Essential of GENETICS

Question 1

Mendel derived the following three postulates, or principles of inheritance…..

Answer 1

1. unit factors in pairs:

genetic characters are controlled by unit factors existing in pairs in individual organisms. Each diploid individual receives one factor from each parent

2. DOMINANCE/RECESSIVENESS

3. SEGREGATION​

Question 2

The testcross.

Answer 2

if a tall plant of genotype DD is testcrossed with a dwarf plant, which must have the dd genotype, all offspring will be tall phenotypically and Dd genotypically. However, as shown in Figure 3–4(b), if a tall plant is Dd and is crossed with a dwarf plant (dd), then one-half of the offspring will be tall (Dd) and the other half will be dwarf (dd).

Therefore, a 1:1 tall/dwarf ratio demonstrates the heterozygous nature of the tall plant of unknown genotype. The results of the testcross reinforced Mendel’s conclusion that separate unit factors control traits.

Question 3

The key to identifying a pedigree that reflects a dominant trait is that all affected offspring will have >>………………….

Answer 3

The key to identifying a pedigree that reflects a dominant trait is that all affected offspring will have a parent who also expresses the trait

Such as: Huntington disease.

Question 4

albinism is inherited as ………………

Answer 4

albinism is inherited as a recessive disease where the pigment melanin is obstructed. Characteristic of a situation
in which a parent has a rare recessive trait, the trait, “disappears, “ in the offspring of the next generation. Assuming excessiveness,

Question 5

Tay–Sachs disease (TSD)

Answer 5

recessive disease

caused by defect in production of enzyme, hexosaminidase A ⇒(normally
found in lysosomes within cells⇒ needed to break down the ganglioside GM2,►lipid storage disorder→

a lipid component of nerve cell membranes

Infants with TSD are unaffected at birth and appear to develop normally until they are about 6 months old.

Then, a progressive loss of mental and physical abilities occurs. Afflicted infants eventually become blind, deaf, mentally retarded, and paralyzed,

Question 6

Huntington ( HTT) disease

Answer 6

a Dominant lethal allele cause ⇒ brain disease.

passed down in families from generation to generation.

caused by a genetic defect on chromosome 4.

⇒Genetic defect caused by triple repeat (CAG repeat).

⇒cause excessive production of

Glutamine► Excitotoxicity►…Neuronal death in basal Ganglion

usually a late onset, typically at about age 40, By that time,
the affected individual may have produced a family, and each of the children has a 50% probability of inheriting the lethal allele.

Question 7

For the following pedigree, predict the mode of inheritance and
the resulting genotypes of each individual. Assume that the
alleles A and a control the expression of the trait.

Answer 7

since there is skip generation = no expression of trait in parent while, present in F1

Question 8

in this figure Draw all possible conclusions concerning the mode of inheritance of the trait expressed in each of the following limited pedigrees.

(Each case is based on a different trait.)

Answer 8

There are two possibilities. Either the trait is dominant,
In which case I-1 is heterozygous, as are II-2 and II-3, or the trait is recessive and I-1 is homozygous and I-2 is heterozygous.
Under the condition of recessiveness, both II-1 and II-4 would be heterozygous; II-2 and II-3 would be homozygous.

Question 9

in this figure Draw all possible conclusions concerning the mode of inheritance of the trait expressed in each of the following limited pedigrees.

(Each case is based on a different trait.)

Answer 9

Recessive: Parents Aa, Aa

Question 10

in this figure Draw all possible conclusions concerning the mode of inheritance of the trait expressed in each of the following limited pedigrees.

(Each case is based on a different trait.)

Answer 10

Recessive: Parents Aa, Aa

Question 11

wild-type allele

Answer 11

normal type allele, The allele that occurs most frequently in a population,

Wild-type alleles are responsible for the corresponding wild-type phenotype and are the standards against which all other mutations occurring at a particular locus are compared.

Question 12

null allele.

Answer 12

if a gene is responsible for the synthesis of a specific enzyme, ¿ a mutation in that gene may ultimately change the conformation of this enzyme and reduce or eliminate its affinity for the substrate.

Such a case is designated as:

a loss-of-function mutation. ⇒If the loss is complete, the mutation has resulted in what is called ► a null allele.

Question 13

Gain-of-function mutations

Answer 13

mutations may enhance the function of the wild-type product.

generally……result in dominant alleles since one copy in a diploid organism is sufficient to alter the normal phenotype.

for example —excess gene product. The result is the creation of a cancerous cell.

Question 14

Neutral mutations

Answer 14

mutation of the gene product presents no change to either
the phenotype or the evolutionary fitness of the organism.

Question 15

triple repeat disorders

Answer 15

Triplet repeat diseases (TRDs)

caused by pathogenic expansions of trinucleotide sequence repeats within coding and non-coding regions of different genes. They are typically progressive, very disabling and frequently involve the nervous system.

Question 16

Codominant pattern inheritance

Answer 16

Codominance, as it relates to genetics, refers to a type of inheritance in which two versions (alleles) of the same gene are expressed separately to yield different traits in an individual

example ABO blood-type.

Question 17

MN Blood Group system in Humans

Answer 17

The MN blood group system is under the control of an autosomal locus found on chromosome 4,

The blood-type is due to a glycoprotein present on the surface of red blood cells, which behaves as a native antigen, Humans are diploid, so three combinations are
possible, each resulting in a distinct blood type:

with two alleles designated LM and LN.

GenotypePhenotype
L^M L^MM
L^M L^NMN
L^N L^N N

Question 18

ABO Blood Group System

Answer 18

In these assignments the IA and IB alleles are dominant
to the i allele but are codominant to each other.

Genotype Antigen Phenotype

I^A I^A A
A
I^A i A

I^B I^B B
B
I^B i ^B

I^A I^B A, B AB

i i Neither O

Question 19

Bombay Phenotype

Answer 19

Named for the city in which it was first discovered,

the “Bombay phenotype” describes individuals whose RBCs lack the H antigen.

Because the A and B antigens cannot be formed without the H antigen precursor.

recessive mutation of FUT1 gene (encoding an enzyme, fucosyl transferase),.

while they may have I^A or I^B alleles.⇔ neither the A nor B
antigen can be added to the cell surface,⇒ thus women of Bombay Phenotype woman in Bombay expressed blood type O,( she is not genetically true) but she was able to pass the I^B allele to her children as shown in the figure.

