First Week of notes

Question 1

Mendelian Laws

Answer 1

1) Law of segregation At meiosis each allele (2 total) of a gene separates so 50% of it goes to a different gamete, 50:50 ratio
Ex: Start with a diploid, double the chromatin so there are 92 chromosomes total (two doubles of 43)
Will ultimately separate via meiosis I and meiosis II into four separate genes

2) At meiosis the segregation of each pair of alleles in >2 genes is independent each 50:50 ratio
Each Gene will independently end up in a gamete

Question 2

Genotype: The DNA sequence (allele(s)) at a particular locus

Answer 2

o Homozygous: 2 identical alleles at a given locus
o Heterozygous: 2 different alleles at a given locus
o Hemizygous: refers mostly to males (XY), who have just a single copy of each X-chromosomal gene

Question 3

Basic Definitions in Mendelian Genetics

Answer 3

• Phenotype: The observed/measured trait
• Dominance: A phenotype that is expressed (observed) in the heterozygous state
• Recessive: A phenotype that is expressed only in homozygotes or hemizygotes.
o either aa or an xY (homozygous aa or an x-recessive with the Y)
• Semi-Dominant: When the heterozygous phenotype is intermediate between the two homozygous phenotypes
• Penetranceingeneticsis the proportion of individuals carrying a particular variant of agene(alleleorgenotype) that also expresses an associated trait (phenotype). Inmedical genetics, the penetrance of a disease-causing mutation is the proportion of individuals with the mutation who exhibit clinical symptoms. For example, if amutationin the gene responsible for a particularautosomal dominantdisorder has 95% penetrance, then 95% of those with the mutation will develop the disease, while 5% will not.
o Penetrance pretty much means how much they will express the gene/phenotype that exists from a genotype (70% penetrance will show the genotype expressed 70% more or will have the disease 70% more)

Question 4

The pedigree (or family tree) is a graphical representation of the family tree 
o National Library of Medicine: The record of descent or ancestry, particularly of a particular condition or trait, indicating individual family members, their relationships, and their status with respect to the trait or condition.

Answer 4

• proband: (sometimes called the ‘propositus’ or ‘index case’) is the affected individual through whom a family with a genetic disorder is ascertained; may or may not be the consultand (see below)
• The consultand: is the individual (not necessarily affected) who presents for genetic evaluation and through whom a family with an inherited disorder comes to attention
• Consanguinity: identifies cases of genetic relatedness between individuals descended from at least one common ancestor (e.g. first-cousin marriages or even silbings

Question 5

Inheritance patterns

Answer 5

1) Autosomal dominant Means that one of the two alleles will being primarily dominant
- Ex: XX, Xx, or xx only takes a single X in XX or Xx to have the gene show
- Will see the gene show “vertically” in an inheritance pattern tree, both males and females will have the trait

2) Autosomal Recessive Much more rare, need to have both of the recessive alleles for the gene to show
Example: AA, Aa, or aa
• Need to have the aa for the gene to show, otherwise the A will run rampant over it
• An Aa will be considered a carrier, as they will not show the phenotype/represent the gene, only will carry the “a” with them

3) X-linked Dominant Because the gene is located on the X chromosome, there is no transmission from father to son, but there can be transmission from father to daughter (all daughters of an affected male will be affected since the father has only one X chromosome to transmit).
-With the XX, Xx, or the Xy, xY, it has to an XX,Xx, or an XY
It is IMPOSSIBLE for a man to give it his son, only can give Y gene
• Man can give it to his daughter
-Children of an affected woman have a 50% chance of inheriting the X chromosome with the mutant allele. X-linked dominant disorders are clinically manifest when only one copy of the mutant allele is present.

4) X linked Recessive The disease will only manifest in males, due to the female giving a single x chromosome, and the male having to give a Y chromosome
- X-linked recessive traits are fully evident in males because they only have one copy of the X chromosome, thus do not have a normal copy of the gene to compensate for the mutant copy. For that same reason, women are rarely affected by X-linked recessive diseases

5) Mitochondrial Disease Will only be seen in ALL offspring if mother has the disease
- This is due to all mitochondrial genes coming from maternal inheritance

Question 6

•Population Genetics–> The study of allele frequencies and how they change in a population

Answer 6

o Four main evolutionary forces affect allele frequencies:
1) natural selection
2) genetic drift
3)mutation
4) gene flow
o Population sampling by phenotype can lead to estimates of allele frequency if the underlying genetic mechanism is known (i.e. dominant vs. recessive // autosomal vs. sex-linked)
♣ It is critical in the case of phenotype, that the disease has a high penetrance
• Remember penetrance is related to how much the allele is expressed (high penetrance= highly expressed gene)

Question 7

Example: Sad Clown disease with basic populations

♣ 100 people in a class, 1 person has the sad clown disease

Answer 7

100 people in a class, 1 person has the sad clown disease
• Prevelance= 1% (only 1 person has the disease)
• Mutation= 1
We are assuming the disease is a new mutation that could be considered autosomal dominant
• If everyone is cc, then one of the alleles mutated into a dominant C
What is the percentage of the mutant allele?
• 100* 2 alleles= 200 total alleles, than there is a .05% mutation rate or 1/200

Question 8

Population rate Equation

Answer 8

p2 + 2pq + q2 = 1 can be used to estimate carrier rates
o P^2= the homozygous dominant (AA)
o Q^2= the homozygous Recesive (aa)
o 2pq = the hertozygous (Aa)

Also p+q=1
o Cystic Fibrosis (CF): Prevalence 1:2500
o q2 = 1:2500 q ≈ 1:50
o Carrier rate = 2pq 2(49/50)(1/50) ≈ 1/25

Question 9

Sometimes it’s helpful to measure allele frequencies and use them to predict genotypes using the Hardy-Weinberg Law

Answer 9

o (alleles) p + q = 1 = p2 + 2pq + q2 (genotypes)
o [p=frequency of common allele; q=rare allele]
o Assumptions:
o Population is large and mating are random
o Allele frequencies remain constant over time because:
1) No appreciable rate of mutation
2) All genotypes are equally fit (equal chance to pass alleles to next generation)
3) No significant immigration/emigration of individuals with different allele frequencies

Question 10

Most useful information for autosomal recessive diseases in population genetics?

Answer 10

Most useful for autosomal recessive diseases where you want to estimate carrier rates

Applied in genetic counseling situations to tell couples their risk of having a child affected with a particular disease
o Used in genetic counseling to predict risks for a couple to have an affected child
o Prevalence of disease is approximated by q2
o From q2 can calculate q, p, and then the carrier frequency 2pq

Question 11

The Basic number info of the human genome

Answer 11

3 x 10^9 bp = haploid human genome sequence
Human genomic DNA is distributed on 46 nuclear chromosomes
23 pairs of human chromosomes:
o 22 autosomes (1-22)
o 1 pair of sex chromosomes (XX or XY)

Each chromosome is believed to consist of a single, continuous DNA double helix
o Chromosome number generally based on size:
Chr 1: 245,203,898 bp
Chr 22: 49,476,972 bp

Question 12

Human genome represents human history, has changed over time as certain areas (such as the Olfactory portion on chromosome 11) have become invalid over time

Answer 12

o Genotype (genome) + Environment = Phenotype 
o Random variation in a highly ordered structure = almost always deleterious consequences

Genetic diseases from variation–> Price to pay for possibility to evolve with changing genome

Question 13

Genome Basics of Chromatin

Answer 13

o There are highly expressed regions = eurochromatin
-Chromosome 19 is highly expressed

o There are poorly expressed regions and chromosomes= heterochromatin

Question 14

Most likely Chromosomes for Trisomies

Answer 14

Chromosome 13,18, and 21 (viable for trisomies)
-Atrisomyis a type ofpolysomyin which there are three instances of a particularchromosome, instead of the normal twoMuch of the chromosomes are stable, unstable regions however are disease associated

-e.g. SMA (Chr 5q13); DiGeorge syndrome (Chr 22q); 12 diseases associated with unstable region on Chr 1 (1q21.1)GC-rich regions (38% of genome), AT-rich regions (54% of genome)

Question 15

Genome predominance A–T and G–C

Answer 15

• The A-T region is the most dominant
• GC-rich regions (38% of genome), AT-rich regions (54% of genome)
• Clustering (i.e. non-random distribution) of GC-rich and AT-rich regions is basis for chromosomal banding patterns (cytogenetics, karyotype analysis)

Question 16

Notice that chromosome size does not necessarily mean how much its going to be expressed

Answer 16

Chromosome 19 is smaller, but THIRD most expressed

Question 17

Finishing the euchromatic sequence of the human genome in 2004

Answer 17

The initial genome sequence (Build 35) contains 2.85 billion nucleotides interrupted by only 341 gaps
o It covers 99% of the euchromatic genome and is accurate to an error rate of 1 event per 100,000 bases
-Many of the remaining euchromatic gaps are associated with segmental duplications and will require focused work with new methods

o The near-complete sequence, the first for a vertebrate, greatly improves the precision of biological analyses of the human genome including studies of gene number, birth and death

Question 18

Composition of the Human genome

Answer 18

o ~1.5% is translated (protein coding)
-A tiny portion (roughly 1.5 percent) is actually translated as protein

o 20-25% is represented by genes (exons, introns, flanking sequences involved in regulating gene expression)
o 50% “single copy” sequences
-One gene/one copy
-Does not have multiple copies like repetitive DNA

o 40-50% classes of “repetitive DNA”
-Three major categories of repetitive DNA (Terminal repeats, Tandem repeats, Interspersed repeats)

