looking at the F2 phenotypic ratio)
— If one gene is involved in the expression of one trait, from the cross of two monohybrids we get 3:1 ratio (with complete dominance) or 1:2:1 (with incomplete dominance)
— if two genes are involved in the expression of one trait, in case of gene interaction, the phenotypic ratio will be 9:3:3:1
— if the phenotypic ratio in F2 is 9:4:3 or 9:7 or 12:3:1 we are in presence of epistasis
Why are some mutations dominant and others recessive?
a+ is the dominant allele the result of which is the wild-type phenotype
a is the recessive amorphic loss-of-function allele whose result is a severe mutant phenotype
ah is the recessive hypomorphic loss-of-function allele whose result is a mild mutant phenotype
aD is the allele that contains the information of a protein which is dominant over the wild-type; it is considered a dominant gain-of-function mutation
In the third case a+aD the aD allele wins over the a+ allele.
Genes that codify for different traits always assort independently?
No, linkage between genes was first discovered in experiments with sweet peas the experiment was carried out by Bateson and Punnet.
They studied 2 different traits in sweet pea plants, the
flower colour and the pollen grain size. They used the same strategies as Mendel which is using plants from several cycles of self-fertilization.
They started with Red flower with Long pollen grains and a White flower with Short pollen grains.
In F1 progeny the outcome was 100% Red flower and Long pollen grains, demonstrating that these traits were dominant. The heterozygous products of F1 progeny were crossed and the
expected ratio was 9:3:3:1 considering that these 2 traits show the same behaviour demonstrated by Mendel. But the outcome was different
The reason of the difference in the ratio is because the genes for flower colour and pollen grain size do not assort independently.
Bateson and Punnett might have come up with this explanation after a testcross instead of an intercross in the F1.
A testcross of F1 plants would directly reveal the types of gametes produced.
Back-cross is using the second parent in double recessive form. If there is a recessive phenotype then it can be known that the genotype is recessive homozygous.
Also, the recessive homozygous parent can only form 1 type of gamete. From looking at the progeny its possible to understand if the genes are independent or in linkage.
RrLl X rrll
(Dominant and heterozygous) ( Recessive and homozygous).
In case of independent genes, the ratio is 1:1:1:1
1.First possibility
Since the 4 classes are present with the same frequency it can be said that gene R and gene L are localized in different pair of homologs.
2.Second possibility
Parent 1 produced gametes RL and rl. The reason is that these 2 genes are very close to one another(linked) that the probability of crossing over is 0 so the progeny can inherit one chromosome carrying RL or rl. The genes are in coupling meaning both dominant alleles are in one chromosome and both recessive alleles are in homolog.
3.Third possibility
Parent 1 produced Rl and rL (linked) but this time they are in repulsion form meaning on one chromosome there is the dominant allele for one gene and the recessive for the second and on the homolog recessive for the one and dominant for the second.
4.Forth possibility
The first 2 classes (the majority) are parental classes but the last two are formed due to crossing over and they can be called recombinant classes.
*** The back cross is carried out. Among 1000 progeny 920 show the parental combination and
the remaining 80 are recombinant. The frequency of the recombinant progeny by the F1
generation is 80/1000=0.08. We can use this frequency (usually called recombination frequency)
to measure the intensity of linkage between genes. Genes that are tightly linked barely
recombine, whereas loosely linked once can recombine often.
*** When the frequencies are equal like 50% to 50% or 25% each there is no recombinant or
parental classes. If this is the case either the genes are in different chromosomes, or they are in
the same chromosome but the distance between them are more then 50cM (centimorgan).
***Back-cross results, in this case parental combinations are: RL and rl so these genes are in
coupling. Both dominant alleles are on the same chromosome. Gametes Rl and rL come from
crossing over events, and they are less frequent.
Genes that are on the same chromosome travel together; However, linked genes can berecombined by crossing over.
Crossing over
It takes place in Meiosis I prophase I.
Firstly in S phase the DNA is duplicated so each chromosome has 2 sister chromatids linked from centromere.
*The distance between two points on the genetic map of a chromosome is the average number of
crossovers between them.