Question 20

Marfan syndrome

Answer 20

Autosomal dominant mutation in the gene encoding the connective tissue protein fibrillin.

this protein is widespread in many tissues in the body. SO

this single gene defect result in multiple body effects.

such as lens dislocation, increased risk of aortic aneurysm, and lengthened long bones in limbs.

Question 21

Explain the pattern pf inheritance in color blindness disease, where the mother in generation I passes the trait to all her sons but to none of her daughters.

Answer 21

X-Linked disease, transmitted from mother to 50% of her sons and non of daughters affected.

e.g.: Duchenne muscular dystrophy.

It normally occurs only in males

Question 22

indicate the pedigree inheritance pattern

Answer 22

This pedigree is consistent with an X-linked recessive trait because

the father would contribute an X chromosome carrying the a mutation to the aa daughter.

The mother would have to be heterozygous Aa.

Question 23

explain the pedigree

Answer 23

X-Linked recessive disorder

This pedigree is consistent with an X-linked recessive trait because the mother could be Aa and transmit her a allele to her one son (a/Y) and her A allele to her other son (A/Y).

Question 24

IS This pedigree (aa mother) consistent with an X-linked mode of inheritance?

Answer 24

This pedigree is not consistent with an X-linked mode of inheritance because the aa mother has an A/Y (unaffected) son.

**dominant pattern of inheritance is more likely

Question 25

Myotonic dystrophy (DM)

Answer 25

genetic disorders that cause progressive muscle loss and weakness.
Dominant disorder
Mutation of genes on chr-19
congenital cause of male infertility–High GnRH, Hihg FSH, High LH and Low T.–
Testicular atrophy —- infertility.

Question 26

X-Linked Dominant disease

Answer 26

X-linked dominant inheritance refers to genetic conditions associated with mutations in genes on the X chromosome.

A single copy of the mutation is enough to cause the disease in both males (who have one X chromosome) and females (who have two X chromosomes)

Question 27

sex-influenced inheritance

Answer 27

Sex-limited and sex-influenced inheritance

occurs when the sex of the organism affects the phenotype controlled by a gene located on an autosome.(not X or Y)

Example: baldness in humans,

Question 28

Lesch–Nyhan syndrome

Answer 28

• *Lesch–Nyhan syndrome**, inherited as an X-linked
• *recessive disease**,

characterized by abnormal nucleic acid metabolism (biochemical salvage of nitrogenous purine bases),

leading to the accumulation of uric acid in blood and
tissues, ►mental retardation, palsy, and self-mutilation of then lips and fingers.

The disorder is due to a mutation in the gene
encoding hypoxanthine-guanine phosphoribosyl transferase (HGPRT).

Newborns are normal for 6-8 months prior
to the onset of the first symptoms.

Question 29

Duchenne muscular dystrophy (DMD)

Answer 29

X-linked recessive disorder

associated with progressive muscular wasting

It is not usually diagnosed until the child is 3-5 years old.

even with treatment the disease is often fatal in the early 20s.

Question 30

pedigrees shown below. Of these combinations

• X-linked recessive,
• X-linked dominant
• autosomal recessive,
• autosomal dominant,

which modes of inheritance can be absolutely ruled out in each case?

Answer 30

For both pedigrees, X-linked recessive and autosomal
recessive remain possible, provided that the maternal
parent is heterozygous in pedigree (b).

Pedigree (a)

Autosomal dominance seems unlikely in pedigree (a), since at least half of the offspring should express a dominant trait expressed by one of their parents. However, while it is true that ♦ if the affected parent carries an ** autosomal dominant gene heterozygously,**

► النتيجة لابد تكون each offspring has a 50% chance of inheriting and expressing the mutant gene, the sample size of four offspring is too small to rule out this possibility.

In pedigree (b)

,► autosomal dominance is clearly possible.

In both cases, rule out X-linked dominance because the female offspring would inherit and express the dominant allele, and they do not express the trait in either pedigree.

Question 31

Hemophilia

Answer 31

Hemophilia is usually an inherited bleeding disorder in which the blood does not clot properly. This can lead to spontaneous bleeding as well as bleeding following injuries or surgery.

Hemophilia → X-linked recessive hereditary disorder

*carrier mother transmit mutation to 50% of her sons(affected), while 50% of daughters will be carrier.

**affected father ⇒ 50% of daughters will be carrier, while no transmition to his sons

Question 32

Cryptorchidism…….

Hermaphrodite……….

Answer 32

cryptorchidism ⇔undescended testicle

hermaphrodite is an individual or organism that has both kinds of reproductive organs and can produce both gametes

note that

Embryos ►by the 5th week of gestation⇒ gonadal ridge tissue can develop to form male or female gonads. thus this period human embryo may considered as hermaphroditic.

Question 33

• arround the 7th week of gestation

cortex and medulla form………………..while Wolffian and Mullerian form ………….

Answer 33

primordial germ cells► migrate to genial ridges,

where an outer cortex and inner
medulla form

The cortex is capable of developing into an ovary, while the medulla may develop into a testis.

Wolffian and Müllerian ducts exist in each embryo.
Wolffian ducts differentiate into other organs of the male
reproductive tract, while

Müllerian ducts differentiate into structures of the female reproductive tract.

Question 34

Identify each constitution of the Y chromosome

Answer 34

• PAR: Pseudoautosomal region
• SRY: Sex-determining region Y
• MSY: Male-specific region of the Y

note the small size of Y:

Y chromosome has ≈ 75 genes,

compared to 900–1400 genes on the X.

Question 35

what is the main differences between MSY and PARs regions on Y chromosomes?

Answer 35

pseudo-autosomal regions (PARs)

that share homology with regions on the X chromosome and synapse and recombine with it during meiosis.
and is critical to segregation of the X and Y chromosomes during male gametogenesis.

Male-specific region of the Y (MSY) ► also called nonrecombining region of the Y (NRY).

about 95% of the Y chromosome, does not synapse or recombine with the X chromosome.

MSY consist of *Euchromatin contain specific genes (sex-determining region Y SRY) while the heterochromatine region Laking genes.

Question 36

To who attribute that? X chromosomes is inactive in the cells of females.

