Genetics Final Flashcards
Describe somatic cells
- Diploid
- One chromo from each parent
- Contain either metacentric chromosomes (centromere in the middle) or acrocentric chromosomes (centromere near one end)
They contain _____ and sex chromosomes, with their full _____ being displayed in pairs of _____ chromosomes. Any deviation from the typical number of chromosomes in somatic cells is termed _____, such as trisomy or monosomy
autosomes
karyotype
homologous
aneuploidy
Describe gametes
- Haploid, sex cells
- Come from either mom or dad
- Do not exhibit homologs until fertilisation
- Karyotype is 1/2 of the amount in somatic cells
Describe zygotes
- The fusion of two gametes during fertilisation
- Diploid, equal contributions from each parent
- Karyotype contains homologous pairs of chromosomes, organised into autosomes
- Aneuploidy may occur during this stage
Describe homologous chromosomes
These are pairs of chromosomes that are similar in size, shape, and gene content, inherited one from each parent. They have the same genes arranged in the same order but may carry different alleles. For instance, a homologous pair may include one metacentric chromosome from the mother and a metacentric or acrocentric chromosome from the father. Homologs align during meiosis and are key in genetic recombination.
Describe non-homologous chromosome
These are chromosomes that do not share the same structure or gene sequence. They belong to different pairs in the karyotype and carry unrelated sets of genes. For example, chromosome 1 and chromosome 2 in humans are non-homologous. Differences in acrocentric or metacentric structures are often seen between these unrelated chromosomes
Describe sister chromatids
These are identical copies of a single chromosome, formed during DNA replication. They are joined at the centromere and contain the same genes and alleles. Sister chromatids are found in a single chromosome prior to cell division and remain together until they are separated during mitosis or meiosis.
Non-sister chromatids
- Belong to homologous chromosomes (i.e., from different parents) and contain the same genes but potentially different alleles.
- During meiosis, non-sister chromatids may undergo genetic recombination, exchanging genetic material to increase variation.
- Non-sister chromatids can also be found in either metacentric or acrocentric homologous chromosomes.
Describe interphase
During this phase, chromosomes exist in an extended form as chromatin, making them difficult to distinguish individually under a microscope. In the G1 phase, the cell grows and performs its normal functions, while also preparing for DNA replication. During the Synthesis phase, DNA is duplicated, resulting in each chromosome being composed of two sister chromatids, which are held together by cohesin proteins at the centromere. By the time the cell reaches G2, the chromosomes have duplicated, and the centrosome, containing a pair of centrioles, has also replicated, preparing the cell for mitosis.
Describe prophase
- First phase of meiosis
- Chromosomes begin to condense going through a process known as condensation
- Each chromosome now consists of two tightly paired sister chromatids
- The nuclear envelope begins to break down, and the centrosomes move to opposite poles of the cell
- The formation of the mitotic spindle—a complex of microtubules originating from the centrosomes—begins.
Describe Prometaphase
- The second phase is marked by the complete disintegration of the nuclear envelope.
- This allows spindle microtubules to interact directly with chromosomes.
- Specialized protein complexes called kinetochores form at the centromere of each chromatid.
- Kinetochore microtubules from opposite spindle poles attach to the kinetochores.
- These microtubules exert forces on the chromosomes, guiding them towards the center of the cell.
- Other microtubules, such as polar microtubules, extend from centrosomes and overlap instead of attaching to chromosomes.
- Astral microtubules radiate outward, anchoring the spindle poles to the cell cortex
Describe metaphase
The chromosomes align at the cell’s equatorial plane, known as the metaphase plate, where they are held under tension by the kinetochore microtubules. This alignment ensures that each chromatid faces opposite poles. The cell undergoes a critical mitotic checkpoint here, where the spindle assembly is checked for proper attachment and tension. If any chromosome is not correctly attached, the process is halted to prevent errors in chromosome segregation, which could lead to aneuploidy.
Describe anaphase
Once all chromosomes are correctly aligned and tension is detected, the cohesin proteins holding the sister chromatids together are cleaved by a protease called separase. This allows the sister chromatids to separate and move toward opposite poles of the cell. The kinetochore microtubules shorten, pulling the chromatids to the poles, while polar microtubules elongate, pushing the centrosomes further apart and helping to elongate the cell. Astral microtubules assist by anchoring the centrosomes to the cell cortex.
Describe telophase
When the chromatids, now considered individual chromosomes, reach the poles of the cell, the nuclear envelope reforms around each set of chromosomes, creating two distinct nuclei. The chromosomes begin to decondense back into their relaxed chromatin state, making them less visible under a microscope. The mitotic spindle disassembles, and the cell prepares to divide its cytoplasm.
Describe cytokinesis
Cytokinesis follows mitosis, dividing the cytoplasm to form two separate daughter cells. A contractile ring composed of actin filaments forms beneath the cell membrane at the site of the cleavage furrow. This ring contracts, eventually pinching the cell into two genetically identical daughter cells, each containing its own nucleus. In cases where cytokinesis is incomplete or absent, such as in a syncytium (a multinucleated cell formed from fusion or incomplete cytokinesis) or coenocyte (a multinucleated cell from nuclear division without cytokinesis), the result is a cell with multiple nuclei.
The centrosomes, each containing a pair of _____, are the primary microtubule-organising centres in animal cells. During interphase, the centrosome replicates, and during mitosis, the two centrosomes migrate to opposite poles of the cell, helping to form the _____. The ____ assist in organising the spindle apparatus, although their exact role remains somewhat debated, as cells can still form spindles in their absence.
centrioles
mitotic spindle
centrioles
Describe kinetochore microtubules
These attach to the kinetochores of the chromosomes and pull the sister chromatids toward opposite poles during anaphase. The shortening of these microtubules generates the force required for chromatid separation.
Describe Polar Microtubules:
These extend from each centrosome but do not attach to chromosomes. Instead, they overlap with the polar microtubules from the opposite spindle pole. As they elongate during anaphase, they help push the centrosomes apart and elongate the cell.
Describe Astral Microtubules
These radiate from the centrosomes toward the cell membrane, helping to anchor the spindle poles in place. They play an essential role in positioning the mitotic spindle and ensuring proper orientation.
The ______ is a protein complex that assembles at the centromere of each chromosome during mitosis
kinetochore
______ ______ attach to kinetochores and generate the force needed to move chromosomes. The attachment of microtubules to kinetochores is dynamic, allowing for adjustments to ensure correct bi-orientation, where each sister chromatid is connected to microtubules from opposite poles.
Kinetochore microtubules
_____ is a protein complex that holds sister chromatids together after DNA replication until ______, when it is cleaved by separase. This ensures that chromatids are segregated accurately to each daughter cell.
Cohesin
anaphase
Describe the metaphase plate and tension
During metaphase, chromosomes align at the metaphase plate. The tension created by the opposing forces of kinetochore microtubules pulling in opposite directions ensures proper alignment. The mitotic checkpoint at this stage verifies that all kinetochores are properly attached to microtubules and that equal tension is applied to each chromatid pair, preventing premature progression to anaphase
Describe anaphase movement
In anaphase, the shortening of kinetochore microtubules pulls sister chromatids apart. At the same time, polar microtubules lengthen, pushing the centrosomes further apart and elongating the cell. Astral microtubules assist by anchoring the centrosomes to the cell cortex, helping stabilise the spindle as it moves chromosomes to opposite poles.`
Describe the contractile rings with cytokinesis
A contractile ring of actin filaments forms at the equator of the cell, beneath the plasma membrane. As this ring contracts, it pulls the membrane inward, creating a cleavage furrow that eventually divides the cell into two genetically identical daughter cells. The process is guided by signals from the spindle midzone and is tightly regulated to ensure the proper partitioning of the cytoplasm and organelles.
Meiosis occurs in ___ ____ and involves two successive divisions: ____ (reductional division) and _____ (equational division). These divisions reduce the chromosome number by half, creating four haploid gametes from a diploid cell.
germ cells
meiosis I
meiosis II
_____ This is the most complex phase of meiosis and involves multiple sub-stages where significant chromosome behaviours occur
Prophase I
Describe the five stages of prophase I
- Leptotene: Chromosomes start to condense.
- Zygotene: Homologous chromosomes pair up (synapsis).
- Pachytene: Crossing-over happens between homologous chromosomes.
- Diplotene: Homologous chromosomes start to separate but remain connected at chiasmata.