Question 19

Current Status of mapping remainder of Euchromatic DNA

Answer 19

Euchromatic regions (more relaxed) and heterochromatic regions (more condensed; more repeat-rich)
o Genome sequencing efforts focused on euchromatic regions
o Heterochromatic regions essentially unsequenced
o Many (>200) sequence gaps still remain even in euchromatic regions

Question 20

Amount of content for Genome

Answer 20

o Less than 1.5% of genome expresses for proteins, another 5% is regulatory region (promoters, enhancers, ect.)
o Roughly half of the genome is Single-Copy or Unique DNA (roughly 50%)
-Single Copy/ Unique DNA DNA whose linear order of nucleotides is unique
• Much of single copy DNA is still mysterious, as only roughly 1.5% of it encodes for proteins
• Each is roughly several kilobases (several thousand base-pairs)
o Other half of Genome (50%) is repetitive DNA–> DNA whose nucleotides are repeated

Question 21

Repetitive DNA

Answer 21

• Repetitive DNA for repeated nucleotide sequences, important to note if the repeated sequences in the same location or are interspersed throughout the chromosome

Tandem Repeats A tandem repeat is a sequence of two or more DNA base pairs that is repeated in such a way that the repeats lie adjacent to each other on the chromosome. Tandem repeats are generally associated with non-coding DNA.
• Tandem repeats are called “tandem” becase the repeated sequences start head to tail of one another
♣ Satellite DNA different types of Tandem Repeats, these are large arrays of repeating nucleotides
• Alpha Satellite DNA repeating DNA in centromeres

Dispersed Repetitive Elements Instead of being on a specific chromosome or several locations, these are repetitive DNA that’s throughout the genome (on multiple chromosomes)
2 Major families of Dispersed repetitive elements
• ALU family a type of dispersed resistive element seen on many chromosomes, they are a repeating 300 base pairs
o At least a million different Alu families
o Make up 10% of the genome

•LINE (Long interspected Nuclear Elements) The second major dispersed component of repetitive DNA, is roughly 6 kb long (6000 base pairs)
o Roughly 850,000 copies per genome and makes up 20% of the genome

Question 22

2 Major families of Dispersed repetitive elements

Answer 22

Dispersed Repetitive Elements Instead of being on a specific chromosome or several locations, these are repetitive DNA that’s throughout the genome (on multiple chromosomes)

• ALU family a type of dispersed resistive element seen on many chromosomes, they are a repeating 300 base pairs
o At least a million different Alu families
o Make up 10% of the genome

• LINE (Long interspected Nuclear Elements) The second major dispersed component of repetitive DNA, is roughly 6 kb long (6000 base pairs)
o Roughly 850,000 copies per genome and makes up 20% of the genome

Question 23

Tandem Repeats

Answer 23

Type of repetitive DNAo Tandem Repeats A tandem repeat is a sequence of two or more DNA base pairs that is repeated in such a way that the repeats lie adjacent to each other on the chromosome. Tandem repeats are generally associated with non-coding DNA.
• Tandem repeats are called “tandem” becase the repeated sequences start head to tail of one another

Satellite DNA different types of Tandem Repeats, these are large arrays of repeating nucleotides
• Alpha Satellite DNA repeating DNA in centromeres

Question 24

SNPs (Single Nucleotide Polymorphism)

Answer 24

o A change in one in every 1000 nucleotide base pairs, will change between an A—T or an G—C
Occur more often in non-coding genome
Can have consequences in coding genome such as changing a codon to a stop codon and such

Question 25

Insertion-Deletion Polymorphism

Answer 25

•Insertion-Deletion Polymorphism any type of insertion or deletion of base-pairs that can range from a single base pair to over a thousand
oMicrosatellite Polymorphism Are strands of DNA that are a few base pairs/alleles that repeat 5-50 times approximately
-Can be used for DNA fingerprinting, as people RARLY have similar microsatellite patterns
-Also called Short Tandem Repeats

Question 26

Copy Number Variants

Answer 26

Copy Number Variants similar to microsatellites, but are variations of larger segments of the genome
o The smaller CNVS can be only several alleles or it can be up to many different alleles
o A copy number variation (CNV) is when the number of copies of a particular gene varies from one individual to the next.
- Following the completion of the Human Genome Project, it became apparent that the genome experiences gains and losses of genetic material. The extent to which copy number variation contributes to human disease is not yet known. It has long been recognized that some cancers are associated with elevated copy numbers of particular genes.

Question 27

Classes of Repetitive DNA

Answer 27

1) Tandem Repeats Tandem: repeats occur in DNA when a pattern of one or more nucleotides is repeated and the repetitions are directly adjacent to each other.
2) Long Terminal Repeats: are identical sequences of DNA that repeat hundreds or thousands of times found at either end of retrotransposons or proviral DNA formed by reverse transcription of retroviral RNA[citation needed. They are used by viruses to insert their genetic material into the host genomes.

o 3) Dispersed Repetitive Elements Across different chromosomes! Can have families of repeat such as Alu or LINE

Question 28

Tandem Repeats

Answer 28

Tandem Repeats Tandem–> repeats occur in DNA when a pattern of one or more nucleotides is repeated and the repetitions are directly adjacent to each other.

• Satellite DNA consists of very large arrays oftandemlyrepeating,non-coding DNA. Satellite DNA is the main component of functionalcentromeres, and form the main structural constituent ofheterochromatin.[1][2]

• Some (a particular pentanucleotide sequence) are found as part of human-specific heterochromatic regions on the long arms of Chr 1, 9, 16 and Y (hotspots for human-specific evolutionary changes)
• On Chromosome 1,9,16, and Y, can major differences in DNA at C-bands not seen in primates

“α-satellite” repeats (171 bp repeat unit) around centromeres
• Type of satellite repeating seen around the Centromere (the center of the chromosome)
• Might be critical for function centromeres during mitosis and meiosis (as the chromosomes split)

Question 29

Human chromosomes that are “evolutionary hotspots”

Answer 29

Some (a particular satellite sequence) are found as part of human-specific heterochromatic regions on the long arms of Chr 1, 9, 16 and Y (hotspots for human-specific evolutionary changes)
-On Chromosome 1,9,16, and Y, can major differences in DNA at C-bands not seen in primates

Question 30

Long Terminal Repeats

Answer 30

are identical sequences of DNA that repeat hundreds or thousands of times found at either end of retrotransposons or proviral DNA formed by reverse transcription of retroviral RNA[citation needed. They are used by viruses to insert their genetic material into the host genomes.

Question 31

Dispersed Repetitive Elements

Answer 31

Dispersed Repetitive Elements Across different chromosomes! Can have families of repeated DNA base pairs
1) Alu family (e.g. of SINEs: Short Interspersed repetitive Elements) - ~300 bp related members - 500,000 copies in genome
• AnAlu elementis a short stretch ofDNAoriginally characterized by the action of the Alu (Arthrobacter luteus)restrictionendonuclease.[1]Alu elements of different kinds occur in large numbers in primategenomes. In fact, Alu elements are the most abundanttransposable elementsin the human genome.
o Remember a Transposable element is a portion of a chromosome that repeats multiple times, while Dispersed Repetitive Elements are across multiple chromosome

2) L1 family (e.g. of LINES: Long Interspersed repetitive Elements) - ~6 kb related members - 100,000 copies in genome
• Long Interspersed Nuclear Elements9 are a group of genetic elements that are found in large numbers in eukaryotic genomes, composing 17% of the human genome (99.9% of which is no longer capable of mobilization).[10]Among the LINE, there are several subgroups, such as L1, L2 and L3. Human codingL1 begin with an untranslated region(UTR) that includes anRNA polymerase IIpromoter, two non-overlappingopen reading frames(ORF1 and ORF2), and ends with another UTR.

Alu’s and L1’s can be of significant medical relevance
• Retrotransposition may cause insertional inactivation of genes
• Repeats may facilitate aberrant recombination events between different copies of dispersed repeats leading to diseases - Non-allelic homologous recombination (NAHR)

Question 32

Insertion-Deletion Polymorphisms

Answer 32

o Minisatellites is a tract of repetitive DNA in which certain DNA motifs (ranging in length from 10–60 base pairs) are typically repeated 5-50 times.[1] Minisatellites occur at more than 1,000 locations in the human genome and they are notable for their high mutation rate and high diversity in the population.

o Microsatellites a tract of repetitive DNA in which certain DNA motifs (ranging in length from 2–5 base pairs) are repeated, typically 5-50 times.[1] Microsatellites occur at thousands of locations in the human genome and they are notable for their high mutation rate and high diversity in the population.

• Shorter versions of minisattelites, they are only di, tri, and quad repeats
• STRPs (Short Tandem Repeat Polymorphisms)

Question 33

Microsatellites and VNTR

Answer 33

oMicrosatellites and their longer cousins, theminisatellites, together are classified asVNTR(variable number of tandem repeats) DNA. The name”satellite”refers to the early observation that centrifugation of genomic DNA in a test tube separates a prominent layer of bulk DNA from accompanying “satellite” layers of repetitive DNA.

VNTR is a location in a genome where a short nucleotide sequence is organized as a tandem repeat. These can be found on many chromosomes, and often show variations in length between individuals. Each variant acts as an inherited allele, allowing them to be used for personal or parental identification.
• Tandem repeats groups of nucleotides right next to each other that repeat over and over again

Question 34

Single Nucleotide Polymorphism

Answer 34

• a mutation in a single nucleotide at a base-pair

-frequency of 1 in ~1000 bp -PCR-detectable markers, easy to score, widely distributed

Question 35

Copy number variations

Answer 35

CNV–> A form of structural variation—are alterations of the DNA of a genome that results in the cell having an abnormal or, for certain genes, a normal variation in the number of copies of one or more sections of the DNA. CNVs correspond to relatively large regions of the genome that have been deleted (fewer than the normal number) or duplicated (more than the normal number) on certain chromosomes.