Answer 36

Barr bady: to stabilize (equivelant) the dosage of genetic information that can be expressed in males and females

• In mammals, female somatic cells randomly inactivate one of two X chromosomes during early embryonic development, a process important for balancing the expression of X chromosome-linked genes in males and females.

Question 37

Lyon hypothesis

Answer 37

Lyon hypothesis

a Barr body inactivation of an X chromosome occurs randomly early in development,

and thereafter all progeny cells have the same

X chromosome inactivated as their progenitor,

According to this hypothesis, female mammals isolate(hide) one X chromosome in each of their cells during the early stages of development.

Question 38

a clone

Answer 38

in vitro culture of cells
derived from a single cell is called a clone

Question 39

X-inactive specific transcript XIST

Answer 39

[X-inactive specific transcript (XIST) ]

is a critical gene for X-inactivation.

“about the mechanism of X-inactivation”

RNA that is transcribed from the XIST gene► RNA product is quite large and does not encode a protein, and thus is not translated.

Question 40

Many reptiles show temperature-dependent effects on sex determination. True or false?

Answer 40

True

Although specific sex chromosomes determine genotypic sex in many reptiles, temperature effects on genes involved in sexual determination affect whether an embryo develops a male or female phenotype.

Question 41

Under what circumstances can a girl have a karyotype of
46, XY?

Answer 41

Female external genitalia in patients with 46,XY

refers to phenotypic females with a male genotype

Disorders of sex development (DSD) due to impaired androgen synthesis or action.

مهم جدا تفاصيل الصورة المرفقة

Question 42

if any, treatments can be given to XO Patient
Will she able to
have children?

Answer 42

Most women with Turner’s syndrome have ovarian dysgenesis; therefore, they are usually infertile, and in very rare cases have spontaneous menses followed by early menopause. Only 2% of the women have natural pregnancies, with high rates of miscarriages, stillbirths and malformed babies.

• ovarian tissue cryopreservation may involved treatment

Question 43

48 XXXX

Answer 43

oogenesis defect** non-disjunctions at both meiotic divisions,

the presence of additional X chromosomes appears to disrupt the delicate balance of genetic information essential to normal female development.

47XXX, 48 XXXX, 49 XXXXX

extra X increase severity

Question 44

true or false

X chromosomes play no role in sex determination, while the Y chromosome causes maleness and its absence causes femaleness?

Answer 44

true

Question 45

Drosophila utilizes a different sex determination
mechanism than mammals, even though it has the same sex-chromosome compositions in males and females?

Answer 45

true

The ratios of X chromosomes to sets of autosomes determine the sex male or female ,……

Question 46

nondisjunction in human female gametes can give rise to Klinefelter and Turner syndrome offspring following fertilization by a normal male gamete.?

Answer 46

Nondisjunction can occur during anaphase of ⇒mitosis, meiosis I, or meiosis II.

During anaphase, sister chromatids (or homologous chromosomes for meiosis I), will separate and move to opposite poles of the cell, pulled by microtubules.

In nondisjunction, the separation fails to occur causing both sister chromatids or homologous chromosomes to be pulled to one pole of the cell.

Mitotic nondisjunction can occur due to the inactivation of either topoisomerase II, condensin, or separase. This will result in 2 aneuploid daughter cells, one with 47 chromosomes (2n+1) and the other with 45 chromosomes (2n-1).

Nondisjunction in meiosis I

occurs when the tetrads fail to separate during anaphase I. At the end of meiosis I, there will be 2 haploid daughter cells, one with n+1 and the other with n-1.

Both of these daughter cells will then go on to divide once more in meiosis II, producing 4 daughter cells, 2 with n+1 and 2 with n-1.

Nondisjunction in meiosis II results from the failure of the sister chromatids to separate during anaphase II.

Since meiosis I proceeded without error, 2 of the 4 daughter cells will have a normal complement of 23 chromosomes. The other 2 daughter cells will be aneuploid, one with n+1 and the other with n-1.

Question 47

Indicate the expected number of Barr bodies in interphase cells of individuals with

• (a) triple X syndrome (XXX),
• (b) XYY syndrome,
• (c) Klinefelter syndrome,
• (d) Turner syndrome, and
• (e) karyotype 48, XXXX.

Answer 47

number of Barr body = n(X) - 1

• a) 3-1=2
• b) 1-1=0
• c) 2-1=1
• d) 1-1=0
• e) 4-1=3

Question 48

Campomelic dysplasia (CMD1) is a congenital human
syndrome featuring malformation of bone and cartilage.
It is caused by an autosomal dominant mutation of SOX9 gene located on chromosome 17. Consider the following observations in sequence, and in each case, draw whatever appropriate conclusions are warranted.

(a) Of those with the syndrome who are karyotypically 46,XY, approximately 75% are sex reversed, exhibiting a wide range of female characteristics.

(b) The nonmutant form of the gene, called SOX9, is expressed in the developing gonad of the XY male, but not the XX female.
(c) The SOX9 gene shares 71% amino acid coding sequence homology with the Y-linked SRY gene.
(d) CMD1 patients who exhibit a 46,XX karyotype
develop as females, with no gonadal abnormalities.

Answer 48

(a) Something is missing from the male-determining system of sex determination at the level of the genes, gene products, or receptors, and so on, and the loss is correlated with CMD1.
(b) The SOX9 gene, or its product, is probably involved in male development. Perhaps it is activated by SRY.
(c) There is probably some evolutionary relationship between the SOX9 gene and SRY. There is considerable evidence that many other genes and pseudogenes are also homologous to SRY.
(d) Normal female sexual development does not require the SOX9 gene or gene product(s)

Question 49

SOX9

Answer 49

SOX9 is a transcription factor essential for both sex and skeletal development.

Transient expression of the Y chromosome gene SRY

Question 50

Describe the major differences between XO individuals in
Drosophila and those in humans.

Answer 50

In humans, the sex is determined by the presence or absence of the SRY gene located on the Y chromosome.

Therefore, XO individuals are females with ovaries, a uterus, and oviducts, but very few ova.

In Drosophila, the sex is determined by the balance of female determinants on the X chromosome(s) and male determinants on the autosomes, mediated by the Sxl gene. The Y chromosome does not determine sex, but is required for sperm production. Therefore, XO individuals are sterile males

In Drosophila it is the balance between the number of X chromosomes and the number of haploid sets of autosomes that determines sex. In contrast to humans, XO Drosophila are males and the XXY complement is female.