- Diakinesis: Chromosomes further condense, preparing for the next stage of meiosis
Describe metaphase I
Homologous chromosome pairs align at the equator of the cell (the metaphase plate). Each homolog is attached to spindle microtubules from opposite poles, ensuring that they will be pulled apart in the next phase.
Describe anaphase I
The homologous chromosomes are pulled apart toward opposite poles of the cell. This is the key step in reductional division, where the chromosome number is halved. Each resulting daughter cell will have one copy of each homologous chromosome.
Describe telophase I
The separated chromosomes arrive at the poles, and the cell may undergo cytokinesis to form two daughter cells. These cells are haploid, but each chromosome still consists of two sister chromatids. This is followed by a short resting phase called interkinesis, without further DNA replication.
Describe meiosis II
This phase is similar to mitosis and is known as equational division, where the sister chromatids are separated.
- Prophase II: Chromosomes condense, and a new spindle apparatus forms.
- Metaphase II: Chromosomes line up at the metaphase plate, with each sister chromatid attached to microtubules from opposite poles.
- Anaphase II: Sister chromatids are pulled apart and move toward opposite poles.
- Telophase II: Chromatids reach the poles, and nuclear membranes reform around each set of chromosomes. Cytokinesis follows, resulting in four haploid gametes.
Errors during meiosis, such as _____ (failure to separate homologs or sister chromatids), can result in _____ like trisomy (e.g., Down syndrome).
nondisjunction
aneuploidies
DESCRIBE INDEPENDENT ASSORTMENT
During metaphase I, homologous chromosome pairs align randomly at the metaphase plate. Each pair can orient itself independently of the other pairs, meaning that either the maternal or paternal homolog can be pulled to a given pole during anaphase I. This random distribution of homologs generates different combinations of chromosomes in the gametes, following Mendel’s Law of Independent Assortment.
Describe crossing over (recombination)
In prophase I, homologous chromosomes exchange genetic material at points of crossing-over (chiasmata). This process shuffles alleles between homologous chromosomes, creating new combinations of genes. As a result, sister chromatids are no longer identical, and when chromatids are separated during anaphase II, each gamete will carry a unique set of genetic material. This genetic recombination contributes significantly to genetic diversity in offspring.
What are the key events of Meiosis I
- Prophase I: Chromosomes condense, homologs undergo synapsis and form tetrads, and crossing-over occurs at chiasmata.
- Metaphase I: Homologous chromosome pairs align at the metaphase plate.
- Anaphase I: Homologs are pulled to opposite poles, reducing the chromosome number by half.
- Telophase I: Two haploid cells are formed, though each chromosome still consists of two sister chromatids
Describe the key events of meiosis II (equational division)
- Prophase II: Chromosomes condense again in both daughter cells.
- Metaphase II: Chromosomes align at the metaphase plate, with sister chromatids facing opposite poles.
- Anaphase II: Sister chromatids separate and move to opposite poles.
- Telophase II: Four haploid gametes are formed after cytokinesis.
Describe the law of segregation
This law states that during the formation of gametes, the two alleles for each gene segregate so that each gamete carries only one allele for each gene. During anaphase I of meiosis, homologous chromosomes are separated, ensuring that each gamete receives only one copy of each chromosome (and hence, one allele of each gene).
In humans, sex is typically determined by the presence of X and Y chromosomes. The sex-determining region of the Y chromosome, or ____ gene, plays a crucial role in male development.
SRY
Describe XY
A typical male has one X chromosome and one Y chromosome. The SRY gene on the Y chromosome initiates the development of male characteristics, making the individual male.
Describe XX
A typical female has two X chromosomes. Since there is no Y chromosome (and thus no SRY gene), the individual develops as female.
Describe XXY (Klinefelter syndrome)
This is a male with an extra X chromosome. Despite the extra X, the presence of the Y chromosome with the SRY gene leads to male development, although affected individuals may have some atypical male characteristics and may be infertile.
Describe X0 (Turner syndrome)
Individuals with only one X chromosome (and no Y chromosome) develop as females. They often have underdeveloped female characteristics and may have health issues related to missing one X chromosome.
Describe XYY
This karyotype results in a male because of the presence of a Y chromosome with the SRY gene. These individuals tend to be phenotypically normal males.
Describe XXX
This results in a female, as there is no Y chromosome. These individuals generally develop as females, though they may be taller than average and have some mild learning difficulties.
Describe sex role reversal in humans in XY females
In some cases, individuals with an XY karyotype develop as females. This can happen if the SRY gene is deleted or non-functional, meaning the signals for male development are absent. Without the action of SRY, the gonads develop into ovaries, leading to a female phenotype despite the presence of a Y chromosome.
Describe sex role reversals
In this case, individuals with two X chromosomes develop as males. This occurs if the SRY gene is translocated onto one of the X chromosomes during meiotic recombination. The presence of SRY triggers male development despite the absence of a Y chromosome.
Sex reversal can also result from mutations in other genes involved in sex determination, such as SOX9 or WNT4, that are _____ of SRY in the sex-determination pathway
downstream
Ignore
Describe the means of sex determination in Birds (ZZ/ZW System):
Birds have a ZZ/ZW system, where males are the homogametic sex (ZZ) and females are the heterogametic sex (ZW). Here, it is the female that determines the sex of the offspring. Unlike mammals, there is no SRY gene involved in determining sex, and the molecular basis of sex determination in birds involves different genes such as DMRT1.
Describe the means of sex determination in Humans and Most Mammals (XX/XY System):
Humans follow the XX/XY system of sex determination. Males are the heterogametic sex, producing sperm with either an X or Y chromosome, while females are the homogametic sex, producing eggs with only X chromosomes. The presence of a Y chromosome, specifically the SRY gene, directs male development. The pseudoautosomal regions (PARs) of the X and Y chromosomes allow them to pair and recombine during meiosis.
Describe the means of sex determination in Insects (e.g., Drosophila)
In fruit flies (Drosophila), sex is determined by the ratio of X chromosomes to autosomes rather than the presence of a Y chromosome. A ratio of 1:1 (two X chromosomes for two sets of autosomes) results in a female, while a ratio of 0.5 (one X chromosome for two sets of autosomes) results in a male. The Y chromosome in flies is not necessary for male development, but it is required for fertility.
Describe the means of sex determination in Reptiles (Temperature-Dependent Sex Determination):
In many reptiles, such as turtles and crocodiles, sex is determined by the incubation temperature of the eggs rather than chromosomes. Higher temperatures may result in one sex (e.g., females), while lower temperatures result in the other sex. There is no involvement of X or Y chromosomes in these cases.
Describe the means of sex determination in Bees and Ants (Haplodiploid System)
In hymenopterans like bees and ants, sex is determined by ploidy. Females are diploid (having two sets of chromosomes), while males are haploid (having one set of chromosomes). Males develop from unfertilised eggs, while females develop from fertilised eggs. There are no X or Y chromosomes involved in this system.
_______ Sex: The sex that produces two different types of gametes concerning sex chromosomes (e.g., XY males in humans).
________ Sex: The sex that produces identical gametes concerning sex chromosomes (e.g., XX females in humans).
Heterogametic
Homogametic
Describe oogenesis
Location: Occurs in the ovaries.
Cells: Starts with oogonia (female germ cells).
Timeline:
* Begins in fetal development.
* Arrested at prophase I until puberty.
* One secondary oocyte produced per menstrual cycle.
Meiosis:
* Resumes at ovulation, completes Meiosis II only if fertilised.
Products:
* One ovum and 2–3 polar bodies (degenerate byproducts).
Cytoplasm:
* Unequal division—most goes to the secondary oocyte.
Lifespan:
* ~400,000 oocytes at birth; ~400 ovulate during the reproductive years.
Long process: Starts in utero, finishes only upon fertilisation.
Oogenesis produces one ____ and up to three ____ ____, while spermatogenesis results in ____ sperm cells.
ovum
polar bodies
four
Oogenesis starts in the ____ stage and arrests at various stages until fertilisation, while spermatogenesis starts at ____ and continues throughout life.
fetal
puberty
Describe spermatogenesis
- Location: Occurs in the testes.
- Cells: Begins with spermatogonia (male germ cells).
- Timeline:
- Starts at puberty, continues throughout life.
- Primary spermatocytes enter meiosis at puberty and develop continuously.
- Meiosis:
- Completed before sperm maturation, producing mature sperm regularly.