•CNV (Copy Number Variations) A genome having copies (normal or abnormal of other sections of the DNA strand
o Primary type of structural vartion seen with the Genome
o CNV loci account for about 12% of the genome
-A huge amount of the genome are CNVs
o CNV is implicated in many different diseases =having too many versions of a gene or an abnormal gene

o CNV is also highly involved with evolutionary changes:
♣ enriched for human specific gene duplications
♣ enriched for genome sequence gaps
♣ enriched for recurrent human diseases
♣ Chromosomes: - 1q21.1; 9p13.3-9q21.12, 5q13.3

Question 36

Gene families and duplication

Answer 36

o Many human genes are members of gene families
o Gene family is composed of genes with high sequence similarity (e.g. >85-90%) that may carry out similar but distinct functions
o Gene families arise through gene duplication, a major mechanism underlying evolutionary change. Rationale: when a gene duplicates it frees up one copy to vary while the other copy continues to carry out a critical function
-Allows innovation, and possibility of staying alive
-Hox family

Question 37

Limitations for Genome Sequencing

Answer 37

o Nextgen DNA Sequencing

• No mammalian genome has been completely sequenced & assembled
• Nextgen sequencing relies on short read sequences Complex, highly duplicated regions are typically unexamined Such regions are implicated in numerous diseases, e.g. 1q21

o Genome-Wide Association Studies

• Missing heritability” for complex diseases: Many large-scale studies implicate loci (e.g. SNPs) that account for only a small fraction of the expected genetic contribution
• Many regions of the genomes are unexamined by available “genome-wide” screening technologies: is this where the “missing heritability” lies?

Question 38

Meiosis: the process where a diploid cell ultimately goes from being a single diploid (2N in genetic content) to becoming haploid (n in genetic content) with 4 gamete cells (2n–>4n—> 2x2n—> 4x 1n)

Answer 38

Key notes for Meiosis and mitosis
o During Meiosis I The two homolog chromosomes will pair and face each other inside the cell
-Homologous chromosomes will split away to separate into other cells
- During prophase I the chromosomes will form stuctures called “bivalents” which are the chromosomes linking up next to each other

•This will cause the chromosomes to form a synaptonemal complex–> Allows the Chiasmata to hook up the chromosomes with one another

During the metaphase I, homologous recombination underwent, allowing for genetic recombination to happen
• Chiasmata–> A physical link in meiosis I that allows DNA material to be exchanged between the homologous chromosomes
Homolog pairs in mitosis rarely have recombination rarely have recombination

Meisos II and Mitosis are more similar, as Sister chromatin will split to separate sides
Meisos II: The homolog chromosomes will split in each cell to form 4 haploid cells with one set of 23 chromosomes
Mitosis The sister chromatid (at 92 total chromosomes) will split into respective sides for normal diploid cells at 46 chromosomes

Question 39

Chiasmata

Answer 39

Chiasmata–> A physical link in meiosis I that allows DNA material to be exchanged between the homologous chromosomes

Question 40

Meiosis II and Mitosis

Answer 40

Meisos II and Mitosis are more similar, as Sister chromatin will split to separate sides
Meisos II: The homolog chromosomes will split in each cell to form 4 haploid cells with one set of 23 chromosomes
Mitosis The sister chromatid (at 92 total chromosomes) will split into respective sides for normal diploid cells at 46 chromosomes

Question 41

Chiasmata

Answer 41

A physical link in meiosis I that allows DNA material to be exchanged between the homologous chromosomes

Question 42

Genetic consequences from meiosis

Answer 42

o Allowed for genetic shuffling roughly 2-3 cross-overs during meiosis I
o Also random segregation based on homologous split, each chromosome has different DNA from recombination and from homolgous split during meisos I

Question 43

Mitosis

Answer 43

• one round of chromosome segregation, resulting in daughter cells identical inchromosomal content to the parental cell
• DNA replication precedes each round of chromosome segregation
• no pairing of homologous chromosomes
-only sister chromatids will pair
-centromeres on paired sister chromatids segregate at each anaphase
• infrequent recombination
• occurs in somatic cells and in germ line precursor cells prior to entry into meiosis

Question 44

Meiosis

Answer 44

• two rounds of chromosome segregation without an intervening round of DNA replication
• parental cells must be diploid and the chromosome number is halved in the resultant cells
• requires the pairing of homologous chromosomes and recombination for its successful completion
o centromeres on paired sister chromatids divide only at anaphase II in a normal meiosis
• occurs only in the germ line

Question 45

Chromosome Identification and nomenclature:

Answer 45

• Chromosomes are numbered based on size, with chromosome 1 being the largest and 22 being the smallest
o No base on amount of expression
o Much of the classification of chromosomes is based around the centromere

Question 46

Human chromosomes are put into three different categories

Answer 46

o 1) Metacentric: The centromere is directly in the center of the chromosome, so that both arms are equal length
o 2) Submetacentric: The centromere is slightly removed or a slight distance from the center
o 3) Acrocentric: The Centromere is near one end of the chromosome

Question 47

Chromosomes are also identified and labeled based on where specific bands are from dyes

Answer 47

o banding patterns observed microscopically after treatment with stains such as Giemsa and DAPI (4’,6-diamino-2- phenylindole. These patterns result from the differential staining of various chromosomal regions (e.g. regions with high G+C, or A+T base compositions, or the presence of heterochromatin)

-G-banding, G banding, or Giemsa banding is a technique used in cytogenetics to produce a visible karyotype by staining condensed chromosomes. It is useful for identifying genetic diseases through the photographic representation of the entire chromosome complement. The metaphase chromosomes are treated with trypsin (to partially digest the chromosome) and stained with Giemsa stain. Heterochromatic regions, which tend to be rich with adenine and thymine (AT-rich) DNA and relatively gene-poor, stain more darkly in G-banding.

Question 48

Names of the Chromosome arms

Answer 48

o p-arm: The smaller arm of the chromosome, known as the P arm due to being the petite arm

• remember the P –arm is PETITE
• q-arm: The larger arm of the chromosome

o Each chromosomes is labeled based on regions, counting OUT form the centromere
-Example: P1, P2, P3, going out on the petitie arm from the centromere
-13 Q2, 13Q4 working outward on the Q (large) arm of the chromosome from the centromere
o Convenient to say proximal or distal part of chromosome arm 13Q proximal or 13 Q distal (proximal or distal to centromere)

Question 49

Nomenclature of chromosome abnormalities: Numerical abnormalities:

Answer 49

o Triploidy 69,XXX; 69 XXY; 69,XYY
o Trisomy e.g. 47,XX,+21
o Monosomy e.g. 45,X
o Mosaicism e.g. 45,X/46,XX

Question 50

Chromosomal Abnormalities

Answer 50

Aneuploidy: Where a cell has an abnormal number of chromosomes, often caused by nondisjunction

Nondisjunction: the failure of chromosomes to separate/ mis-segregate during metaphase of meiosis or mitosis, ultimately daughter cell will have too many or too few chromosomes

• Based around the mechanics of the chiasmata during metaphase I,
• Different errors can arise based on distance of recombination from the centromere (too close or too far from centromere can lead to issues

• Crossover too distal from centromere: Can increase the chance spindle attachment error
• Crossover to proximal to centromere: Can lead to entanglement, greater chances of disjunction
More likely crossover= greater chance of Nondisjunction

o Mosonomy–> Cell is lacking one chromosome

• Example: 45,X
• Is usually lethal at an embryonic example, save for 45,X which Lead’s to Turner’s syndrome

o Trisomy–> Have an entire extra copy of a chromosome

• Example: 47XX, +21 (extra chromosome at 21)
• Not as lethal as a Mosonomy
• Yet often leads to spontaneous abortions
• Down’s syndrome is most common chromosomal disorder from Trisomy

Question 51

Aneuploidy

Answer 51

Aneuploidy–> Where a cell has an abnormal number of chromosomes, often caused by nondisjunction

Question 52

Nondisjunction

Answer 52

o Nondisjunction the failure of chromosomes to separate/ mis-segregate during metaphase of meiosis or mitosis, ultimately daughter cell will have too many or too few chromosomes
-Based around the mechanics of the chiasmata during metaphase I,

Different errors can arise based on distance of recombination from the centromere (too close or too far from centromere can lead to issues
• Crossover too distal from centromere: Can increase the chance spindle attachment error
• Crossover to proximal to centromere: Can lead to entanglement, greater chances of disjunction

More likely crossover= greater chance of Nondisjunction

Question 53

Maternal Age effect

Answer 53

As women age, there is a greater chance of aneuploidy of the chromosomes

o “two hit theory” as women age, several events may be causing aging

1) first event is lack of recombination, possibly due to lack of chiasma or misslocalization of the chromosomes
2) second event is ability of oocytes to successfully complete chromosome segregation in the presence of unfavorable recombination events is thought to diminish over time

o Second possibility of maternal age effect The degradation of cohesion complexes during meisos I will lead to eventual aneuploidy

Cohesion Complex is a protein complex that regulates the separation of sister chromatids during cell division, either mitosis or meiosis. Aging will lead to degredation of these proteins, and premature termination
• With aging, this complex is possibly degrading over time, allowing the Chiasmata separate towards the ends of the chromosomes
o The chiasmata “terminalizing” towards the end will lead to aneuploidy

Question 54

Cohesion Complex

Answer 54

Cohesion Complex is a protein complex that regulates the separation of sister chromatids during cell division, either mitosis or meiosis. Aging will lead to degredation of these proteins, and premature termination