Question 51

A group of scientists developing an XX zygote in vitro are curious to see the impact of certain chemicals on the development of the said organism. They incubate the zygote with the help of testosterone and some transcription factors, which are usually produced by the activity of the Y chromosome. They discover that the zygote develops into a sterile female with masculinized reproductive organs. Explain why this happens.

Answer 51

The zygote develops masculinized characteristics because of the hormones and transcription factors that are produced by the activity of the Y chromosome after fertilization.

Even though the zygote is a female genetically, it has masculinized reproductive organs.

Question 52

Nondisjunction during the first and second meiotic divisions

Answer 52

Question 53

female with Turner syndrome (45,X) also expresses the X-linked trait hemophilia, as did her father.
Which of her parents underwent nondisjunction during meiosis, giving rise to the gamete responsible for the syndrome?

Answer 53

لاحظ X-Linked disease

If the father had hemophilia, it is likely that the Turner syndrome individual inherited the X chromosome from the father and no sex chromosome from the mother.

دليل انه الكروموسوم اكس الموجود مرتبط بالهيموفيليا من الاب بالتالي الكروموسوم الغير موجود بسبب الام

If nondisjunction occurred in the mother, either during meiosis I or meiosis II, an egg with no X chromosome can be the result. See the text for a diagram of primary and secondary nondisjunction.

Question 54

Cri du chat syndrome in humans

Answer 54

• results from the deletion of a small terminal portion in short arm of chromosome 5
• considered a case of segmental deletion (partial monosomy)
• genetic constitution designated as 46,5p-, meaning that the individual has all 46 chromosomes but that some or all of the p arm (the petite, or short, arm) of one member of the chromosome 5 pair is missing.
• the missing part contains the TERT gene.

Question 55

copy number variants (CNVs)

Answer 55

• Duplications of portions of genes, most often involving thousands of base pairs, occur on a regular basis.
• Differences in the number of large DNA sequences, are termed copy number variants (CNVs) and are found in both coding and noncoding regions of the genome.
• When individuals in the same species are compared.
• example: CNVs could be defined as regions of DNA at least 1 kb in length (1000 base pairs) that display at least 90% sequence identity.

Question 56

Why CNVs are of major interest in genetics?

Answer 56

CNVs appear to have both positive and negative impacts with many diseases in which the genetic basis is not yet fully understood.

For example, an association has been reported between CNVs and autism, the well-known neurodevelopmental disorder that impairs communication, behavior, and social interaction.

Question 57

………….is a type of chromosomal aberration in which a segment of a chromosome is turned around 180 degrees within a chromosome.

Answer 57

chromosome inversion

Question 58

types of inversion

Answer 58

• paracentric inversion.
• pericentric inversion.

If the centromere is not part of the rearranged chromosome segment كسر نهايات الكروموسوم وتبادلها, it is a paracentric inversion, as shown in Figure .

If the centromere is part of the inverted segment, it is
described as a pericentric inversion.

Question 59

differences between paracentric and pericentric inversion

Answer 59

if inversion NOT include the centromere ► Paracentric

f inversion include the centromere⇒ Pericentric inversion

Organisms with one inverted chromosome and one noninverted homolog are called ►inversion heterozygotes.

Question 60

Translocation

Answer 60

• movement of a chromosomal segment to a new location in the genome.
• Reciprocal translocation: exchange of segments between two nonhomologous chromosomes.
• in reciprocal translocation genetic information is not lost or gained.
• just only a rearrangement of genetic material
• In contrary if, non reciprocal translocation occurs also between nonhomologous chromosomes genetic information may lost or gained.

Question 61

Robertsonian translocation.

Answer 61

A Robertsonian translocation is one in which, effectively, the whole of a chromosome is joined end to end with another. This type of translocation involves only chromosomes 13, 14, 15, 21 and 22 ????

→because the ends of their short arms have similar repetitive DNA sequences that induces their fusion.

احد اهم الامثلة

familial Down syndrome. also called Translocation DS.

Question 62

true or false; Fragile-X Syndrome (FXS)

• Normal individuals have 6 - 54 CGG Repeats
• Carriers 55- 230 repeats
• More than 230 repeats lead to expression of (FXS)

Answer 62

true

the most common form of inherited mental retardation. This syndrome affects about 1/4000 males and 1/8000 females. All males bearing this X chromosome exhibit the syndrome, while about 60% of females bearing one affected chromosome exhibit the syndrome.

In addition to mental retardation, affected males and females have characteristic long, narrow faces with protruding chins and enlarged ears.
A gene that spans the fragile site is now known to be
responsible for this syndrome. Named FMR1 where In
FMR1, the trinucleotide sequence CGG (remember HD) is repeated in an untranslated area adjacent to the coding sequence of the gene (called the “upstream” region).

Question 63

Regarding FXS complete ⇒

(over 230 repeats) occurs only during the transmission of the gene by the …………… parent, not by the ………… parent.

Thus, a “carrier” male may transmit a stable chromosome to his ……………, who may subsequently transmit an unstable chromosome with an increased number of repeats to ………… Their grandfather was the source of the original chromosome.

Answer 63

(over 230 repeats) occurs only during the transmission of the gene by the maternal parent, not by the paternal parent.

Thus, a “carrier” male may transmit a stable chromosome to his daughter, who may subsequently transmit an unstable chromosome with an increased number of repeats to her offspring. Their grandfather was the source of the original chromosome.

Question 64

The Link between Fragile Sites and Cancer??

Answer 64

• fragile sites on autosomal chromosomes linked to lung cancer.

FHIT gene (standing for fragile histidine triad) and FRA3B located fragile sites on the p arm of chromosome 3.

As conclusion:

Fragile sites in human mitotic chromosomes on the X chromosome is associated with the most common form of inherited mental retardation,

while other autosomal sites have been linked to various forms of cancer.

Question 65

How can you explain that not all Down syndrome cases are due to nondisjunction?

Answer 65

Familial Down syndrome also called Translocation DS

inherited Down syndrome are due to a translocation between chromosomes 14 and 21.

bath parents are phenotypically normal, yet the translocation leads to trisomy 21 during meiosis.

Question 66

A couple planning their family are aware that through the past three generations on the husband’s side a substantial number of stillbirths have occurred and several malformed babies were born who died early in childhood.