- Products:
- Four sperm cells per spermatogonium.
- No polar bodies.
- Cytoplasm:
- Equal division—results in four spermatids of similar size.
- Lifespan:
- Virtually infinite—hundreds of millions of sperm produced daily.
- Process: Takes 64–72 days to complete and occurs continuously throughout a male’s reproductive life
Oogenesis involves ____ cytoplasmic division, whereas spermatogenesis has ___ divisions.
unequal
equal
Describe Oogonia (2n):
Oogonia are the female germ cells in the ovaries that give rise to oocytes, or egg cells. They are diploid cells (2n) formed during fetal development and undergo mitosis to increase in number. As development progresses, some oogonia differentiate into primary oocytes, which enter prophase I of meiosis and become arrested in this stage until puberty. Oogonia play a crucial role in female reproductive development, as they are the precursors to the eggs needed for fertilization.
Describe Primary Oocyte (2n):
A primary oocyte is a diploid (2n) cell that forms from oogonia during female fetal development. It is characterized by being arrested in prophase I of meiosis, where it remains until puberty. Each primary oocyte is surrounded by a layer of granulosa cells, forming a structure known as a primordial follicle. Upon maturation during the menstrual cycle, primary oocytes complete meiosis I, resulting in a secondary oocyte and a polar body, which is typically discarded. Primary oocytes are essential for female fertility, as they ultimately develop into the eggs that can be fertilized by sperm.
Describe Secondary Oocyte (n):
A secondary oocyte is a haploid (n) cell that results from the completion of meiosis I by a primary oocyte. During this process, the primary oocyte divides asymmetrically, producing one large secondary oocyte and a smaller polar body, which usually degenerates. The secondary oocyte enters meiosis II but is arrested in metaphase II until fertilization occurs. If fertilization takes place, the secondary oocyte will complete meiosis II, resulting in a mature ovum and another polar body. The secondary oocyte is the cell that is released during ovulation and is capable of being fertilized by sperm.
Describe the Ovum (n)
- Autosome complement: 22 autosomes
Sex chromosome complement: X - Stage: The mature egg cell, formed after fertilisation, following the completion of meiosis II. The second polar body is also formed during this stage.
Describe Spermatogonia (2n):
- Autosome complement: 44 autosomes
- Sex chromosome complement: XY
- Stage: Diploid germ cells in the testes, undergo mitosis to produce primary spermatocytes.
- Primary Spermatocyte (2n):
- Autosome complement: 44 autosomes
- Sex chromosome complement: XY
- Stage: Undergoes meiosis I to produce secondary spermatocytes.
Describe Secondary Spermatocyte (n):
- Autosome complement: 22 autosomes
- Sex chromosome complement: X or Y
- Stage: Formed after the first meiotic division. Each secondary spermatocyte contains either an X or a Y chromosome, along with 22 autosomes. Undergoes meiosis II.
Describe Spermatids (n):
- Autosome complement: 22 autosomes
- Sex chromosome complement: X or Y
- Stage: Formed after the second meiotic division. These haploid cells develop into mature sperm.
Describe sperm
- Autosome complement: 22 autosomes
- Sex chromosome complement: X or Y
- Stage: The final, mature male gametes. Each sperm will carry either an X or a Y chromosome, determining the sex of the future offspring.
A trait can be inferred to be _______ (most commonly X-linked) if it follows a particular inheritance pattern observed across generations. These patterns can be analysed by examining the results of genetic crosses.
sex-linked
____ ____ ____work with fruit flies (Drosophila) provided key evidence for sex-linked traits. He observed a _____ _____ pattern, where X-linked traits alternate between sexes in subsequent generations.
Thomas Hunt Morgan’s
crisscross inheritance
What happened in thomas hunt morgan’s experiments
For example, when Morgan crossed white-eyed male flies (a recessive X-linked trait) with red-eyed females (dominant), he found that all female offspring had red eyes and all male offspring had white eyes in the F1 generation. In the F2 generation, females could carry the recessive allele and pass it to their sons, which created the characteristic crisscross inheritance
Describe crisscross inheritance
Crisscross inheritance occurs when a trait is passed from an affected male to his daughters, who are carriers (if heterozygous), and from those daughters to their sons. This pattern suggests X-linked inheritance.
Describe hemizygous males
Hemizygous males are individuals, typically referring to male organisms with one X chromosome and one Y chromosome (such as in humans and many other species). Because they have only one copy of the X chromosome, any allele present on that chromosome will be expressed, whether it is dominant or recessive. This is in contrast to females, who have two X chromosomes and can be homozygous or heterozygous for X-linked traits. Hemizygosity makes males particularly susceptible to expressing X-linked recessive traits, such as color blindness or hemophilia, since they lack a second X chromosome that could potentially mask the recessive allele.
a trait can be inferred to be sex-linked if?
- The trait predominantly affects males in the family.
- Male-to-male transmission does not occur (since males pass their Y chromosome, not the X chromosome, to their sons).
- There is a crisscross inheritance pattern with daughters acting as carriers, passing the trait to their sons.
Describe nothing
Describe how Morgans work provided a link between chromosomes and genes
Morgan’s work provided the first experimental evidence that genes reside on chromosomes, by showing that the inheritance of certain traits is linked to the inheritance of specific chromosomes. His experiments with sex-linked traits demonstrated that chromosomes are not just carriers of genetic material but also the location of individual genes.
Later studies on ______ in Drosophila showed that incorrect segregation of sex chromosomes (i.e., failure of chromosomes to separate properly during meiosis) leads to abnormal chromosomal compositions and distinct phenotypes. This provided further evidence linking genes to chromosomes
nondisjunction
For example, when nondisjunction occurred, offspring with abnormal numbers of X or Y chromosomes (e.g., XXY or X0) displayed phenotypes that correlated with the inheritance of extra or missing sex chromosomes, proving that chromosomes carry the genetic instructions for trait expression.
Thus, Morgan’s discovery of ___ ____ and the demonstration of _____ offered definitive proof that genes are located on chromosomes, a fundamental concept in genetics
sex linkage
nondisjunction
Describe how x-linked recessive traits are determined from pedigree analysis
- Males (hemizygous) express any allele on their single X chromosome.
- Affected males inherit the trait from carrier mothers (heterozygous).
- Females express the trait only with two recessive alleles (one from each parent).
- Affected males cannot pass the X chromosome to sons (fathers pass Y).
- Affected males have unaffected daughters (carriers) who can pass the trait to their sons.
What steps would you use in pedigree analysis of an X-linked recessive trait(s)
- Identify Affected Individuals: Start by marking all affected individuals (usually males) in the pedigree.
- Trace Maternal Lines: Look for affected males to trace back to their mothers, identifying carrier mothers.
- Check Female Progeny: Examine daughters of affected males; if they are unaffected, they may still be carriers.
- Evaluate Sons of Carriers: Assess the next generation by tracking sons of carrier females for the expression of the trait.
- Look for Patterns: Identify crisscross patterns of inheritance and absence of male-to-male transmission.
Describe how x-linked dominant traits are determined from pedigree analysis
- Both males and females can be affected, but females may show milder symptoms because of X-chromosome inactivation (more on this below).
- Affected males pass the trait to all of their daughters (since daughters inherit their father’s X chromosome) but none of their sons.
- Affected females pass the trait to about half of their children, regardless of sex.
- Affected females may show more variability in the severity of symptoms due to random inactivation of one X chromosome.
What steps would you use in pedigree analysis of an X-linked dominant trait(s)
Identify Affected Individuals:
- Mark all individuals exhibiting the trait (typically shaded shapes in the pedigree).
Determine Gender of Affected
Individuals:
- Note that both males and females can be affected, but affected males will have affected daughters.
Look for Male-to-Female Transmission:
- Affected males will pass the trait to all of their daughters (but not to their sons).
Assess Female Progeny:
- Affected females can pass the trait to about 50% of their offspring (both sons and daughters). Look for affected children of affected mothers.
Check for Unaffected Parents:
- Unaffected individuals who have affected children may indicate that the trait is inherited from a carrier mother, especially if the mother is not affected herself.
Examine Generational Patterns:
- Observe that the trait typically appears in every generation due to the dominant nature of the allele. There should be no skipping of generations.
Consider Incomplete Penetrance:
- Be aware that some individuals may carry the allele but not express the trait visibly.
Look for Carrier Females:
- If a female is affected and has an unaffected child, the child may be a carrier.