With aging, this complex is possibly degrading over time, allowing the Chiasmata separate towards the ends of the chromosomes
o The chiasmata “terminalizing” towards the end will lead to aneuploidy

Question 55

Common Human Aneuploidy Syndromes:

Answer 55

o Trisomy 21 – Down syndrome Characteristic facies, short stature, hypotonia, moderate intellectual disabilities Cardiac anomalies, leukemia in infancy.
o Trisomy 18 – Edwards syndrome Small for gestational age, small head, clenched fingers, rocker-bottom feet
o Trisomy 13 – Patau syndrome Characteristic facies, severe intellectual disabilities Congenital malformations – holoprosencephaly, facial clefts, polydactyly, renal anomalies
o 45, X -Turner syndrome Short stature, webbed neck, edema of hands and feet, broad shield-like chest, narrow hips, renal and cardiovascular anomalies, gonadal dysgenesis (failure of ovarian maintenance).
o 47, XXY – Klinefelter syndrome Tall stature, hypogonadism, elevated frequency of gynecomastia, high frequency of sterility, language impairment

Question 56

Mosaicism

Answer 56

Mosaicism–> Is two genetically different cells in tissue that derived from the same single zygote.
o Can be different cells throughout an organism that have different genetic makeup
o Condition typically derives from nondisjunction of a single cell that is prenatal (before birth), sometimes postnatal

Each cell from the nondisjunction can continue to divide and create cells that have different genotypes
o Example: one of the milder forms of Klinefelter syndrome, called 46/47 XY/XXY mosaic wherein some of the patient’s cells contain XY chromosomes, and some contain XXY chromosomes. The 46/47 annotation indicates that the XY cells have the normal number of 46 total chromosomes, and the XXY cells have a total of 47 chromosomes.

Question 57

Chromosomes rearrangement

Answer 57

•Chromosomes rearrangement will happen with double strand DNA breaks (DBSs), these can happen to environment issues such as UV light or radiation, but will happen most often during mitosis or meiosis during chromosome interaction
o Structural rearrangement can be inherited, and lead to further rearrangement during meiosis
o Two major types of rearrangements exist: Balanced and unbalanced

Question 58

Homologous Recombination

Answer 58

Is a type of genetic recombination in which nucleotide sequences are exchanged between two similar or identical molecules of DNA. It is most widely used by cells to accurately repair harmful breaks that occur on both strands of DNA, known as double-strand breaks. Homologous recombination also produces new combinations of DNA sequences during meiosis

Question 59

Balanced Rearrangement

Answer 59

Balanced Rearrangement–> with balanced rearrangement, there will be no loss or gain of total genetic material with rearrangement of chromosomes, however there are possible issues that can arise, especially for the offspring

Different Balanced Rearrangement:

• Inversion (paracentric and paricentric)
• Reciprocal translocation (alternate and adjacent 1)
• Robertsonian Translocation

Question 60

Unbalanced Rearrangement

Answer 60

Unbalanced Rearrangement the chromosome set has additional or missing material. Phenotypes ofthese individuals are likely to be abnormal. Duplication of genetic material in gametes can lead to partial trisomy after fertilization with a normal gamete, while deletions lead to partial monosomy.

Contiguous gene syndromes/ microduplications & microdeletions —>also known as a contiguous gene deletion syndrome is a clinical phenotype caused by a chromosomal abnormality, such as a deletion or duplication that removes several genes lying in close proximity to one another on the chromosome.

Question 61

Contiguous gene syndromes/ microduplications & microdeletions

Answer 61

•	Contiguous gene syndromes/ microduplications & microdeletions   also known as a contiguous gene deletion syndrome is a clinical phenotype caused by a chromosomal abnormality, such as a deletion or duplication that removes several genes lying in close proximity to one another on the chromosome. The combined phenotype of the patient is a combination of what is seen when any individual has disease causing mutations in any of the individual genes involved in the deletion.
o	Example diseases:
o	del(4p16.3) Wolf-Hirschhorn syndrome facial clefting prominent ears microcephaly, intellectual disabilities
o	dup(11p15.5) (paternal) Beckwith-Wiedemann syndrome overgrowth omphalocele predisposition
o	del(15q11-q13) (paternal) Prader-Willi syndrome hypotonia, hypopigmentation hypogenitalism, obesity
o	del(15q11-q13) (maternal) Angelman syndrome seizures intellectual disabilities

Question 62

Inversion

Answer 62

Inversion during a double strand DNA break, the arm of the chromosome (p or q) will invert to the other side (the segment of the DNA will flip opposite) there are 2 major kinds: Pericentric Inversions and Paracentric Inversions

o Pericentric Inversions–> pericentric inversions include the centromere during the flip
o Paracentric inversion–>no centromere, only the chromosome arm portion is flipping

If crossover happens during paracentric inversion, can lead to two or no centromeres on chromosome, ultimately ruining chromosome

o Typically, both inversions can often have normal genetic expression, as long as none of the genes were compromised
o However, during a paracentric inversion, it is possible that crossover during an inversion can lead to a chromosome having 2 centromeres or no centromeres, ultimately causing the chromosome to degrde
-2 centromeres Dicentric
-no centromeres Acentric

Question 63

Pericentric Inversions

Answer 63

pericentric inversions include the centromere during the flip

Question 64

Paracentric inversion

Answer 64

no centromere, only the chromosome arm portion is flipping

If crossover happens during paracentric inversion, can lead to two or no centromeres on chromosome, ultimately ruining chromosome

Question 65

Reciprocal Translocation

Answer 65

Reciprocal Translocation—> Reciprocal translocations are usually an exchange of material between non-homologous chromosomes. Such translocations are usually harmless and may be found through prenatal diagnosis. However, carriers of balanced reciprocal translocations have increased risks of creating gametes with unbalanced chromosome translocations, leading to miscarriages or children with abnormalities.

o breakage and rejoining of non-homologous chromosomes, with a reciprocal exchange of the broken segments.
o have a good chance of producing unbalanced gametes
-can lead to abortion or infertile males

2 major types of reciprocal translocation:
Alternate Segregation= healthy
Adjacent 1 Segregation= not healthy

Question 66

Alternate Segregation

Answer 66

Alternate Segregation–> Centromeres of the homologous chromosomes go to opposite ends, leading to the arms of the chromosomes going to opposite sides, this is the healthy version producing normal chromosomes

• Leads to gametes with Normal Chromosomes and balanced translocation
• P arm on one end, the Q arm on the other end

Ex: In the alternate segregation pattern, the two translocation chromosomes (T1 and T2) go to one pole, while the two normal chromosomes (N1 and N2) move to the opposite pole. Both kinds of gametes resulting from this segregation (T1, T2, and N1, N2) carry the correct haploid number of genes; and the zygotes formed by union of these gametes with normal gamete will be viable.

Question 67

Adjacent 1 Segregation

Answer 67

Adjacent 1 Segregation—> Centrioles of the chromosomes go to the same end, leading chromosomes being adjacent, this is the undesired translocation that is unbalanced for the genome

• Forms non-homologous centromeres
• Will lead to unbalanced genome: results in trisomy and monosomy for chromosomes

Question 68

Robertsonian Translocation

Answer 68

will cause the long arms of two acrocentric chromosomes to form, resulting in the loss of the tiny short arms of the chromosome (often these short arms have rDNA or repeating DNA that’s nonexpressive)

Even though there is a decrease in DNA, the Robersonian Translocation is usually considered Balanced, this is due to the acrocentric chromosomes still carrying functional DNA
-However, the offspring of Rob. Translocation will often have monosomies and trisomies

Question 69

Acrocentric Chromosome

Answer 69

Acrocentric Chromosome—> an an acrocentric chromosome the p arm contains genetic material including repeated sequences such as nucleolar organizing regions, and can be translocated without significant harm, as in a balanced Robertsonian translocation.

Question 70

Clinical Cytogenics

Answer 70

The different ways to read and analyze chromosomes. Allows to ultimately understand why there are aberration in the chromosome number and chromosome shape, and the diseases that come from it.

o Regular tools for the such analysis: G-bands, FISH, microarray

Question 71

Cancer Cytogenetic

Answer 71

o Specimens for cytogenetic studies
-Bone Marrow—> for cancer

Blood

• Unstimulated cancer detecting
• Constitutional

Tissue
• Lymph node
• Solid tumor
• POC

Fluids
• CNS cancer detecting
• Amniotic fluid

Question 72

Chromosome counting

Answer 72

o Hyperdiploid–> Has more than 46 chromosomes
• Example: 55 chromosomes, prognosis is favorable
-Can have detrimental health effects, but is treatable
-Use FISH to detect problem genes or chromosomes early, and treat with

o Hypodiploid Have less than 46 chromosomes
• Example: A 36 Chromosome count (monosomy)
-Not good prognosis Necessary Genes gone from chromosomes

Question 73

FISH fluorescence in situ hybridization

Answer 73

o	is acytogenetictechnique that usesfluorescent probesthat bind to only those parts of the chromosome with a high degree of sequencecomplementarity. It was developed by biomedical researchers in the early 1980s[1]and is used to detect and localize the presence or absence of specificDNAsequencesonchromosomes.
o	Types of Probes (know these ones)
Centromere Cen
Used for Enumeration-leukemias
o Example: Cen 4, 8, 10, 17

Dual Fusion DF, F
• Used for Translocation
o Ex: BCR, ABL PML,RARalpha (cancer)

• Deletion-Leukemias
o p53

Question 74

Chromosomal Translocation leading to Cancer

Answer 74

Chromosomal Translocation–> leading to cancer Cancer can be caused by translocation in two manners (two genes coming together to become hyper-expressed, or create some new gene that creates a novel rogue protein)