Morgan’s Experiment
Morgan studied Drosophila Melanogaster (fruit fly). He studied Body colour and Wing size.
In parent 1 he chooses long wings (dominant) and black body (recessive) in parent 2 short wings (recessive) and grey body (dominant).
In F1 progeny all of them were carrying both dominant
traits Long wings and Grey body double heterozygote. Considering the parental combination from parent 1 it has inherited Lb from parent 2 lB.
Genotype of P1: LLbb
Genotype of P2: llBB
Genotype of F1: LlBb
In the second part of the experiment F1 male (LlBb) was crossed with a female (llbb) with shortwings and black body (both recessive traits). The outcome was long wings black body (Llbb) and short wings grey body(llBb). The male produced 2 type of gametes LB and lb.
It created a question, why there isn’t 4 classes of gametes later it has been discovered that only in drosophila
spermatogenesis no cross over events takes place .
Second Experiment
Morgan used F1 female long wings/grey body LlBb (VvBb)with male llbb(vvbb) and 4 classes of gamete were present Vb,vB,VB,vb. Vb (Vvbb) and vB(vvBb) are parental combination.
VB(VvBb) and (vvbb) are recombinant.
With this experiment Morgan identified parental and recombinant combinations also he was able to calculate the distance between these 2 genes.
In 1936 Bell and Haldane discovered the first case of linkage in humans. The genes responsible for haemophilia and colour blindness are both located on X chromosome.
Increasing the crossing over probability
*The greater the distance between the genes the higher the probability of crossing over
***Unequal Crossover occurs due to high homology meaning that they have very similar sequences. This unequal crossover event can be the basis of triplet expansion in DNA that can
give rises to diseases
*Hot Spot Regions: DNA sequences of high susceptibility to mutation due to;
-Some inherent instability
-Tendency toward unequal cross over.
-Chemical predisposition to single nucleotide substitutions (gene mutation).
-Region where mutations are observed with greater frequency.
The distance on the genetic map do not correspond exactly to physical distance along the chromosome.
How to calculate the distance among genes on a chromosome?
Firstly, crossing two plants AABB x aabb
F1 will be heterozygote for both characters AaBb
A back-cross is carried out AaBb x aabb
Then from the outcome the distance can be calculated;
50+50(recombinants)/1000(the number of total outcomes)x100=10cM
The distance from A to B is 10cM.
To evaluate the distance between A and C crossing is carried out:
AACC x aacc
F1 progeny will be AaCc
Back-cross AaCc x aacc
Distance from A to C is: 40+40/1000x100=8cM
To evaluate the distance between B and C crossing is carried out:
BBCC x bbcc
F1 progeny will be BbCc
Back-cross BbCc x bbcc
10+10/100x100=2cM
So, the distance from A to B is 10cM from A and C is 8cM from B to C is 2cM.
Inheritance of X-linked traits
In the beginning of the 19th century, biologists didn’t know anything about genes and where they were located.
They suspected that genes were situated on chromosomes, but they didn’t have any clear proof. Researchers needed to find a gene that could be unambiguously located on a chromosome. This goal required that the gene was defined by a mutant allele and the chromosome was morphologically distinguishable.
Morgan was particularly lucky because he found the gene that codes for the color of drosophila, located on the X-chromosome.
Thus, he worked on Drosophila melanogaster, a fruit fly that is a very useful animal for research because it is not expensive and has a short reproduction cycle. Furthermore, in Drosophila cells, only 4 pairs of homologs are present: 3 pairs of autosomes and 1 pair of sex-chromosome, such that females are XX, and males are XY. Also, it is possible to understand the sex of drosophila morphologically i.e just by looking at the shape of the abdomen: in females, the abdomen is light and pointed while in males, it is dark and round.
Through careful experiments, Morgan was able to show that the eye color mutation was inherited along with the X chromosome, suggesting that the gene for eye color was physically located on that chromosome. Later, one of his students, Calvin B. Bridges, obtained definitive proofs for this chromosome theory of heredity. He also discovered the meiotic event of non-disjunction.