Analyze for Crisscross Patterns:
- Unlike X-linked recessive traits, X-linked dominant traits do not exhibit the same crisscross inheritance pattern. Instead, affected males will directly affect their daughters.
Describe autosomal recessive traits
- Both males and females are equally likely to be affected.
- Affected individuals typically have unaffected parents who are carriers (heterozygous).
- The trait may appear to “skip” generations as it reappears when two carriers have children.
Describe autosomal dominant traits
- Both males and females are equally likely to be affected.
- The trait usually appears in every generation, with affected individuals passing it on to about half of their offspring, regardless of sex.
- Male-to-male transmission can occur, which helps distinguish autosomal dominant traits from X-linked traits.
Describe y-linked traits
- Only males are affected, since only males inherit the Y chromosome.
- The trait is passed from an affected father to all of his sons.
- There is no female involvement in the transmission of Y-linked traits.
How can Human Cells Compensate for the X-Linked Gene Dosage Difference in XX and XY Nuclei
- Females have two X chromosomes (XX), while males have only one (XY). If both X chromosomes in females were fully active, it would result in dosage imbalance—females would produce twice as many X-linked gene products as males. To prevent this imbalance, dosage compensation mechanisms exist to equalise gene expression between the sexes.
- A main mechanism of this is x-inactivation
Describe x-inactivation
- In females, one of the two X chromosomes in each cell is randomly inactivated early in development. This process, called X-chromosome inactivation, ensures that females, like males, have only one functional X chromosome per cell.
- The inactivated X chromosome becomes a dense, inactive structure known as a Barr body. Barr bodies are visible in the nucleus and can be used to determine the number of inactivated X chromosomes.
- X-inactivation occurs randomly in each cell, so some cells in a female express genes from one X chromosome, and other cells express genes from the other X chromosome. This creates a mosaic pattern of gene expression in females.
Describe the mechanisms of x-chromosome inactivation
- The process is initiated in a region of the X chromosome known as the X-inactivation centre (XIC).
- A key gene in this region, XIST, produces an RNA transcript that coats the X chromosome to be inactivated, leading to its silencing.
- Once inactivated, the X chromosome remains inactive in all descendant cells, creating a clone of cells with the same X chromosome inactivated. This results in tissue mosaicism for X-linked genes.
Describe sex-limited traits
traits that are expressed in only one sex, despite being determined by autosomal or X-linked genes. For example, milk production in mammals is a sex-limited trait that is only expressed in females, even though both sexes carry the genes for it.
Describe sex-influenced traits
Sex-influenced traits are autosomal traits where the expression of the trait differs between males and females due to hormonal differences. For instance, male-pattern baldness is a sex-influenced trait where the dominant allele results in baldness in males but has little effect in females unless they are homozygous for the trait.
What are the four groups of genital precursor cells?
- Bipotential gonads (genital ridge)
- Mullerian ducts
- Wolffian ducts
- External genitalia (cloaca)
Describe bipotential gonads
- Initially, all embryos have bipotential gonads, meaning these structures can develop into either testes or ovaries depending on genetic and hormonal cues.
- In XY individuals, the SRY gene (Sex-determining Region Y) on the Y chromosome triggers the gonads to differentiate into testes around week 6-7 of development. In the absence of the SRY gene (in XX individuals), the gonads develop into ovaries.
Describe Mullerian ducts
- These ducts are precursors to the female internal reproductive organs, including the fallopian tubes, uterus, and upper part of the vagina.
- In XY individuals, the testes produce Anti-Müllerian Factor (AMF), a hormone that causes the Müllerian ducts to regress, preventing the development of female reproductive structures.
- In XX individuals, without AMF, the Müllerian ducts develop into the female internal reproductive organs.
Describe wolffian ducts
- These ducts are the precursors to the male internal reproductive organs, including the epididymis, vas deferens, and seminal vesicles.
- In XY individuals, testosterone produced by the testes stimulates the development of the Wolffian ducts into these male reproductive structures.
- In XX individuals, the lack of testosterone causes the Wolffian ducts to degenerate.
Describe external genitalia
- The external genitalia initially develop from a common structure called the cloaca, which later differentiates into either male or female genitalia.
- In XY individuals, DHT (dihydrotestosterone), a more potent form of testosterone, is critical for the masculinisation of the external genitalia, forming the penis and scrotum.
- In XX individuals, the absence of DHT leads to the development of female external genitalia, including the clitoris and labia.
Describe How Mutant Alleles of Genes in XY Individuals Can Lead to Intersexuality
In XY individuals, mutations in genes involved in sex differentiation can lead to conditions where individuals have both male and female characteristics or ambiguous genitalia. This can result in intersexuality, where the typical pathways for male development are disrupted.
Describe Complete Androgen Insensitivity Syndrome (CAIS):
- Mutations in the androgen receptor (AR) gene result in cells’ inability to respond to androgens like testosterone and DHT.
- XY individuals with a functional SRY gene produce normal testosterone and DHT levels, but their bodies cannot utilize these hormones.
- The Wolffian ducts do not develop into male internal reproductive organs, and external genitalia follow a female pathway, leading to typically female external genitalia.
- However, they lack a uterus and fallopian tubes due to Anti-Müllerian Factor (AMF) causing Müllerian duct regression.
- Individuals with CAIS are genetically male (XY) but phenotypically female, often remaining undiagnosed until puberty when menstruation fails to occur.
Describe 5-Alpha-Reductase Deficiency:
- In this condition, there is a mutation in the gene encoding the enzyme 5-alpha-reductase, which converts testosterone into DHT.
- DHT is necessary for the masculinisation of the external genitalia in XY individuals. Without it, individuals may be born with ambiguous genitalia or female-appearing external genitalia.
- However, at puberty, when testosterone levels rise, some masculinisation may occur, leading to the development of more male-typical characteristics like increased muscle mass, voice deepening, and growth of facial hair.
- These individuals have testes and may develop male internal reproductive organs, but the lack of DHT during early development can result in ambiguous genitalia at birth.
Describe mutations in the SRY gene and how that impacts genital morphology/development
- If there is a mutation in the SRY gene, the bipotential gonads may fail to develop into testes, despite the presence of an XY karyotype.
- This can lead to an individual developing female-typical internal and external genitalia, even though they are genetically male.
- This form of sex reversal occurs because, without SRY function, the typical male differentiation pathway is not initiated.
Describe Androgen Receptor (AR) Mutations and Partial Androgen Insensitivity Syndrome (PAIS):
- In PAIS, mutations in the androgen receptor lead to partial responsiveness to androgens.
- The degree of intersexuality can vary depending on how much androgen signalling is disrupted. Individuals may have ambiguous genitalia or predominantly male or female genitalia with atypical features.
- Testosterone may stimulate partial development of the male reproductive structures, but without full androgen signalling, the individual may not fully masculinise.
Describe gene-linkage
- Refers to the situation where two or more genes are located close together on the same chromosome, leading to their tendency to be inherited together during meiosis. This happens because genes that are physically close to each other are less likely to be separated by the process of crossing over during recombination. Therefore, linked genes do not follow Mendel’s law of independent assortment and tend to be passed down as a set
_____ gametes are those that carry the same allele combinations as found in the parents, reflecting no crossing over between the linked genes. In contrast, _____ gametes arise when crossing over occurs between the gene loci, resulting in new allele combinations that are different from the original parental alleles. _____ indicate genetic recombination, which shuffles the genetic information between homologous chromosomes.
Parental
recombinant
Recombinants
In an unlinked scenario, independent assortment results in the classic __:__:__:__ ratio in the F2 generation. However, if the genes are linked, this ratio is distorted. You’ll find more progeny with _____-type phenotypes and fewer with _____ phenotypes. The deviation from the 9:3:3:1 ratio suggests that the genes are not assorting independently, thus indicating linkage.
9:3:3:1
parental
recombinant
Describe testcrosses
- Crossing a heterozygote with a homozygous recessive individual
- Useful for detecting linkage
- If the genes are unlinked, the testcross would produce equal numbers of the four possible phenotypes (1:1:1:1). However, if the genes are linked, the result will be skewed, with more offspring showing parental-type combinations and fewer showing recombinant combinations. The proportion of recombinant offspring reflects the distance between the genes, with fewer recombinants indicating tighter linkage.