The wife has studied genetics and urges her husband to visit a genetic counseling clinic, where a complete karyotype-banding analysis is performed.
Although the tests show that he has a normal complement of 46 chromosomes, banding analysis reveals that one member of the chromosome 1 pair (in group A) contains an inversion covering 70% of its length.

The homolog of chromosome 1 and all other chromosomes show the normal banding sequence.

(a) How would you explain the high incidence of past stillbirths?
(b) What can you predict about the probability of abnormality/normality of their future children?

(c) Would you advise the woman that she will have to bring each pregnancy to term to determine whether
the fetus is normal?

If not, what else can you suggest?

Answer 66

chromosomal abnormality is expected for offspring and PGT-is highly recommended

Question 67

How do we know that the extra chromosome causing Down
syndrome is usually maternal in origin?

Answer 67

chromosomal analysis and incidence of DS correlated to mother advanced age

Question 68

Compare partial monosomy with haploinsufficiency.

Answer 68

partial monosomy deletion of a small terminal portion of chromosome e.g chri du chat.

haploinsufficiency. a single copy of a gene is inactivated or deleted and the remaining functional copy of the gene is not adequate to produce the needed gene product to preserve normal function.

Haploinsufficiency occurs when the single wildtype
copy of the gene does not produce enough gene product
to bring about a wild-type phenotype.

In humans, Marfan syndrome is an example of a disorder caused by haploinsufficiency—in this case as a result of a loss-of-function mutation in one copy of the FBN1 gene.

Question 69

define autopolyploidy/allopolyploidy

Answer 69

autopolyploidy

When identical complete sets of chromosomes are added to the (parent species) diploid genome.

Allopolyploidy

(from the Greek word allo, meaning “other”or “different”). a diverse genetic origin, created

Question 70

Define autotetraploid/amphidiploid

Answer 70

autotetraploid (4n) are theoretically more likely to be found in nature.

amphidiploid

Allopolyploidy or (amphidiploidy) Multiples of closely related genomes

Question 71

What are the possible reasons behind translocations?

Answer 71

Chromosomes can break spontaneously or due to chemicals, and can abnormally fuse with other nonhomologous chromosomes, leading to a loss or rearrangement of the genetic material.

Telomere shortening can be another reason why chromosome ends fuse with each other.

Question 72

Why do human monosomic most often fail to survive prenatal development?

Answer 72

gametes lacking a single chromosome are functionally
impaired to a serious degree or that the embryo dies so
early in its development that recovery occurs infrequently.

Question 73

9. A couple goes through multiple miscarriages, and fetal karyotyping indicates the same trisomy every time. This trisomy is not compatible with life. Obviously, neither of the parents has this trisomy. Can you explain this situation?

Answer 73

the addition of a large autosome to the diploid complement in both Drosophila and humans has severe effects and is usually lethal during development.

►No autosomal trisomy, survive to term EXCEPT:

• Down syndrome; 47, 21+.
• Patau syndromes ; 47,13+.
• Edwards syndromes ; 47,18+.

Trisomy caused by nondisjunction during gamete miosis which may related to mother age while patrents karyotype not involved in.

Question 74

In a sample of 1000 patients with Down syndrome, a geneticist discovers that:

95% of them are trisomic, while 5% have diploid number of chromosomes. Explain this discrepancy.

Answer 74

Down syndrome are either due to nondisjunction during meiosis in one of the parents, resulting in trisomy 21,

or due to a 14/21, D/G translocation in one of the parents (familial Down syndrome). The latter results in a zygote that has 46 chromosomes but three copies of chromosome 21.

Question 75

Cytosine may also be named 2-oxy-4-amino pyrimidine. How would you name the other four nitrogenous bases, using this alternative system?

Answer 75

• Uracil: 2,4-dioxypyrimidine
• Thymine: 2,4-dioxy-5-methylpyrimidine
• Adenine: 6-aminopurine
• Guanine: 2-amino-6-oxypurine
• Cytosine: 2-oxy-4-amino pyrimidine.

Question 76

What component of the nucleotide is responsible for the absorption of ultraviolet light? How is this technique important in the analysis of nucleic acids?

Answer 76

The nitrogenous bases of nucleic acids (nucleosides, nucleotides, and single- and double-stranded polynucleotides) absorb UV light maximally at wavelengths of 254 to 260 nm.

One can often determine the presence and concentration of nucleic acids in a mixture.

Since proteins absorb UV light maximally at 280 nm, this is a relatively simple way of dealing with mixtures of biologically important molecules.

UV absorption is greater in single-stranded molecules (hyperchromic shift) than in double-stranded structures.

Therefore, by applying denaturing conditions, one can easily determine whether a nucleic acid is in the single- or double-stranded form.

In addition, A-T rich DNA denatures more readily than G-C rich DNA. Therefore, one can estimate base content by denaturation kinetics.

Question 77

DNA Semiconservative replication

Answer 77

Genetic continuity between parental and progeny cells is maintained by semiconservative replication of DNA, as
predicted by the Watson–Crick model.

■■ Semiconservative replication uses each strand of the parent double helix as a template, and each newly replicated
double helix includes one “old” and one “new” strand of DNA.

Question 78

frameshift mutations

Answer 78

the loss or addition of a single nucleotide causes all of the subsequent three-letter codons to be changed.

These are called frameshift mutations because the frame of triplet reading during translation is altered.

A frameshift mutation will occur when any number of bases are added or deleted, except multiples of three, which would reestablish the initial frame of reading.

Question 79

10. Why are frameshift mutations likely to be more detrimental than point mutations, in which a single pyrimidine or purine has been substituted?

Answer 79

Frameshift mutations are likely to change more than one
amino acid in a protein product because as the reading frame is shifted, a different set of codons is generated.

In addition, there is the possibility that a nonsense triplet could be introduced, thus causing premature chain termination.

Question 80

11. In which phases of the cell cycle would you expect double-strand break repair and nonhomologous end joining to occur and why?

Answer 80

DNA polymerases are involved in DNA repair and replication through regions of the DNA template that contain damage or distortions.

One pathway involved in double-strand break repair is homologous recombination repair.

DSB repair usually occurs during the late S or early G2 phase of the cell cycle, after DNA replication, a time when sister chromatids are available to be used as repair templates.