1) Two genes can translocate next to one another, and ultimately lead to an enhancer region or promoter to skyrocket in activity
- Leads to extreme expression that can be cancerous, such as drastic increase in transcription factor

2) Genes can translocate together, and form a new sequence that creates a novel protein
- This novel Protein can start causing chaos

Question 75

Derivative Chromosome

Answer 75

Derivative Chromosome—> Aderivative chromosome(der) is a structurally rearrangedchromosomegenerated either by a rearrangement involving two or more chromosomes or by multiple aberrations within a single chromosome (e.g. an inversion and a deletion of the same chromosome, or deletions in both arms of a single chromosome)

o Acute Myeloid Leukemia (chromosome 15 and 17 rearrangement)
o Chronic Myeloid Leukemia (chromosome 9 and 22)

Question 76

Acute Myeloid Leukemia

Answer 76

oMore frequently seen in adults, rarely in youth
-Have Auer rods you can see in microscopes

o Cause of Acute Myeloid Leukemia
The Translocation of Chromosome 15 and Chromosome 17 will create a new genetic sequence

The PML section of chromosome 15 and the RARA region of chromosome 17 will translocate, leading to the creation of the PML/RARA transcription factor
-which encodes a putative transcription factor, Differentiation of myeloid hematopoietic precursors past the promyelocytic stage is reportedly inhibited by the actions of a PML-RARA fusion protein.

Vitamin A can combat and ruin the structure of the PML-RARA protein

Question 77

Chronic Myeloid Leukemia

Answer 77

o Presentation:

• Night sweats, fatigue, weight loss,anemia
• Lab:Peripheral blood smear shows lobulated large cells

o Function: Chromosome 9 and 22 have a translocation, leading to the production of a BCR/ABL protein
-The ABL gene from Chromosome 9 will translocate and mix with the BCR gene with chromosome 22

o ThePhiladelphia chromosomeorPhiladelphia translocationis a specific abnormality of chromosome22, which is unusually short, as an acquired abnormality that is most commonly associated withchronic myelogenous leukemia(CML).[1]It is the result of a reciprocal translocationbetween chromosome 9 and chromosome 22, which is specifically designated t(9;22)(q34;q11). This gives rise to afusion gene,bcr-abl

Question 78

Chromosomal Microarray (CMA)

Answer 78

Chromosomal Microarray (CMA) is a molecularcytogenetictechnique for the detection of chromosomalcopy number changeson agenomewide and high-resolution scale.[14]Array CGH compares the patient’s genome against a reference genome and identifies differences between the two genomes
o	Determine size and location of deletions/duplications & are supported by multiple levels of SNP data
o	 Determine size and location of deletions/duplications & are supported by multiple levels of SNP data
Cannot detect low level mosaicism or balanced chromosome rearrangements

Laboratory reporting thresholds are
• Deletions ≥ 200kb
• Duplications ≥ 400kb
o CMA cannot detect very low level mosaicism, heterodisomy, or balanced chromosome rearrangements.

Genomic Content of these Regions
• Copy number variants: common CNV’s not reported
• Genes/functions & discussion of known phenotypes
o Runs of homozygosity or ROH is evaluated at ≥5 Mb with a reporting threshold of ~10 Mb.

Question 79

Runs of homozygosity

Answer 79

ROH are regions of the genome where the copies inherited from our parents are identical. This creates a run of homozygous variants, from tens of thousands to millions of letters in length. The two DNA copies are identical because our parents have inherited them from a common ancestor at some point in the past, recently in the case of a cousin marriage, but in fact we all carry ROH, because going far enough back in time we are all related.

ROH may also be important medically. This is because they allow certain variants, called recessive variants to be expressed. Recessive variants only have their effect when present on both copies of an individual’s genome, for example in a run of homozygosity. Recessive variants cause many genetic diseases such as cystic fibrosis, phenylketonuria and Tay-Sachs disease

Question 80

Gleevec

Answer 80

Gleevec is atyrosine-kinase inhibitorused in the treatment of multiple cancers, most notablyPhiladelphia chromosome-positive (Ph+)chronic myelogenous leukemia(CML).[1]
o Molecular antagonist: binds at ATP binding site in abl tyrosine kinase and bcr/abl tyrosine kinase
-In order to survive, cells need signaling through proteins (signal cascade) to keep them alive. Some of the proteins in this cascade use aphosphategroup as an “on” switch. This phosphate group isadded by a tyrosine kinase enzyme.

Question 81

Two major categories of structural rearrangement for Chromosomes:

Answer 81

o Balanced: genome has normal complement of chromosomal material
-Might need different techniques to confirm this, DNA band measuring might show balance, while it may be unbalanced when analyzed with micro-array or DNA sequence analysis
o Unbalanced: if there is any additional or missing chromosomal material

Question 82

Unbalanced Rearrangements

Answer 82

Unbalanced Rearrangements most likely to happen due to deletions or duplications of the genes, happens in 1 out of 1600 live births
-Often leads to partial trisomy or partial monosmy

oDeletions involves a loss of chromosome segment, resulting in chromosomal imbalance
-Clinical consequence is haploid insufficiency–>The inability of a single chromosome to carry out the necessary work of two chromosome genes

Deletions can happen at end of the chromosomes (terminal) or somewhere within the arm (interstitial)
o Marker Chromosomes a structurally abnormal chromosome in which no part can be identified. The significance of a marker is very variable as it depends on what material is contained within the marker.

It is essentially a partial trisomy. However, sometimes the marker is composed of inactive genetic material and has little or no effect.

Question 83

Balanced Rearrangement

Answer 83

Balanced Rearrangement Happens in 1 out 500 youths, typically there are no consequences due to all necessary genes being present, regardless of being rearranged

Major types
Translocations (Reciprocal) –> Alternate and Adjacent
Inversions
Robertsonian Translocations

Question 84

Translocations

Answer 84

Translocations Involves the exchange of chromosome segments between two chromosomes
o Reciprocal Translocations This the type of breakage or recombination between non-homologous chromosomes
Usually the exchange is reciprocal, so the total chromosome number is unchanged, also usually no phenotypical effect

o Where each chromosome goes will decide what type of segregation it undergoes
Alternate Segregation —>Will produce balanced chromosome translocation
Adjacent 1 and Adjacent 2—> will have unbalanced and possibly abnormal gametes

Question 85

Quadrivalent

Answer 85

The cross structure that the 2 non-homologous chromosomes will form

Question 86

Robertsonian Translocation

Answer 86

Robertsonian translocation The most common translocation in human species, where to acrocentric chromosomes will fuse near the centromere regions, losing the short arms

Acrocentric chromosomes have tiny short arms (p-arms) that have almost no genes on them
Robertsonian Translocations are Non-Reciprocal, and will have karyotypes of 45, due to 2 chromosomes having their long arms fuse together
• Can be fine for parents, but puts the progeny at risk for trisomy

Question 87

Inversions

Answer 87

Inversions occurs when a single chromosome undergoes two breaks and is reconstructed with the segment being inverted (flipped around) Two Types:

Paracentric: Both breaks occur in one arm (does not include centromere)
Paricentric: A break in each arm that includes a centromere

Question 88

Epigenetics

Answer 88

Epigenetics is the study, in the field of genetics, of cellular and physiological phenotypic trait variations that are caused by external or environmental factors that switch genes on and off and affect how cells read genes instead of being caused by changes in the DNA sequence. Hence, epigenetic research seeks to describe dynamic alterations in the transcriptional potential of a cell.

mitotically and meiotically heritable variations in gene expression that are not caused by changes in DNA sequence.

Reversible, post-translational modifications of histones and DNA methylation are examples of epigenetic mechanisms that alter chromatin structure, thereby affecting gene expression.
• Histones are often acetyled or demethlyated for greater expression of a gene
• While deacetlyation or methylation of histone will repress a gene quite often

Key note They will NOT alter the underlying DNA sequence
• This differs them from genetic mechanisms

Question 89

Modification at the Histones

Answer 89

Typically the core histones of the nucleosomes (H2,H3,H4)
Typically these are modifications at the N-tails that carry a positive charge (DNA binding)
The histones can have a variety of modifications, such as acetylation, methylation, and phosphorylation (often at the Lysine tail of histone)

Can think of this as epigenetic factors binding to histone tails
Epigenetic modifications influence the histones
-Acetyl added= more open chromosome for expression
-Histone Deacetylase will repress
-Methyl on histone tail typically repressed

Question 90

DNA Methylation

Answer 90

Typically the CpG sites are methylated to silence a region of genes, with the methylation happening on the cytosine to form a 5-methylcytosine

CpG These are NOT Base pairs, but rather these are cytosine and guanines that line up next to one another, “CpG” is shorthand for “—C—phosphate—G—”,
• Typically, 70-80% of cytosines in CpG of mammals are methylated
• Recognized by protein MECP2, allowing a protein complex to bind and help silence the gene

MECP2—> MeCP2 protein binds to forms of DNA that have been methylated. The MeCP2 protein then interacts with other proteins to form a complex that turns off the gene. Methylation is a chemical alteration made to a “cytosine” (C) when it occurs in a particular DNA sequence
• Some believe that it is involved in activation in some cases
Rarely DNA Methylation will lead to increased expression

Resetting the CpG methylation It appears that extensive demethylation occurs during Germ cell development and in early stages of embryonic development
-Appears to need to reset chromatin environment to obtain totipotent and pluripotency (want cells to be able to differentiate during development)

Question 91

CpG

Answer 91

These are NOT Base pairs, but rather these are cytosine and guanines that line up next to one another, “CpG” is shorthand for “—C—phosphate—G—”,
• Typically, 70-80% of cytosines in CpG of mammals are methylated
• Recognized by protein MECP2, allowing a protein complex to bind and help silence the gene

Question 92

MECP2

Answer 92

MeCP2 protein binds to forms of DNA that have been methylated. The MeCP2 protein then interacts with other proteins to form a complex that turns off the gene. Methylation is a chemical alteration made to a “cytosine” (C) when it occurs in a particular DNA sequence

• Some believe that it is involved in activation in some cases
• Rarely DNA Methylation will lead to increased expression

Question 93

DNA methylation is established in the gametes

Answer 93

o During Gametogenesis (the creation of four haploid cells) the Methylation/Silencing Sites are RESET, then they are methylated based on being from Paternal or Maternal

• Without reset, possibility of passing on chromosomes that have a silenced or double of an important gene ( The father can accidentally give a maternal chromosome)
• With the reset, the gene is already imprinted and will be methylated again

Question 94

DNA methylation reset during:

Answer 94

DNA methylation is reversible so that it can be reset during gametogenesis to transmit the appropriate sex-specific imprint to progeny.