Recombination occurs during ____ I of meiosis through a process called crossing over.
prophase
Describe crossing over
- A genetic process that occurs during prophase I of meiosis, where homologous chromosomes exchange segments of DNA. This reciprocal exchange of genetic material results in new allele combinations, contributing to genetic diversity in offspring. The sites of crossing over are called chiasmata, and the process plays a crucial role in the proper segregation of chromosomes during meiosis.
Describe chiasmata
- Crucial for the proper segregation of homologous chromosomes during meiosis I. They are the physical manifestations of crossing over and help hold homologous chromosomes together after they synapse (pair). During anaphase I, the tension at the chiasmata ensures that the homologous chromosomes are pulled apart correctly towards opposite poles, reducing the risk of chromosomal abnormalities such as nondisjunction.
Describe the recombination frequency (RF)
The recombination frequency (RF) is the proportion of recombinant offspring compared to the total offspring in a genetic cross, and it reflects the physical distance between two loci on a chromosome. The further apart two loci are, the more likely a crossover event will occur between them, resulting in a higher RF
Describe map distance
Map distance is measured in centiMorgans (cM), where 1 cM equals a 1% recombination frequency. A higher RF indicates that the loci are further apart, while a lower RF suggests they are closely linked.
What is the value cap on recombination frequency?
The maximum recombination frequency (RF) between any two genes is limited to 50% due to the nature of gene linkage and independent assortment. When genes are unlinked (located on different chromosomes or far apart on the same chromosome), they assort independently, leading to an RF close to 50%. However, for linked genes, even with maximal recombination events (double crossover events, DCO), the proportion of recombinant gametes cannot exceed 50%, as there will always be a significant number of non-crossover (NCO) or single crossover (SCO) gametes produced. This limitation ensures that any genetic mapping using RF remains valid
Describe two-point testcrosses
A two-point testcross involves crossing an individual heterozygous for two genes (e.g., AaBb) with a homozygous recessive individual (aabb). The offspring can display either parental phenotypes (matching the parents) or recombinant phenotypes (new combinations resulting from crossing over).
How do you interpret two-point testcrosses?
The proportion of recombinant offspring reveals the recombination frequency (RF) between the two loci, which is a measure of how frequently crossing over occurs between them. For example, if 10% of the offspring are recombinants, the two genes are 10 map units (centiMorgans) apart. This distance allows us to position genes relative to each other on a chromosome. The smaller the RF, the closer the genes are; the larger the RF, the further apart they are.
Explain How a Testcross Can Provide Evidence for or Against Linkage
A testcross involves crossing an individual exhibiting a dominant phenotype (whose genotype is unknown) with a homozygous recessive individual. If the unknown parent is homozygous for the dominant alleles of two linked genes, the offspring will show predominantly parental phenotypes, with few or no recombinants. Conversely, if the unknown parent is heterozygous and the genes are unlinked, we would expect a 1:1:1:1 ratio of phenotypes among the progeny, indicative of independent assortment. Therefore, the presence of a preponderance of parental classes of gametes in the testcross progeny supports the hypothesis of linkage, while a more even distribution of phenotypes suggests the genes are assorting independently.
What do synaptonemal complexes do?
Synaptonemal complexes are protein structures that form between homologous chromosomes during prophase I of meiosis, facilitating their pairing and alignment. These complexes help stabilize the connection between homologs, ensuring accurate recombination and crossing over at chiasmata. By promoting close interaction between chromosomes, synaptonemal complexes play a crucial role in the proper segregation of chromosomes during meiosis
Describe reciprocal exchanges
Reciprocal exchanges refer to the process during meiosis where segments of DNA are exchanged between non-sister chromatids of homologous chromosomes. This exchange occurs during crossing over, leading to new combinations of alleles on the chromosomes. As a result, reciprocal exchanges enhance genetic diversity in gametes, contributing to the variation seen in offspring
Differentiate between non-crossover (NCO), crossover (SCO) and double crossover (DCO) configurations
- Non-crossover (NCO) configurations refer to gametes that retain the original combinations of alleles from the parents, showing no recombination.
- Single crossover (SCO) configurations result from one crossover event between homologous chromosomes, producing gametes with new allele combinations for the genes involved.
- Double crossover (DCO) configurations occur when two separate crossover events happen between the same homologous chromosomes, leading to a more complex combination of alleles, which may restore some parental combinations while introducing new ones for the intervening genes.
Why do two-point crosses?
- By performing multiple two-point crosses, the RF between pairs of loci can be compared. However, this approach has limitations, such as not accounting for double crossovers (DCOs), which can skew the results and underestimate the distance between genes.
Describe three-point testcrosses
Three-point testcrosses involve a parent heterozygous for three loci crossed with a homozygous recessive individual. This approach provides more accurate mapping because it allows for the identification of the gene in the middle. To analyze the data, researchers count the offspring phenotypes and calculate RFs for the pairs of loci. The formula for calculating RF for three loci involves considering single crossovers (SCOs) and double crossovers (DCOs).
Why do three-point crosses
By determining which phenotypic classes are less frequent, one can identify the gene that is in the middle. The correction for double crossovers is essential, as these events can confound the estimated distances. This refinement results in a more accurate genetic map and often reveals nonuniform crossover frequencies among different gene pairs.
Describe nonuniform crossover frequencies
Nonuniform crossover frequencies refer to the variation in the likelihood of crossover events occurring at different regions along a chromosome during meiosis. Certain areas, often influenced by chromatin structure and gene density, may experience higher or lower rates of recombination. This variation can lead to uneven genetic mapping distances and affect the accuracy of genetic maps, as some loci may be more likely to undergo crossover than others.
Explain How a Genetic Map (in Map Units) Is Related to Actual Physical Distance (in Base Pairs of DNA)
Genetic maps, measured in map units (m.u.) or centimorgans (cM), represent relative distances based on recombination frequencies. While 1% RF corresponds to 1 m.u., the relationship between genetic map distances and actual physical distances (in base pairs) is not linear. This discrepancy arises from nonuniform crossover frequencies, where some regions of a chromosome may have higher or lower rates of recombination than others. Thus, while genetic maps provide a useful framework for understanding gene order and distance, they do not directly translate into precise base pair measurements. However, genetic maps can be correlated with physical reality using techniques like genome sequencing.
Describe the Relationship Between Linkage Groups and Chromosomes
Each chromosome represents a linkage group, and the genes within that group are subject to recombination events, which can alter their inheritance patterns. The organization of linkage groups helps researchers identify chromosomal locations of genes and understand genetic relationships in multiple-factor crosses, which can further establish linkage groups.
Describe chromosomal interference
refers to the phenomenon where the occurrence of one crossover event affects the likelihood of another crossover occurring nearby. It can complicate the interpretation of genetic maps.
Describe the coefficient of coincidence (C)
- A measure used to assess the extent of interference. It is calculated using the formula:
Describe second-division segregation patterns
Second-division segregation in ascospores occurs during Meiosis II when recombination (crossing-over) happens between a gene and its centromere. This recombination causes homologous alleles to remain together through Meiosis I, and only during Meiosis II do the sister chromatids, now carrying different alleles, separate.
This results in a 2:2:2:2 or 4:4 arrangement of alleles in the ascus, depending on the pattern of recombination. For example, a gene with alleles A and a may produce a tetrad with alternating ascospores (e.g., A, a, A, a) or two pairs of ascospores (e.g., AA, aa). The key feature of second-division segregation is that recombination between the gene and centromere leads to the separation of alleles later, during the second meiotic division
Describe the kind of mutations that occur in germ-line cells
Mutations in germ-line cells (cells giving rise to sperm or eggs) can be passed to offspring, allowing these mutations to become heritable. Germ-line mutations can also include reverse mutations (reversions), restoring function to a mutated gene, though these are rare.
Explain the fluctuation tests
The fluctuation test, developed by Luria and Delbrück, provided evidence that spontaneous mutations arise from random events rather than being directed by environmental pressures. They cultured bacteria in multiple small samples and observed significant variation in mutation rates across colonies, supporting the idea that mutations occur at random, independent of external selective pressures.
Describe pyrimidines and purines
Purines: A & G
Pyrimidines: C & T
Describe replica plating
Replica plating, an experiment developed by Joshua and Esther Lederberg, also demonstrated the randomness of mutations. In this technique, bacterial colonies were transferred from one plate to another with a selective agent (e.g., antibiotic). Colonies that survived the antibiotic were shown to already contain mutations before exposure, indicating that spontaneous mutations are pre-existing, rather than induced by the environment.