Because an undamaged template is used during repair synthesis, homologous recombination repair is an accurate process.

Question 81

human ß-thalassemia

Answer 81

A number of different types of mutations in the HBB gene can cause human ß-thalassemia, a disease characterized by various levels of anemia. Many of these mutations occur within introns or in upstream noncoding sequences. mutations in these regions often lead to severe disease, although they may not directly alter the coding regions of the gene.

The HBB gene provides instructions for making a protein called beta-globin.

Beta-globin is a component (subunit) of hemoglobin . Hemoglobin consists of four protein subunits, typically two subunits of beta-globin and two subunits of another protein called alpha-globin.

Question 82

Some mutations that lead to diseases such as Huntington disease are caused by the insertion of trinucleotide repeats. Describe how the process of DNA replication could lead to expansions of trinucleotide repeat regions.

Answer 82

Replication slippage is a process that generates small
deletions and insertions during DNA replication.

While it can occur anywhere in the genome, it is most prevalent in regions already containing repeated sequences. Thus, repeated sequences are hypermutable.

Question 83

Cystic fibrosis (CF) is a severe autosomal recessive disorder in humans that results from ………?

Answer 83

Cystic fibrosis (CF) is a severe autosomal recessive disorder in humans that results from

a chloride ion channel defect in epithelial cells.

responsible gene (CFTR, or cystic fibrosis transmembrane regulator)

• Male CF carrier and affected persons may have OA

Question 84

Microarrays

Answer 84

Microarrays, also known as gene chips, consist of a glass microscope slide onto which single-stranded DNA molecules are attached, or “spotted,” using a computer-controlled high-speed robotic arm called an arrayer.

Arrayers are fitted with a number of tiny pins. Each pin is immersed in a small amount of solution containing millions of copies of a different single-stranded DNA molecule.

A single microarray can have over 20,000 different spots of DNA (and over 1 million for exon-specific microarrays), each containing a unique sequence that serves as a probe for a different gene.

Probes for entire genomes are available on microarrays, including the human genome.

Question 85

what’s the important of studying DNA microarrays??

Answer 85

DNA microarrays or gene chips have been valuable for transcriptome analysis by studying expression patterns for thousands of genes simultaneously.

Question 86

what technique used for the protein content of a cell?

Answer 86

Proteomics methods such as mass spectometry are valuable for analyzing proteomes—the protein content of a cell.

Question 87

What are gene microarrays? How are microarrays used?

Answer 87

Most microarrays, known also as gene chips, consist of a
glass slide that is coated, using a robotic system, with single-stranded DNA molecules.

Some microarrays are coated with single-stranded sequences of expressed sequenced tags or DNA sequences that are complementary to gene transcripts.

A single microarray can have as many as 20,000 different spots of DNA, each containing a unique sequence. Researchers use microarrays to compare patterns of gene expression, in tissues under different conditions or to compare gene expression patterns in normal and diseased tissues.

In addition, microarrays can be used to identify pathogens. Microarray databases allow investigators to compare any given pattern to others worldwide.

Question 88

We usually think of enzymes as being most active at around 37°C, yet in PCR the DNA polymerase is subjected to multiple exposures of relatively high temperatures and seems to function appropriately at 70–75°C. What is special about the DNA polymerizing enzymes typically used in PCR?

Answer 88

Taq polymerase is from a bacterium called Thermus aquaticus, which typically lives in hot springs. It is heat stable like some other enzymes used in PCR that are isolated from thermal vents in the ocean floor.

Question 89

What is the difference between a knockout animal and a transgenic animal?

Answer 89

A knockout animal has a piece of DNA missing, whereas a
transgenic animal usually has a piece of DNA added.

Question 90

What is third-generation sequencing (TGS)?

Answer 90

Third-generation sequencing (TGS) Technologies based on high-throughput methods that sequence a single-stranded DNA molecule.

Question 91

NGS technologies

Answer 91

• DNA sequencing technologies are changing rapidly. Next-generation and third-generation methods produce large amounts of fairly accurate sequence data in a short time.
• The technology is used to determine the order of nucleotides in entire genomes or targeted regions of DNA or RNA.

Question 92

In a sequence encompassing 99.4% of the euchromatic regions of human chromosome 1, Gregory et al. identified 3141 genes.

• (a) How does one identify a gene within a raw sequence of bases in DNA?
• (b) What features of a genome are used to verify likely gene assignments?
• (c) Given that chromosome 1 contains approximately 8 percent of the human genome, and assuming that there are approximately 20,000 genes, would you consider chromosome 1 to be “gene rich”?

Answer 92

(a) To annotate a gene, one identifies gene-regulatory
sequences found upstream of genes (promoters, enhancers,ىand silencers), downstream elements (termination sequences), and in-frame triplet nucleotides that are part of the coding region of the gene. In addition, 59 and 39 splice sites that are used to distinguish exons from introns as well as polyadenylation sites are also used in annotation.
(b) Similarity to other annotated sequences often provides
insight as to a sequence’s function and may serve to substantiate a particular genetic assignment. Direct sequencing of cDNAs from various tissues and developmental stages aids in verification.
(c) Taking an average of 20,000 for the estimated number of genes in the human genome and computing the percentage represented by 3141 gives 15.7 percent. It appears as if chromosome 1 is gene rich.

Question 93

Compare and contrast whole-genome shotgun sequencing to a map-based cloning approach.

Answer 93

• Whole-genome shotgun sequencing involves randomly cutting the genome into numerous smaller segments.
• Overlapping sequences are used to identify segments that were once contiguous, eventually producing the entire sequence.
• Difficulties in alignment often occur in repetitive regions of the genome.

Map-based sequencing relies on known landmarks (genes,
nucleotide polymorphisms, etc.) to orient the alignment of
cloned fragments that have been sequenced. Compared to
whole-genome sequencing, the map-based approach is somewhat cumbersome and time consuming.

Whole-genome sequencing has become the most common method for assembling genomes, with map-based cloning being used to resolve the problems often encountered during whole-genome sequencing.

Question 94

. List and describe three major goals of the Human Genome Project?