Here is a picture showing the inheritance of the X-chromosome. The blue dot spots the X chromosome of the male carrying a mutation. Crossing a male and female yields females with exclusively the male X-chromosome. This is because males inherit only the Y-chromosome from the father and the X-chromosome from the mother.
Then the F1 progeny are crossed between each other. The blue dotted chromosome is inherited in the F2 progeny by both sexes, and if this dot represents a recessive mutation, in males, 50% express this mutation and 50% are free of any mutation, while in females, 50% carry the mutation but don’t necessarily express the disease, and 50% don’t have any mutation. This is called criss-cross inheritance: the mutated gene is transmitted from the father to the daughter, and then the character is inherited to the second generation through the carrier of the first generation.
Morgan experiments
Morgan had in his lab many boxes containing drosophila, and they all presented red eyes, which is the wild-type eye color, but one day, he found one male drosophila showing white eyes. In order to understand the phenomenon, he organized crossings. He first crossed a red-eyed female to the white-eyed male. He obtained 100% of progeny with red eyes, regardless of gender. This result confirmed that red eyes were dominant over white eyes.
Then, in F2 progeny, he obtained a phenotypic ratio of 3:1. A quarter of progeny that showed white eyes were exclusively males. This pattern suggested that the inheritance of eye color was linked to the sex chromosomes.
Then, he carried out a reciprocal cross: he crossed white-eyed females with wild-type males. Mendel also carried out this kind of reciprocal cross, and always obtained the same results because the traits he studied were located on autosomes.
In the F1 progeny, Morgan obtained 50% red-eyed females, and 50% white-eyed males. Intercrossing the F1 progeny yielded a ratio of 1-1-1-1: half the progeny of each sex had white eyes, and the other half had red eyes. Thus, Morgan’s hypothesis that the gene for eye color was linked to the X-chromosome was confirmed.
Events of meiotic nondisjunction
One of Morgan’s students, C.B. Bridges, provided evidence of the chromosome theory by showing that exceptions to the rules of inheritance could also be explained by chromosome behavior.
Bridges performed one of Morgan’s experiments on a larger scale.
He crossed a white-eyed female with a red-eyed male and examined many F1 progenies.
His experiments in F1, besides the expected progeny, white eyed-males and red-eyed females, also showed white-eyed females and red-eyed males. He explained the appearance of these unexpected phenotypes considering nondisjunction events at meiosis.
If this event occurs, the female will produce gametes, 50% carrying 2 X-chromosomes and 50% empty.
When these eggs are fertilized, we obtain:
XXX female that usually dies
XO-chromosome male
XXY white-eyed female
YO-chromosome that dies
Bridges continued experiments by crossing the exceptional XO red-eyed male with the exceptional XXY white-eyed female to determine how they might have arisen.
This crossing gave no progeny: it was clear that the males were sterile.
Later, he decided to use and repeat the cross carried out by Morgan on a larger scale again. He crossed exceptional females with normal red-eyed males.
Besides the expected white-eyed males and red-eyed females progeny, he obtained white-eyed females and red-eyed males.
He inferred that XXY flies were female and XO flies were male and confirmed the chromosome constitutions of these exceptional flies by direct cytological observation.
Because the XO animals were male, Bridges concluded that in Drosophila, the Y-chromosome has nothing to do with the determination of the sexual phenotype.
However, because the XO males were always sterile, he realized that this chromosome must be important for male sexual function.
This proves the nondisjunction events at meiosis and these kinds of events are at the basis of the alteration of chromosome number in humans too. In fact, it gives rise to trisomy (more on this later).
Reminder on meiosis (important for the written test):
During the first meiotic division, homologous chromosomes separate, and during the second meiotic division, sister-chromatids separate. So at the end of meiosis, there are balanced gametes containing 1 set of chromosomes constituted by one DNA molecule (=monochromatidic).
The nondisjunction during the first meiosis gives rise to 100% unbalanced gametes: 50% shows the double amount of the studied chromosome and 50% doesn’t contain the studied chromosome.
If the nondisjunction event occurs during the second meiosis, then it gives rise to 50% of normal gametes and 50% of unbalanced gametes.