Describe depurination
- A natural process that can produce mutations by damaging DNA
- This process involves the loss of a purine base (adenine or guanine) from the DNA, creating an apurinic site where the base is missing. If unrepaired, this can lead to a mutation during replication.
Describe deamination
- A natural process that can produce mutations by damaging DNA
- Deamination removes an amino group from cytosine, converting it to uracil. This alteration can cause a mispairing during DNA replication, resulting in mutations.
Describe Base tautomerization
- Base tautomerization occurs when DNA bases temporarily shift into alternative chemical structures, called tautomers, which differ slightly in the positioning of hydrogen atoms and bonding patterns. These rare tautomers can alter base pairing properties, leading to unusual pairings during DNA replication.
- For example, a tautomeric form of adenine may pair with cytosine instead of its usual partner, thymine. This mispairing introduces an error in the DNA sequence, creating a base mispair that, if not corrected, can propagate through future rounds of replication.
- When the DNA is replicated, the presence of these tautomers can result in permanent mutations, as the mispair is incorporated into one of the new DNA strands. This spontaneous change, although rare, is a source of genetic variation and can contribute to mutation rates in cells.
While DNA polymerase has a ______ function to correct most errors, some mistakes persist, allowing ______ and ______ mutations to be passed on.
proofreading
Base tautomerization
Slipped mispairing
Two types of mutagens:
- ________: These are chemicals that mimic normal DNA bases and can be incorporated into DNA during replication, leading to mispairing and mutations.
- _______: These chemicals insert themselves between DNA base pairs, distorting the double helix and often causing insertions or deletions.
- ______ agents - add an –OH group, altering base structure
- ______ agents - add ethyl or methyl groups
- ______ agents - remove amine (-NH2) groups
Base analogs
Intercalators
Hydroxylating
Alkylating
Deaminating
Describe the Ames test
The Ames test is a widely used assay to identify carcinogens—substances that could cause cancer—by testing their ability to induce mutations in bacteria. In this test, bacteria with a specific mutation (often in the gene required for histidine synthesis) are exposed to a suspected mutagen. These bacteria are then grown on petri plates without histidine, where only revertants (cells that have undergone a reverse mutation restoring histidine synthesis) can grow.
If the mutagen induces a high rate of reversion mutations, more colonies will appear, suggesting that the substance is likely a carcinogen. The Ames test relies on the correlation between mutagenicity and carcinogenicity, providing a preliminary screening for substances that may damage DNA.
Describe base excision repair
Damaged bases are removed and replaced by enzymes like DNA glycosylase, which identifies and removes the damaged base, leaving an AP (apurinic/apyrimidinic) site. An enzyme called AP endonuclease then makes a nick in the DNA backbone at the AP site, allowing DNA polymerase and ligase to fill in the gap.
Describe nucleotide excision repair
Nucleotide excision repair (NER): The UvrA, UvrB, and UvrC proteins work together to detect and remove large distortions in the DNA, such as those caused by UV light (e.g., thymine dimers).
Describe homologous recombination
This method of repair uses a homologous sequence as a template, resulting in accurate repair without losing genetic information. It occurs during certain cell cycle phases (S and G2) when a sister chromatid is available. HR is precise and is used for genome editing because it maintains sequence fidelity.
Describe nonhomologous end-joining
This repair mechanism directly joins the broken ends of DNA without needing a homologous template, which can lead to errors, such as small insertions or deletions. It is active throughout the cell cycle, making it a faster but less accurate method than HR.
Explain how methyl-directed mismatch repair can distinguish which strand to repair when replication errors occur.
In methyl-directed mismatch repair (MMR), the cell differentiates between the newly synthesized strand and the template strand based on methylation patterns. Adenine methylase methylates adenine residues in the parent DNA strand, while the new strand remains unmethylated for a short period after replication. The MutS protein detects mismatches, MutL recruits MutH, and MutH cleaves the unmethylated (new) strand at the site of the error. DNA polymerase then synthesizes the correct sequence, and DNA ligase seals the strand. This method allows for accurate correction by identifying the newly made strand as the one to repair.
Describe sloppy DNA polymerases
- Sloppy DNA polymerases are specialised enzymes used by cells in error-prone repair systems when DNA damage is too severe for high-fidelity repair. In the SOS system, these polymerases can bypass DNA lesions that would normally block replication, allowing the cell to continue dividing despite extensive damage.
- However, because sloppy DNA polymerases lack proofreading abilities, they often introduce mutations during repair, making this system a risky but necessary last resort. This approach allows the cell to survive DNA damage that might otherwise be lethal, albeit with an increased chance of accumulating mutations.
Deletions remove DNA from the genome, leading to missing genes or regulatory regions. In ________, carrying the deletion on only one chromosome, there can still be detrimental effects due to _______—when a single gene copy does not produce enough protein for normal function.
heterozygotes
haploinsufficiency
Describe deletion loops
- Deletion loops occur during meiosis when homologous chromosomes align for recombination, and one chromosome contains a deletion—a missing segment of DNA. The loop forms as the normal chromosome attempts to pair with its homologous counterpart, creating a looped-out region where the deleted segment cannot align.
- These loops can lead to improper chromosome segregation, resulting in gametes that lack certain genetic information. In heterozygous individuals, where one chromosome has a deletion and the other does not, deletion loops are important for understanding the impact of deletions on gene dosage and potential phenotypic consequences in offspring.
Explain how researchers use deletions to locate genes.
Researchers use deletions to locate genes by uncovering recessive mutant alleles that are otherwise masked by a dominant allele on the other chromosome. When a deletion overlaps a region containing a recessive mutation, the mutation’s phenotype becomes visible. This uncovering method allows researchers to map gene locations based on the phenotypes observed in deletion heterozygotes with known deletions. Using deletions to locate genes is an effective approach to narrow down gene regions in genetic mapping.
Describe tandem duplications
Tandem duplications involve the replication of a segment of DNA adjacent to the original sequence, resulting in two or more copies located next to each other on the same chromosome. This type of duplication often occurs in regions with repetitive sequences and can increase gene dosage, potentially leading to enhanced expression of the duplicated genes.
Describe nontandem duplications
Nontandem duplications occur when a segment of DNA is duplicated and inserted into a different location within the genome, which can be on the same chromosome or a different chromosome. This type of duplication may disrupt normal gene function or regulation by placing the duplicated gene under the control of different regulatory elements, potentially leading to unique phenotypic effects.
Duplications increase the risk of ____ ___-___ during meiosis, where misaligned duplications may result in further copy number changes. This process is observed in ___ and double-___ mutant types in Drosophila, where duplications cause abnormal eye morphology due to gene dosage effects.
unequal crossing-over
Bar
Bar
Describe Bar mutant types
Bar mutants are a type of genetic mutation in Drosophila (fruit flies) characterized by a narrow, elongated eye shape due to a deletion in the bar locus on the X chromosome. This mutation leads to a reduced number of photoreceptor cells in the compound eyes, resulting in the distinct bar-shaped appearance.
Describe double-Bar mutant types
Double-Bar mutants exhibit an even more pronounced phenotype than Bar mutants, often caused by a homozygous condition or the presence of two closely linked mutations that further enhance the reduction of photoreceptor cells. This results in an even narrower and more elongated eye shape, reflecting a greater degree of gene dosage effect on eye development.
Explain why breakpoints of inversions determine whether the inversions have phenotypic effects.
Inversions reorganise the DNA sequence of a chromosome by flipping a segment, but their phenotypic effects depend on whether the inversion’s breakpoints disrupt essential genes or regulatory regions. In pericentric inversions, which include the centromere, or paracentric inversions, which do not, genes interrupted by the breakpoints may lose function, leading to phenotypic consequences. However, if breakpoints do not disrupt genes, the inversion may have no phenotypic effect, though it can still affect fertility and crossover events.
Describe pericentric inversions
Pericentric inversions are chromosomal rearrangements where a segment of a chromosome is reversed, including the centromere. This inversion changes the order of genes, potentially altering their expression and function.
During meiosis, pericentric inversions can complicate homologous chromosome pairing, forming inversion loops. If crossovers occur within these loops, it can produce gametes with duplications and deletions, leading to genetic imbalances that may affect fertility or viability in offspring.
Describe paracentric inversions
Paracentric inversions are chromosomal rearrangements where a segment of a chromosome is reversed without including the centromere. This inversion alters the order of genes within the inverted region, potentially affecting gene expression and function.