After DNA Methylation is established in Gametes Will be maintained in cells upon fertilization…
Somatic Cells will maintain the DNA methylated sites for the organism, except with reset during gametogenesis

Question 95

How is Methylation propagated with DNA Replication

Answer 95

o With DNA replication, it is semiconservative so there is a parent strand (old) and a daughter strand (nascent)
o The Parent strand will already have the necessary Methylated/silencing sites
o The daughter strand will need to rely on the enzyme Methyl Transferase to come in and add methyl groups

Methyl Transferase–> Will add methyl groups to the Cytosine of CpG, critical for Methyl propagation in somatic cells, especially for the Daughter strands in replication

Question 96

Methyl Transferase

Answer 96

Will add methyl groups to the Cytosine of CpG, critical for Methyl propagation in somatic cells, especially for the Daughter strands in replication

Question 97

Why is erasure and Resetting of the Methylation Patterns so critical during Gametogenesis?

Answer 97

o With normal erasure Makes sure that the haploid/gamete cells will give the correct, established methylated genes that need to be passed down
oWithout erasure/reset the haploid/gametes can have random chromosomes from the paternal or maternal line that are methylated

-Giving a random methylated gamete for fertilization can lead to Organisms having two silenced genes that are necessary, or having two genes that are possibly over-active

•“Embryos with no active copies, or two active copies, of imprinted genes would be produced at high frequencies”

Question 98

Prader Willi and Angelman Syndrome

Answer 98

o Both diseases are based on disorders involving Chromosome 15

o Prader Willi excessive eating, short stature, hypogonadism, a degree of intellectual disability
o Angelman Syndrome short stature, severe intellectual disability, seizures, spasticity

• Prader Willi and Angelman Syndrome are based on genetic regions that are methylated on chromosome 15

Prader Willi= PWS gene region
• Has five regions that are only expressed on the paternal chromosomes

Angelman Syndrome= AS gene region
• Has maternal-only expression of 2 genes

Question 99

Mechanism of Prader Willi Syndrome (How it’s Caused) The PWS gene is NOT being expressed: either due to deletion of gene, both PWS genes being silenced, or a mutation of PWS center

Answer 99

1) Deletetion of Paternal 15q 11-13 (70% of cases)
Majority of cases for Prader Willi is deletion of a gene on the 15 chromosome
Maternal’s PWS Gene is silenced, and the paternal PWS gene is silenced
• Not enough expression

2) Maternal Uniparental Disomy (28 % of cases)
There are two mother copies of the PWS gene, they are both silenced
Not enough gene expression with two silenced PWS gene

3) Mutation of imprinting center on PWS paternal allele (2 or less %)
- The imprinting Center is NOT working, the paternal genes are silenced

Question 100

• Mechanism causing Angelman Syndrome the AS gene is NOT being expressed either due to deletion of gene, both AS genes being silenced, or different mutations of AS center

Answer 100

1) Deletion of Maternal Gene 15q11 (70% of cases)
Missing the necessary genes to maternally express the AS region

2) Paternal Uniparental disomy (4%)
Having two paternal regions of AS, which are both silenced so no expressing of the necessary AS genes

3) Mutation of imprinting Center on AS maternal gene (8%)
The imprinted center is NOT working, no expression of the maternal AS region

4) Mutation of UBE3A on maternal sign (8%)

Question 101

How these breaks happen in Prader Willi and Angelman Syndome

Answer 101

Normally there is proper alignment and a normal recombinant event
There are common breaks due to uneven crossing over in the Chromosome 15
-Misalignment followed by recombination, can lead to deletions predominantly

Question 102

Prader Willis can result in from uniparental disomy

Answer 102

o Maternal Chromosmes 15 are disomic (there are two) while there is a normal single paternal 15 chromsome
2 maternal chromosomes + 1 paternal chromosome= 3 chromosome
1 needs to be kicked out to avoid trisomy

o With one chromosome being kicked out, particularly the male chromosome, will lead to two maternal chromosome 15s
Trisomy Rescue Kick out an extra chromosome to avoid trisomy, can lead to diesease

-With two maternal 15 chromosomes ,these will both have PWS gene silenced leading to Prader Willi syndrome

Question 103

The three major causes of Down’s syndrome:

Answer 103

1) Trisomy 21 An extra copy of the third chromosome
♣ is in 95% of the patients
♣ Risk increases drastically with mother in mid 30’s, more likely of disjunction

2) Unbalanced Translocation unbalanced translocation between chromosome 21 and some other acrocentric chromosome
♣ 3-4% of patients with down’s syndrome
♣ Important to check karyotype of parents

3) Mosaic Trisomy 21 mixture of cells that have trisomy 21 and other cells with Down’s Syndrome
♣ Due to many of the cells not being Trisomy, much of the phenotype will be normal
• Depends on what cells have Trisomy

Question 104

Prenatal Counseling Down’s Syndrome

Answer 104

With increasing age, there is greater chance of nondisjunction
o 1st trimester screening detection rate of roughly 82-87%
♣ Ultrasound measurement of nuchal folds + b-hCG (human chorionic gonadotropin) + PAPP-A (pregnancy-associated plasma protein A)

o 2nd trimester Screening detection rate of 80%
♣ quad screen - b-hCG (human chorionic gonadotropin), AFP (a-fetoprotein), unconjugated estriol, and inhibin level

o 1st Trimester screening + 2nd trimester screening= 95% detection rate
♣ good to combine both trimester tests

o Suspicion of DS based on 1st or second trimester screening can by confirmed by chromosome analysis via amniocentesis or CVS (chorionic villus sampling)

Question 105

Summary of the three major Trisomy

Answer 105

Down Syndrome (Trisomy 21) is the most common chromosomal abnormality seen in liveborn infants, with estimated incidence of 1/700 births. There are 3 kinds of trisomy which may be seen in liveborn infants: Trisomy 13, 18, and 21. The birth defects associated with Trisomies 13 and 18 are often life-threatening, and most babies with these diagnoses do not survive to see their 1st birthday. Each of these trisomies has their own unique appearance (phenotype) and associated anomalies.

Question 106

Physical Features Down’s Syndrome

Answer 106

o Growth parameters are usually normal
o midfacial hypoplasia (what does this mean?) smaller face
o upslanting palpebral fissures, epicanthal folds
(The slanted eye shape)
o small ears
o large-appearing tongue due to the midfacial hypplasia
o low muscle tone, increased joint mobility
o short fingers, transverse palmar crease, Vth finger incurving (clinodactyly), increased space between toes 1 and 2

Question 107

Common medical problems

for Down’s Syndrome

Answer 107

o Cardiac Issues Down’s Syndrome
♣ Seen in approximately 50% of patients with DS
♣ All types of anomalies may be present, but Atrioventricular Canal is common to DS
• Atrioventricular Canal A canal that forms between the different chambers of the heart, is a serious heart defect, common in Down’s Syndrome
♣ Echocardiogram in the newborn period is recommended

o Gastrointestinal Down’s Syndrome
approx 10-15% of infants may have structural anomalies
• Esophageal atresia
• Duodenal atresia
• Hirschsprung’s
–>Very important that the infant swallows Amniotic Fluid for development, this is difficult for Down’s Syndrome
♣ Many children with DS have functional GI issues
• Feeding problems – very common
• constipation - very common
• GERD - very common, gastro reflux
• Celiac Dz (recommended screen is TTG + IgA)

o Ophthalmologic (eye) problems Down’s syndrome
blocked tear ducts
myopia (near-sightedness)
lazy eye
Nystagmus – often indicative of vision problems
Cataracts – may present in the newborn period or develop during infancy; checking for a light reflex is important at every visit.

o Ear, Nose and Throat Problems Down’s Syndrome
♣ chronic ear infections
♣ Deafness – both sensorineural and conductive
♣ chronic nasal congestion
♣ enlarged tonsils and adenoids
♣ obstructive apnea – preschool age is a common time to present with this problem; it is also an issue in older children who develop obesity

o Endocrine Problems (autoimmune disorders) Down’s Syndrome
Thyroid dz – most commonly hypothyroidism which may be congenital or acquired
• Typically Hypothyroidism in Down’s syndrome
Insulin Dependent Diabetes
Alopecia Areata
reduced fertility (but normal puberty)

o	Orthopedic Problems
♣	hips
♣	joint subluxation
•	Asubluxationmay have different meanings, depending on themedicalspecialty involved. It implies the presence of anincomplete or partialdislocationof ajointororgan
♣	atlantoaxial subluxation

o Hematologic Issues
♣ Myeloproliferative disorder in the newborn
• Myeloproliferative disorders is the name for a group of conditions that cause blood cells – platelets, white blood cells and red blood cells – to grow abnormally in the bone marrow.
♣ increased risk of leukemia – 12-20x
♣ Iron deficiency anemia

Question 108

Developmental and Psychological Issues

Down’s Syndrome

Answer 108

o hypotonia effects gross motor development
-is a state of lowmuscle tone(the amount of tension or resistance to stretch in a muscle), often involving reduced muscle strength.
o Spectrum of intellectual disability – average is mild-moderate disabilities
♣ IQ for normal person is between 90-110, anything bellow 70 IQ is intellectual disability
Down’s syndrome= Roughly a IQ of 50
o speech problems
♣ importance of sign language
♣ Difficult with face being small for tongue
o Seizures, especially infantile spasms

o depression
o early Alzheimer’s 21 trisonomy carries the Beta-cutting gene that creates the beta cut of amyloid protein
o Autism – 1/10 patients with DS

Question 109

Lysosome Storage Diseases

Answer 109

Lysosomes are the recycling centers of cells - they break down unwanted substances into simpler ones so that the cell can use them to make new material or expel them. If the enzyme that breaks down the waste product is lacking, waste builds up in the cells, eventually undermining its proper function, this can lead to serious health problems. There are at least 40 known lysosomal storage diseases.