Answer 94

The main goals of the Human Genome Project are to establish, categorize, and analyze functions for human genes. As stated in the text:

• To analyze genetic variations between humans, including the identification of single-nucleotide polymorphisms (SNPs)
• To map and sequence the genomes of several model organisms used in experimental genetics, including E. coli, S. cerevisiae, C. elegans, D. melanogaster, and M. musculus (the mouse)
• To develop new sequencing technologies, such as high throughput computer-automated sequencers, to facilitate genome analysis
• To disseminate genome information, both among scientists and the general public

Question 95

BLAST searches and related applications are essential for analyzing gene and protein sequences.

Define BLAST, describe basic features of this bioinformatics tool, and provide an example of information provided by a BLAST search.

Answer 95

One initial approach to annotating a sequence is to compare the newly sequenced genomic DNA to the known sequences already stored in various databases. The National Center for Biotechnology Information (NCBI) provides access to BLAST (Basic Local Alignment Search Tool) software, which directs searches through databanks of DNA and protein sequences.

A segment of DNA can be compared to sequences in major databases such as GenBank to identify matches that align in whole or in part.
One might seek similarities to a sequence on chromosome 11in a mouse and find that or similar sequences in a number of taxa.

BLAST will compute a similarity score or identity value to indicate the degree to which two sequences are similar. BLAST is one of many sequence alignment algorithms (RNA-RNA, protein-protein, etc.) that may sacrifice sensitivity for speed.

Question 96

It can be said that modern biology is experiencing an “omics” revolution.

What does this mean? Explain your answer.

Answer 96

A number of new subdisciplines of molecular biology will
provide the infrastructure for major advances in our understanding of living systems. The following terms identify specific areas within that infrastructure:

• proteomics—proteins in a cell or tissue
• metabolomics—enzymatic pathways
• glycomics—carbohydrates of a cell or tissue
• toxicogenomics—toxic chemicals
• metagenomics—environmental issues
• pharmacogenomics—customized medicine
• transcriptomics—expressed genes

Many other “-omics” are likely in the future.

Question 97

Exome sequencing is a procedure to help physicians identify the cause of a genetic condition that has defied diagnosis by traditional means.

The implication here is that exons in the nuclear genome are sequenced in the hopes that, by comparison with the genomes of nonaffected individuals, a diagnosis might be revealed.

• (a) What are the strengths and weaknesses of this approach?
• (b) If you were ordering exome sequencing for a patient, would you also include an analysis of the patient’s mitochondrial genome?

Answer 97

(a, b) With exon sequencing, one may get lucky and find an
issue within a gene that has relevance.

Given the multitude of genetic variations among individuals, this might be similar to finding a needle in a haystack.ابرة ف كوم قش

Many significant genomic functions are regulated outside the exon pool. Introns and mitochondria are highly variable among individuals, but may have some relevance in some health-related conditions.

Question 98

In order to vaccinate people against diseases by having them eat antigens (such as the cholera toxin) or antibodies expressed in an edible vaccine, the antigen must reach the cells of the small intestine. What are some potential problems of this method?

Answer 98

Antigens are usually quite large molecules, and in the process of digestion, they are sometimes broken down into smaller molecules, thus becoming ineffective in stimulating the immune system.

Some individuals are allergic to the food they eat, testifying to the fact that all antigens are not completely degraded or modified by digestion. In some cases, ingested antigens do indeed stimulate the immune system (e.g., oral polio vaccine) and provide a route for immunization.

Localized (intestinal) immunity can sometimes be stimulated by oral introduction of antigens, and in some cases, this can offer immunity to ingested pathogens.

Question 99

A couple with European ancestry seeks genetic counseling before having children because of a history of cystic fibrosis (CF) in the husband’s family. ASO testing for CF reveals that the husband is heterozygous for the Δ508 mutation and that the wife is heterozygous for the R117 mutation.

You are the couple’s genetic counselor. When consulting with you, they express their conviction that they are not at risk for having an affected child because they each carry different mutations and cannot have a child who is homozygous for either mutation. What would you say to them?

Answer 99

Since both mutations occur in the CF gene, children who possess both alleles will suffer from CF.

With both parents heterozygous, each child born will have a 25 percent chance of developing CF.

Question 100

What limits the use of differences in restriction enzyme sites as a way of detecting point mutations in human genes?

Answer 100

Using restriction enzyme analysis to detect point mutations
in humans is a tedious ممل trial-and-error process.

Given the size of the human genome in terms of base sequences and the relatively low number of unique restriction enzymes, the likelihood of matching a specific point mutation, separate from other normal sequence variations, to a desired gene is low.

Question 101

Dominant mutations can be categorized according to whether they increase or decrease the overall activity of a gene or gene product.

Although a loss-of-function mutation (a mutation that inactivates the gene product) is usually recessive, for some genes, one dose of the normal gene product, encoded by the normal allele, is not sufficient to produce a normal phenotype.

In this case, a loss-of-function mutation in the gene will be dominant, and the gene is said to be haplo-insufficient. A second category of dominant mutation is the gain-of-function mutation, which results in a new activity or increased activity or expression of a gene or gene product. The gene therapy technique currently used in clinical trials involves the “addition” to somatic cells of a normal copy of a gene. In other words, a normal copy of the gene is inserted into the genome of the mutant somatic cell, but the mutated copy of the gene is not removed or replaced. Will this strategy work for either of the two aforementioned types of dominant mutations?

Answer 101

• In the case of haplo-insufficient mutations, gene therapy holds promise; however, in “gain-of-function” mutations, in all probability the mutant gene’s activity or product must be compromised.
• Addition of a normal gene probably will not help unless it can compete out the mutant gene product.

Question 102

What are the different genetic markers that genome-wide association studies (GWAS) employ? How can scientists use this data to calculate the disease risk associated with each variation?

Answer 102

• A GWAS analyzes SNPs, specific differences in genes, CNVs, or changes in the epigenome, such as methylation patterns in particular regions of a chromosome.
• By determining which SNPs, CNVs, or epigenome changes co-occur in individuals with the disease, scientists can calculate the disease risk associated with each variation.

Question 103

What is a genome?

Answer 103

A genome is the set of all the DNA molecules that are characteristic of an organism. Each DNA molecule forms one chromosome in a cell of the organism

Question 104

Arrange the following events in the correct temporal sequence during eukaryotic cell division, starting with the earliest:

• (a) condensation of the chromosomes,
• (b) movement of chromosomes to the poles,
• (c) duplication of the chromosomes,
• (d) formation of the nuclear membrane, (e) attachment of microtubules to the kinetochores, and
• (f) migration of centrosomes to positions on opposite sides of the nucleus.