During meiosis, paracentric inversions can complicate homologous chromosome pairing, leading to inversion loops. If crossovers occur within these loops, they can produce acentric fragments (without a centromere) and dicentric chromosomes (with two centromeres), causing genetic instability and potentially resulting in nonviable gametes or reduced fertility
Summarise the effects of translocations on crossing-over and fertility
In translocation heterozygotes, alternate segregation patterns can produce viable gametes, but adjacent-1 and adjacent-2 segregation patterns often result in semisterility due to unbalanced gametes. Robertsonian translocations (fusion of two acrocentric chromosomes) can also lead to reduced fertility due to uneven chromosome pairing. Translocations may therefore reduce fertility and cause pseudolinkage between genes on different chromosomes.
Describe balancer chromosomes
Balancer chromosomes are specially engineered chromosomes used in genetic research to maintain specific mutations in a population while preventing recombination. They typically contain multiple inversions and may carry dominant markers that help identify individuals with the desired mutations.
By suppressing recombination between the balancer chromosome and the homologous chromosome during meiosis, balancer chromosomes ensure that genetic variants are preserved across generations. This technique is particularly useful in Drosophila genetics for studying gene function and maintaining mutant alleles without loss through recombination.
Define transposable element.
A transposable element (TE), also known as a “jumping gene,” is a DNA sequence that can change its position within the genome through transposition. Transposable elements can move from one genomic location to another, and this movement may disrupt genes, alter gene expression, and contribute to genetic diversity and instability.
Explain how cells prevent excess transposable element mobilisation
Cells employ mechanisms to limit the activity of TEs and maintain genome stability. piRNAs (PIWI-interacting RNAs) play a crucial role in limiting transposable element movement by silencing TE expression, particularly in germ cells. By targeting TE RNA for degradation or preventing translation, piRNAs help suppress TE activity. This regulatory system helps protect the genome from excessive TE mobilisation, which could otherwise lead to harmful mutations and chromosomal instability.
Describe trinucleotide repeats
_______ repair guanine
modifications
_______ repairs thymine dimers
Alkyltransferases
Photolyase
Describe Muller’s lethal tagging test
- Muller’s lethal tagging test is a genetic technique used to locate mutations that cause lethality. By inducing random mutations in an organism’s genome, researchers can tag specific gene locations with lethal mutations, effectively “marking” those regions.
- This method allows for the identification of genes essential for survival, as only non-lethal mutations will pass through generations. Muller’s lethal tagging is particularly useful in mapping essential genes and studying genetic linkage.
Describe immunoglobulins
Where are immunoglobulins coded for in the human genome
Describe V(D)J recombination
Describe the structure of DNA
Describe the role and function of DNA polymerase
Directionality:
- DNAP reads the template strand 3’ to 5’ and synthesises the new strand 5’ to 3’ by adding nucleotides to the 3’-OH end.
Attachment Requirement:
- DNAP requires a free 3’-OH group to attach nucleotide bases, so it cannot start synthesis alone
Deoxynucleotide triphosphates (dNTPs):
- These include dATP, dCTP, dGTP, and dTTP, which provide both the building blocks and the energy required for synthesis. DNAP catalyses the incorporation of each dNTP by forming phosphodiester bonds between the 3’-OH of the growing strand and the incoming dNTP.
Primer:
- Typically a short RNA segment, the primer provides the necessary 3’-OH group that initiates DNA synthesis. DNAP extends the primer by adding nucleotides, forming the new DNA strand.
Describe Sanger Sequencing: Chain Termination Method
- Preparation:
A single-stranded DNA template, DNA polymerase, normal deoxynucleotide triphosphates (dNTPs), and specially modified dideoxynucleotide triphosphates (ddNTPs) are needed. The ddNTPs lack a 3’-OH group, so once incorporated, they prevent further strand extension. - Reaction Mixture:
The DNA template is mixed with a primer that binds to a known starting point, along with DNA polymerase, dNTPs, and a small amount of each ddNTP (ddATP, ddTTP, ddCTP, ddGTP) labelled with fluorescent dyes for detection. - Chain Termination:
DNA polymerase begins to synthesise a complementary strand, incorporating dNTPs as usual. When a ddNTP is randomly incorporated, the strand elongation stops. This creates DNA fragments of varying lengths, each ending at a ddNTP. - Separation and Detection:
The fragments are then separated by capillary electrophoresis, which sorts them by length. Each ddNTP has a unique fluorescent label, allowing the sequence of bases to be detected as they pass through a laser. - Reading the Sequence:
The order of fluorescent signals corresponds to the DNA sequence, with each colour indicating a specific base. The sequence of the original DNA template is then reconstructed.
Describe shotgun sequencing
- DNA Fragmentation:
The genome is randomly broken into thousands or millions of smaller DNA fragments using either mechanical methods, like sonication, or enzymatic digestion. This fragmentation ensures that different parts of the genome are sequenced multiple times, creating overlapping regions that are essential for accurate reconstruction. - Sequencing Fragments:
Each DNA fragment is sequenced individually using high-throughput sequencing techniques. The overlap between fragments means that each section of the genome is covered by multiple fragments, which increases accuracy and helps detect errors or inconsistencies in the sequencing process. - Assembly of Overlapping Sequences:
Specialised bioinformatics software is used to align and match the overlapping ends of the fragments. By finding these overlaps, the software assembles the short sequences back into a continuous sequence, reconstructing the original genome accurately.
Describe next-gen sequencing
Describe Illumina sequencing
- DNA Preparation and Fragmentation:
The DNA sample is fragmented into short pieces, typically around 200–300 base pairs. Adaptors, which are short, synthetic DNA sequences, are then attached to each fragment’s ends to allow attachment to the sequencing flow cell. - Flow Cell Attachment and Amplification:
The flow cell is a specialised glass surface with millions of sites for DNA attachment. The DNA fragments bind to complementary sequences on the flow cell and are then amplified in place to form clusters of identical DNA molecules, ensuring sufficient signal strength for detection. - Sequencing by Synthesis:
Fluorescently labelled nucleotides are introduced along with DNA polymerase. As the polymerase adds each complementary nucleotide to a growing DNA strand, the fluorescent label is detected by a camera. Each of the four nucleotides (A, T, C, G) has a distinct fluorescent colour, allowing for precise detection. - Data Collection and Base Calling:
The flow cell is imaged after each nucleotide incorporation, capturing the fluorescence from each cluster. A computer records and processes these images to determine the sequence of nucleotides for each fragment. - Assembly and Analysis:
The short sequences are aligned and assembled using bioinformatics tools to reconstruct the original DNA sequence. This step involves piecing together overlapping sequences from different fragments to generate a complete and accurate sequence
What’re the pros and cons of Using Next-gen Sequencing
Describe SMRT Sequencing (PacBio)
- DNA Preparation:
The DNA is fragmented and circularised by adding adaptors, creating a SMRTbell template. This closed-loop structure allows DNA polymerase to continuously read the template in multiple passes, which improves sequencing accuracy. - Zero-Mode Waveguides (ZMWs):
The sequencing takes place in tiny wells called zero-mode waveguides (ZMWs), each containing a single DNA molecule bound to a DNA polymerase. ZMWs are engineered to detect fluorescent signals from individual molecules, enabling single-molecule sequencing. - Sequencing by Synthesis:
Fluorescently labelled nucleotides are introduced into the ZMWs. As DNA polymerase incorporates each nucleotide into the growing strand, a unique fluorescent signal is emitted and captured by a detector in real time. Each base (A, T, C, G) has a different coloured label, allowing accurate identification. - Real-Time Data Collection:
The signal from each incorporated nucleotide is recorded continuously, allowing the instrument to capture the sequence in real time. The duration of each fluorescent signal also provides additional information about the chemical environment and can help distinguish between different DNA modifications. - Assembly and Analysis:
The sequencing data from individual molecules is processed to generate long reads, often tens of thousands of base pairs. Multiple passes over the circular SMRTbell template allow for error correction, resulting in highly accurate long-read data.