Question 110

Gaucher disease background

Answer 110

In normal function body, most of the erythrocytes are removed via macrophage cells
♣ 90% of the cells containing the phagocytic macrophages reside in the Liver and in the Spleen

o Red blood Cells are broken down in the vacuoles of the macrophages a portion of the red blood cells is lipid that must be broken down
♣ The macrophages have the extensive enzymatic machinery to break down these lipids and part or this machinery is the enzyme Glucocerebrosidase

o Without the Glucocerebrosidase enzyme, cells will accumulate with the glucose lipid Glucocerbroside Form the cigarette looking cells called Gaucher Cells
♣ These macrophages are then called Gaucher cells which are stored/stuck mainly in liver, spleen and bone marrow.

o Without the Glucocerebrosidase enzyme, the cells will accumulate with glucocerebroside

Question 111

Patients lack enzyme called Glucocerebrosidase

Answer 111

o is an enzyme with glucosylceramidase activity that is needed to cleave, by hydrolysis, the beta-glucosidic linkage of the chemical glucocerebroside, an intermediate in glycolipid metabolism. It is localized in the lysosome
♣ Mutations in the glucocerebrosidase gene cause Gaucher’s disease, a lysosomal storage disease characterized by an accumulation of glucocerebrosides. A related pseudogene is approximately 12 kb downstream of this gene on chromosome 1. Alternative splicing results in multiple transcript variants encoding the same protein
♣ Leads to accumulation of glucocerebroside in macrophage cells

Question 112

Three different Types of Gaucher Disease (I, II, and III)

Answer 112

Gaucher Disease Type I
The most common type accounting for approximately 9 in every 10 cases; most prevalent in the Ashkenazi Jewish population. Patients are usually diagnosed in their late twenties or early thirties, although the disease can become apparent at any age
♣ Common Symptoms:
• Includes hepatomegaly (enlarged liver), fatigue, anemia
• Weakened bones, osteopenia (thinned bones) and osteoporosis
♣ Remember, can become apparent at any age (usually around 6-80 years old)

Gaucher Disease Type 2
The acute neuropathic form is the most severe and rarest form of GD, with an incidence of 1 in 100,000 live births. Typically, signs and symptoms appear when the infant is three to four months old. There are brainstem abnormalities and serious and rapidly progressing brain damage. Most patients with this form do not survive beyond three years of age.
♣ Symptoms: All those possible in Type 1 (anemia, fatigue, hepatomegaly, osteoporosis) plus
• Mental retardation
• Apnea - breathing stops temporarily during sleep
• Dementia, Seizures, Rigidity
♣ Remember, most patients die in infancy, usually discovered within the first year or two

Gaucher Disease Type 3
This form is rarer than Type 1, but more common than Type 2. Signs and symptoms usually appear during childhood or teenage years and progresses more slowly that Type 2. Brain damage is possible, but not as severe as in Type 2.
• Brain damage is possible, but not as severe as in Type 2. Signs and symptoms may include:
• All those possible in Type 1, plus. Mental retardation Dementia, Convulsions
♣ Usually discover 2-60 years old

Question 113

Diagnosed by checking levels of glucocerebrosidase – activity should be above 30%

Answer 113

The doctor identifies some signs and symptoms and suspects Gaucher’s disease, or if there is a family history, tests to check levels of the enzyme glucocerebrosidase will be recommended. The enzyme activity is also one way that may help to distinguish the three types of Gaucher disease. Genetic tests are also possible.

Blood test - in order to determine whether levels of glucocerebrosidase are too low. Carriers have lower-than-normal levels, but higher than those with Gaucher’s.
• Blood testing is ultimately seeing if Glucocerebrosidase Activity is below 30%

Genetic testing - the test looks for the four most common genetic mutations - N370S, L444P, 84gg and IVS2[+1] - as well as some less common ones. This test is not 100% reliable because we probably have not yet identified all the genetic mutations associated with Gaucher’s disease.

Prenatal screening - women who are carriers of the of one faulty GBA gene and want to know whether their fetus has Gaucher’s disease can have:
• Amniocentes - cells in the amniotic fluid are tested
• Chorionic villus sampling - tissue from the placenta is tested

Question 114

Gaucher Disease Genetic

Answer 114

Gaucher disease turned out to be an “autosomal recessive” genetic disease. This means that both of the glucocerebrosidase genes a person inherits - one from the mother and one from the father – must be mutated for the person to have the disease.
-Although Gaucher carriers will not have symptoms of the disease themselves, the odds are 50:50 that the “Gaucher gene” will be passed on to each of their children.

Gaucher disease prevalent in Ashkanazi jewish population
o There is a 1/15 chance of people in the AJ population being carriers

Question 115

Guacher Disease Treatment

Answer 115

• Primary treatment method is giving people the enzyme replacement therapy
o For those with type-I and most type-III, enzyme replacement treatment with intravenous recombinant glucocerebrosidase can decrease liver and spleen size, reduce skeletal abnormalities, and reverse other manifestations.[20][21] This treatment costs about US$200,000 annually for a single person and should be continued for life. The rarity of the disease means dose-finding studies have been difficult to conduct, so controversy remains over the optimal dose and dosing frequency.[16] Due to the low incidence, this has become an orphan drug in many countries, meaning a government recognizes and accommodates the financial constraints that limit research into drugs that address a small population.

Question 116

Pharmacogenetics:

Answer 116

The genomic approach to pharmacogenetics, is concerned with the assessment of common genetic variants in the aggregate for their impact on the outcome of drug therapy.
-Variable response due to multiple loci across the genome

The Codeine Metabolism example
o Baby found dead due to not only taking codeine (which turns into morphine from CYPD2, but also mom’s breast milk was delivering excessive amounts of morphine

Question 117

Pharmacokinetics:

Answer 117

• the rate at which the body absorbs, transports, metabolizes, or excretes drugs or their metabolites.
o Describes absorption, distribution, metabolism and excretion of drugs (commonly referred to as ADME)
Genetic Examples: Cytochrome P450, glucuronyltransferase, thiopurine methyltransferase

o Phase I : polar group added (exposed) solubilize
o Phase II : conjugation reaction (sugar/ acetyl group) detoxify

Question 118

Pharmacokinetics is broken down further into two basic ways that drugs are metabolized through biotransformations:

Answer 118

o Phase I (simplified): attach a polar group onto the compound to make it more soluble; usually a hydroxylation step
o Phase II (simplified): attach a sugar/acetyl group to detoxify the drug and make it easier to excrete

Question 119

The CYP450 Complex:

Answer 119

The CYP450 genes The cytochrome P450 (CYP450) genes encode important enzymes that are very active in the liver and to a lesser extent in the epithelium of the small intestine. CYP450 enzymes metabolize a wide number of drugs. See Figure (18-3) of the CYP genes involved in Phase I metabolism.

o 3 main families ( CYP1, CYP2, CYP3)
♣ 6 main genes (CYP1A1, CYP1A2, CYP2C9, CYP2C19, CYP2D6, and CYP3A4) Phase I for ~90% of common drugs
• The CYP450 genes will metabolize many products that arrive in the liver and the small intestines via Phase 1 pharmacokinetics
o Phase 1 Adding a hydroxyl or polar group to make a reactant soluble for the body

o While most CYP genes are important in the rate of inactivation of a drug, in some cases the CYP gene(s) is required to activate a drug.
♣ example of this is CYP2D6 activity being necessary to convert codeine (inactive, almost no analgesic effect) to morphine (active with a potent analgesic effect).