- DNA Preparation and Loading:
Long DNA molecules are prepared for sequencing, with each strand passed through an adaptor complex that guides it into the nanopore. Unlike other methods, nanopore sequencing can handle long DNA fragments, often tens of thousands of base pairs, which allows for sequencing entire genes or large genomic regions without fragmentation. - Nanopore Detection:
The DNA molecule is threaded through a biological nanopore embedded in a membrane, with an electric current applied across the membrane. As the DNA bases (A, T, C, G) pass through the nanopore, each base disrupts the electrical current in a unique way. - Signal Detection and Base Calling:
Changes in current are recorded as each base moves through the pore, generating a unique signal profile. Software then translates these electrical signals into the corresponding DNA sequence, identifying bases in real time as the DNA passes through. - Data Processing and Analysis:
The raw signal data is processed into DNA sequences using advanced base-calling algorithms. Nanopore sequencing allows for direct sequencing of native DNA strands, which can include modified bases, providing insights into epigenetic modifications without extra processing.
Describe Sequencing Read Output: Quality Checking
Describe reference assembly
- Sequencing and Read Generation:
The DNA sample is sequenced, resulting in many short reads. These reads are fragmented pieces of DNA that collectively cover the entire genome of the sample organism. - Alignment to Reference Genome:
Each read is aligned to the reference genome—a known, closely related genome—using specialised software. The software places each read at the location on the reference genome where it best matches, based on sequence similarity. - Consensus Building:
Once aligned, the reads are combined to create a consensus sequence. This consensus represents the sample genome while incorporating any differences, such as mutations or structural variants, between the sample and the reference genome. - Variant Identification:
Differences between the sample and reference genome, such as single nucleotide polymorphisms (SNPs), insertions, or deletions, are identified. This allows researchers to study genetic variations in the sample genome relative to the reference.
Describe Assembling Reads:
de novo Assembly
- Sequencing and Read Collection:
The DNA of the organism is fragmented and sequenced, producing a large set of short reads. These reads are overlapping fragments that cover different parts of the genome, and they serve as the raw data for assembly. - Overlap Detection and Contig Formation:
Specialised software analyses the reads to find overlapping sequences, which indicate where fragments originate from adjacent regions of the genome. These overlaps are used to join reads together into contigs (continuous sequences). Contigs represent longer stretches of the genome, but may still contain gaps. - Scaffolding:
Contigs are further organised into scaffolds using paired-end reads (reads that come from opposite ends of the same DNA fragment) or long reads. These scaffolds approximate the order and orientation of contigs, effectively bridging gaps between them to form longer, more complete segments of the genome. - Gap Filling and Error Correction:
Additional techniques, such as sequencing at higher coverage or using long-read sequencing, help close gaps between scaffolds. Error correction software is applied to refine the assembly by correcting any sequencing errors or resolving ambiguous regions.
Describe annotations
Describe orthologs and paralogs
Describe comparative analysis in RNA
Describe variations of ploidy
Describe the Creation and use of monoploid plants
Describe the formation of a tetraploid organism
Describe meiosis of a triploid organism
Describe the Generation of tetraploid (4x) cells
Describe how in a tetraploid, pairing of chromosomes as bivalents generates balanced gametes
Describe polyploids in agriculture
Describe the Creation of the allopolyploid Triticale
- Hybridisation of Wheat and Rye:
Triticale is initially produced by crossbreeding wheat and rye. Wheat, a hexaploid (six sets of chromosomes, 2n = 6x = 42), is crossed with rye, a diploid (two sets of chromosomes, 2n = 2x = 14). The resulting offspring have three sets of chromosomes from wheat and one set from rye, leading to partial sterility due to mismatched chromosomes. - Chromosome Doubling:
To overcome sterility, the chromosomes are doubled using colchicine, a chemical that disrupts normal cell division and allows chromosome duplication. This treatment produces a fertile octoploid (2n = 8x = 56) with four sets of chromosomes from wheat and two from rye, allowing successful chromosome pairing during meiosis and restoring fertility.
Describe how a Fertile Triticale can be created from infertile F1 hybrid Triticale
How do short hybridisation probes distinguish single-base
mismatches
Describe ASO chips as microarrays
- Up to 4 million
loci can be
genotyped
simultaneously for
approximately
$100 - Companies like
23andMe and Ancestry
use microarrays to
access SNPs in your
DNA to infer
ethnogeographic
background and
predictors of health
Describe the general principles of cloning
Describe plasmid vectors
Describe Restriction enzyme cloning
Restriction enzyme cloning is a technique used to insert a DNA fragment into a vector by cutting both with the same restriction enzymes. These enzymes recognise specific DNA sequences and create cuts, often leaving sticky ends (overhangs) or blunt ends that facilitate the joining of the DNA fragment and vector. The cut fragment and vector are then ligated (joined) with DNA ligase, creating a recombinant DNA molecule that can be introduced into host cells for replication or expression. This method is widely used for gene cloning, enabling researchers to insert specific genes into vectors for study or practical applications.
Describe gateway cloning (BP and LR)
Gateway cloning is a recombination-based method that allows for the efficient transfer of DNA fragments between vectors without the need for restriction enzymes. It uses two main steps, BP and LR reactions, to move the gene of interest through the cloning system.
BP Reaction: The gene of interest, flanked by specific recombination sites (attB), is mixed with a donor vector containing compatible attP sites. Through recombination catalysed by BP Clonase, an entry clone is created with attL sites flanking the gene, allowing it to be used in downstream applications.
LR Reaction: The entry clone (with attL sites) is mixed with a destination vector containing attR sites. LR Clonase catalyses recombination between attL and attR sites, transferring the gene of interest into the destination vector, now ready for expression in a desired system.
Describe TOPO cloning
TOPO cloning is a rapid method of cloning DNA that uses a topoisomerase enzyme to insert a DNA fragment into a vector without the need for restriction enzymes or ligase. In this method, the vector is pre-linearised and modified with a topoisomerase enzyme, which remains covalently attached to the vector ends, allowing the DNA fragment to be efficiently joined. Typically, PCR-amplified DNA with single 3’-A overhangs pairs with complementary T-overhangs on the TOPO vector, enabling quick and directional insertion. This technique is commonly used for cloning PCR products directly, as it requires fewer steps and is time-efficient.
Describe Gibson Assembly cloning
Gibson Assembly is a cloning method that allows for the seamless joining of multiple DNA fragments in a single reaction without the need for restriction enzymes. In this method, DNA fragments with overlapping ends are combined with a mix of enzymes that includes an exonuclease to chew back the ends, a DNA polymerase to fill in gaps, and DNA ligase to seal the nicks. The exonuclease creates complementary overhangs on adjacent fragments, allowing them to anneal and be joined into a continuous sequence. Gibson Assembly is widely used for constructing large DNA molecules or assembling complex gene constructs with high efficiency.
Describe Type IIS Assembly (Golden Gate & MoClo)
Golden Gate: In this method, Type IIS enzymes (e.g., BsaI or BsmBI) are used to generate sticky ends on both the vector and insert DNA fragments. The fragments are then ligated in a single reaction, with the overhangs ensuring correct orientation and positioning. This process allows for the efficient assembly of multiple DNA parts in a one-pot reaction.
MoClo: The Modular Cloning (MoClo) system builds on Golden Gate by using a set of standardised vectors and plasmids for constructing complex genetic circuits. It simplifies the assembly of multi-part constructs, enabling high-throughput cloning of large DNA assemblies with minimal error.
Describe Cloning for Whole Genome Sequencing
Describe genomic libraries
Describe the injection of plasmids without genomic integration as an approach for integrating foreign DNA in an organism
Describe the injection of transgene DNA followed by random genome integration as an approach for integrating foreign DNA in an organism
Describe the P-element mediated genome integration of plasmids as an approach for integrating foreign DNA in an organism
Describe the Viral vector or agro-bacterium mediated transgenesis as an approach for integrating foreign DNA in an organism
Describe CRISPR
Describe conditional knockout
Describe Pharming
Drugs produced by pharming
Describe GM crops
Describe Somatic cell nuclear transfer
Describe in and ex vivo gene therapy
Describe gene therapy with viral vectors
Describe CRIPSR repair of sickle cell anemia
Describe the basic structure of a gene
How do transcription factors influence accessibility of DNA
- Chromatin Remodeling: They recruit complexes that alter the position or removal of nucleosomes, opening up the DNA.
- Histone Modifications: They recruit enzymes that add or remove chemical marks (like acetyl or methyl groups) to histones, making the DNA more or less accessible.
- DNA Looping: They can bring distant regulatory regions (enhancers) closer to the gene’s promoter, facilitating transcription.
4.Coactivators/Corepressors: They interact with other proteins that either open (coactivators) or close (corepressors) the chromatin structure.