Question 120

How mutations affect the rate of enzymes in the body

Answer 120

Mutation that leads to no activity or less enzyme activity: (Left Graph)
o	Framing/ frameshift
o	Splicing
o	 nonsense mutation (early stop), 
o	Missense mutation

Mutations that lead to excessive activity (Enzymes are clearing out all drugs): on right graph
o Increase Allele copies or extra genes of the enzyme
♣ More production of that enzyme
o Missense mutation (some cases)

Question 121

Sex Chromosome Aneuploidy

Answer 121

Disorders that don’t have the normal number of X or Y chromosomes
X chromosome aneuploidy

• 46,XX
• 47,XXX
• 47,XXY
• 48,XXYY

Important Note: No matter how many X chromosomes, there is still only one active X chromosome, the rest are inactivated

Question 122

Turner Syndrome

Answer 122

Turner Syndrome  45,XO
Signs at birth
•	Prenatal cystic hygroma
•	Webbed neck
•	Puffy hands & feet
•	Heart defects like coarctation of the aorta
•	Short stature
•	Normal intelligence

♣ Infertility due to non-functioning ovaries
♣ Hormone dysfunction Need hormone therapy
♣ Distinctive traits such as low set ears, broad chest
♣ Occurs in 1/2,500 newborn girls

Question 123

Kleinfelter Syndrome

Answer 123

Kleinfelter Syndrome 47, XXY  male has an extra X chromosome
♣	Can be seen in childhood
•	Learning disabilities
•	Delayed speech and language
•	Tendency towards being quiet
♣	Tall stature
♣	Small testes
♣	Reduced facial and body hair
♣	Infertility
♣	Hypospadias
♣	Gynecomastia
♣	Occurs in 1/500 – 1/1000 newborn boys

Question 124

Jacobs Syndrome

Answer 124

o	Jacobs Syndrome 47,XYY
An extra Y disorder
♣	Learning disabilities
•	Speech delays
•	Developmental delays
♣	Behavioral and emotional difficulties
•	Autism spectrum disorders
♣	Tall stature
♣	Occurs in 1/1000 newborn boys

Question 125

Triple X Syndrome

Answer 125

o	Triple X Syndrome 47,XXX - Extra X chromosome in females
♣	May have tall stature
♣	Increased risk of
•	Learning disabilities
•	Delayed speech
•	Delayed motor milestones
•	Seizures
•	Kidney Abnormalities
♣	Occurs in 1/1000 newborn girls

Question 126

Chromosomes are the major determining factor for gonads and Sex

Answer 126

o Y-Chromosome will allow male creation
♣ Penis, seminal vesicles & prostate gland
• Also musculature, hair, vocal cartliage

o X-chromosomes without a Y chromosome= female creation
♣ Of the two X chromosomes, only one is active
♣ Vagina, cervix, uterus, fallopian tubes & mammary glands

o Both ovaries and testicular tubes/structures are available during fetal state, one will remain and the other set will degenerate

Question 127

Embryology of Dimorphic Human Reproductive Organs

Answer 127

o 4th week of conception Primordial germ cells form in wall of yolk sac
o 5th week of conception Coelomic epithelium becomes genital ridge
o 6th week of conception Primordial germ cells migrate to the dorsal mesentary of the hindgut and enter the undifferentiated gonad
Epithelial cells of gonadal ridge proliferate and form primitive sex cords

• 7th week con conception = Where the differentiation Begins
o 7th week for males Differentiation of genital ridge into:
• Sertoli cells - eventually produce sperm
• Leydig cells – interstitial cells
♣ 8th week of conception
• Leydig cells begin producing testosterone
• Sertoli cells begin producing Anti-Mullerian Hormone (AMH)
♣ Primitive sex cords differentiate into:
• Testis cords & rete testis, eventually to become seminipherous tubules during puberty
o Female 7th and 8th week of conception
♣ In the absence of Y chromosome (the SRY gene) & in the presence of 2X chromosomes
• Primitive sex cords dissociate into irregular clusters
♣ Medullary (primitive) cords regress and cortical (secondary) cords are formed:
• Destined to become follicular cells of the ovary
• Follicular cells will eventually surround an oogonium which together are the primary ovarian follicle

Question 128

7th week for males and on

Answer 128

o 7th week for males Differentiation of genital ridge into:
• Sertoli cells - eventually produce sperm
• Leydig cells – interstitial cells

8th week of conception
• Leydig cells begin producing testosterone
• Sertoli cells begin producing Anti-Mullerian Hormone (AMH)

Primitive sex cords differentiate into:
• Testis cords & rete testis, eventually to become seminipherous tubules during puberty

Question 129

Female 7th and 8th week of conception

Answer 129

Female 7th and 8th week of conception
In the absence of Y chromosome (the SRY gene) & in the presence of 2X chromosomes
• Primitive sex cords dissociate into irregular clusters

Medullary (primitive) cords regress and cortical (secondary) cords are formed:
• Destined to become follicular cells of the ovary
• Follicular cells will eventually surround an oogonium which together are the primary ovarian follicle

Question 130

Genital ducts for males and females:

Answer 130

There are initially two pairs of ducts in the fetus (both males and females)
Mesonephric (Wolffian) duct Become male structures
• Wolfes= Males RAAAWWWWRRRR

Paramesonephric (Mullerian) duct Become female structures

Signaling for both ducts:
SRY Gene and SOX9 gene on Y chromosome create
Anti-Mullerian Hormone Leads to Anti-Mullerian Hormone
o Causes regression of paramesonephric duct

X chromosome produces the WNT4 protein and the DHH gene to allow female differentiation
o Allows Paramesonephric duct development and ovaries

Question 131

Mesonephric (Wolffian Duct)

Answer 131

Mesonephric (Wolffian Duct) Results in Male Structure
o 4major genes for male differentiation

SRY gene and SOX9 Gene on the Y-chromosome
• Both transcription factors
• Responsible for production of Anti-Mullerian Horomone (aka Mullerian Inhibitory Substance - MIS)
• Causes regression of the paramesonephric duct

FGF9 Gene
• Chemotactic factor causes tubules from mesonephric duct to penetrate the gonadal ridge
• Essential for differentiation of the testis

SF1/ NR5A1
• Stimulates differentiation of the Sertoli & Leydig cells

o Under the influence of testosterone, mesonephric ducts elongate to form the:
♣ Epidymis
♣ Seminal vesicles
♣ Vas deferens

Question 132

SRY gene and SOX9

Answer 132

SRY gene and SOX9 Gene on the Y-chromosome
• Both transcription factors
• Responsible for production of Anti-Mullerian Horomone (aka Mullerian Inhibitory Substance - MIS)
• Causes regression of the paramesonephric duct

Question 133

FGF9 Gene

Answer 133

• Chemotactic factor causes tubules from mesonephric duct to penetrate the gonadal ridge
• Essential for differentiation of the testis

Question 134

Paramesonephric (Mullerian) Ducts

Answer 134

Lead to Female structure though several genes and proteins:

WNT4 protein
• Extracellular signaling factor responsible for differentiation of the ovary
• Inhibited by SOX9 from the Y chromosome
• Is coactivated by RSPO1 gene

DHH gene
• A nuclear hormone receptor
• Up-regulated by WNT4
• Down-regulates SOX9

RSPO1 gene
• Coactivator of the WNT pathway

o	Under the influence of estrogens (from maternal and placental sources), the following structures are formed
♣	Uterus
♣	Cervix
♣	Broad ligament
♣	Fallopian Tubes
♣	Upper 1/3 of the vagina

Question 135

WNT4 protein and DHH gene

Answer 135

WNT4 protein
•	Extracellular signaling factor responsible for differentiation of the ovary
•	Inhibited by SOX9 from the Y chromosome
•	Is coactivated by RSPO1 gene
DHH gene
•	A nuclear hormone receptor
•	Up-regulated by WNT4
•	Down-regulates SOX9

Question 136

Development of External Gentiles

Answer 136

At 3 weeks, originating from mesenchymal cells in the primitive streak, cells migrate to form a genital tubercle and genital swellings
o Both originate from the urogenital sinus

Male Structures:
• Androgen exposure (in this case Dihydrotestosterone) from the testis results in the formation of the following:
o Penis
o Scrotum
o Location of the urethral opening at the tip of the penis

Female Structures: 
•	Estrogen exposure resulting from maternal and placental sources results in the formation of the following:
o	Clitoris
o	Labia majora and minora
o	Lower 2/3 of the vagina

Question 137

Androgen Insensitivity Syndrome (AIS)

Answer 137

o Has the chromosome characteristics: 46, XY (normal)
o AIS is an X-linked gene, Autosomal Recessive
o Mutation causes abnormality of the androgen receptor
♣ Even though the body makes androgens (testosterone), it doesn’t necessarily recognize or respond to it
♣ Phenotypes range from mild under-virilization (Partial AIS) to full sex reversal (Complete AIS)
o Used to be called testicular feminization

Question 138

5-Alpha Reductase Deficiency

Answer 138

5-Alpha Reductase Deficiency
46, XY
o X-linked gene, AR
o Mutation causes decreased ability of the body to convert testosterone to dihydrotestosterone
o Phenotype shows undervirilized male with increased virilization at the time of puberty

Question 139

Disorders associated with the SRY gene (on the Y chromosome)

Answer 139

o 46, XY or 46, XX
o Y-linked gene
o Deletion or absence of the gene results in full 46, XY sex reversal and a phenotypically normal female
o Ectopic presence of the SRY gene in a 46, XX individual results in a phenotypically normal male
o Mutations in the SRY gene in a 46, XY individual results in decreased or absent production of Anti Mullerian hormone & under virilization of a male

Question 140

Denys-Drash & Frasier Syndrome

Answer 140

o	Sex reversal with 46, XY
o	Due to mutations in the WT1 gene
o	Both cause different types of chronic kidney disease
♣	Diffuse mesangial sclerosis
♣	Focal segmental glomerulosclerosis

o Increased risk for Wilms Tumor
o WT1 – transcription factor for SRY gene
♣ SRY gene is not functioning properly (the Y gene)

Question 141

Congenital Adrenal Hyperplasia

Answer 141

o Ambiguous genitalia in 46, XX
o 21-hydroxylase deficiency

o Complicated by salt wasting in the first few weeks of life and with times of metabolic stress
♣ Decreased sodium and chloride
♣ Increased potassium