Answer 104

(c), (f), (a), (e), (b), (d).

Question 105

In human beings, the gene for b-globin is located on chromosome 11, and the gene for a-globin, which is another component of the hemoglobin protein, is located on chromosome 16. Would these two chromosomes be expected to pair with each other during meiosis? Explain your answer.

Answer 105

Chromosomes 11 and 16 would not be expected to pair with each other during meiosis; these chromosomes are heterologues, not homologues.

Question 106

The following pedigree shows the inheritance of a dominant trait.

What is the chance that the offspring of the following matings will show the trait:

• (a) III-1 × III-3
• (b) III-2 × III-4

Answer 106

• (a) zero
• (b) 1/2

Question 107

The following pedigree shows the inheritance of

a recessive trait. Unless there is evidence to the contrary, assume that the individuals who have married into the family do not carry the recessive allele.

What is the chance that the offspring of the following mating will show the trait:

(a) III-1 × III-12;
(b) II-4 × III-14;
(c) III-6 × III-13;
(d) IV-1 × IV-2?

Answer 107

(a) (1/2) × (1/4) = 1/8;
(b) (1/2) × (1/2) × (1/4) = 1/16;
(c) (2/3) × (1/4) = 1/6;
(d) (2/3) × (1/2) × (1/2) × (1/4) = 1/24

Question 108

Rules of inheritance autosome recessive trait

Answer 108

• appear in bath sexes male and female with equal frequency
• skip generation
• affected offspring born to unaffected parents
• if both parents are heterozygous► 25% of spring will be affected
• more frequently among consanguine marraige.

Question 109

Rules of inheritance Autosomal Dominant trait

Answer 109

• bath sex with equal affect
• no skip generation
• affected offspring must have affected parent or have a new mutation.
• if 1 parent hetero- and the other is unaffected 50% of children will be affected

Question 110

Rules of inheritance X-LINKED recessive trait

Answer 110

• More males than females affected
• unaffected mother transmitt the trait to 50% her sons
• skip generation
• never pass by father to sons
• All daughters of affected father are carrier.
• example : Hemophilia X-LINKED recessive

Question 111

inheritance X-LINKED Dominant trait

Answer 111

• both male and female are affected
• often more female affected than male
• NO skip generation
• affected sons must have affected mother
• affected daughter must have either mother or father affected
• affected mother transmit the trait to 50% of sons and 50% of daughter
• affected father will pass the trait to all daughter and Never pass to sons

Question 112

rules of inheritance Y linked Dominant

Answer 112

• only males are affected
• NO skip gheneration

Question 113

Mitochondrial gene disorder inheritance

Answer 113

trait is inherited from mother only

all children are at risk to be carrier

Question 114

Beckwith-Wiedemann syndrome?

Answer 114

20% ≈ of cases of Beckwith-Wiedemann syndrome (over growth) are caused by a genetic imprinting change known as: paternal uniparental disomy (UPD).

• Paternal UPD. over expression of IGF2 growth factor

active copies of paternally inherited genes rather than one active copy from the father and one inactive copy from the mother.

Question 115

Angelman syndrome

&

Prader-Willi syndrome

Answer 115

Angelman syndrome is a genetic disorder. It’s usually caused by imprinting defect with a gene located on mother chromosome 15 called the ubiquitin protein ligase E3A (UBE3A) gene.

note’ mother imprinting defect

Prader-Willi syndrome is a genetic disorder, defect in imprinting in the genes located in a particular region of paternal chromosome 15

note’ father imprinting defect

Question 118

Crown-rump length (CRL)

Answer 118

An ultrasound measurement that is used during pregnancy. The baby is measured, in centimeters, from the top of their head (crown) to the bottom of their buttocks (rump).

usually identified at 6 -7 weeks’ gestation.

CRL grows approximately 10 mm per week from weeks 8 to 12 and a simple rule to obtain GA is the following:

GA (week) = CRL (cm) + 6.5.

Question 119

A CPI is critical identification points.

a. Yes
b. No

Answer 119

a)

Question 120

A high Mg2+/Ca2+ ratio is used in IVF culture media because:

a. embryos decompact in Ca2+/Ma2+ free media
b. most of the rise in Ca2+ came from intracellular stores.
c. fertilisation media has a low Mg2+ since a Ca2_ spike is part of the acrosome reaction
d. a high Ca2+ results in metabolic perturbations and decreased development
e. a high Ca2+/Mg2+ ratio may lead to an excessive uptake of Ca2+

Answer 120

all true

Question 121

A high Na+/K+ ration may result in:

a. decreased development
b. no discernible effect
c. increased development

Answer 121

a)

Question 122

A question of independence.

a. gametes and embryos cannot survive in an isotonic solution without nutrients
b. gametes and embryos can survive in an isotonic solution without nutrients
c. gametes and embryos can survive in an isotonic solution but only for a short time

Answer 122

c)

Question 123

A question of survival.

a. gametes and embryos carry no nutrients they need to implant
b. gametes and embryos carry few the nutrients they need to implant
c. gametes and embryos carry all the nutrients they need to implant

Answer 123

b)

Question 124

A rapid rise in ATPase activity occurs at:

a. hatching
b. morula stage
c. 4-8 cell stage
d. fully expanded stage
e. early blastocyst stage
f. early compaction
g. expanding stage

Answer 124

g)

Question 125

A serious adverse event:

a. is any event associated with staffing activities
b. is any event associated with clinics activities
c. is any event associated with ART treatment

Answer 125

c)

Question 126

DNA repair

The activated p53 arrests replication during the S phase to facilitate DNA repair.

The activated BRCA1 protein, in conjunction with BRCA2, mRAD51, and other nuclear proteins, is involved in repairing the DNA.

Answer 126

true

Question 127

RCA1, BRCA2 genes

Answer 127

RCA1, BRCA2 :

[oncogenes, Breast cancer A1,2]

these genes called the Tumor-Suppressor Genes

DNA repair Point mutations of RCA1, BRCA2 genes cause: ►Breast, ovarian and prostate cancers

Question 128

Genetic tests that detect mutations in the BRCA1 and BRCA2 oncogenes are widely available.

These tests reveal a number of mutations in these genes—mutations that have been linked to familial breast cancer.

Assume that a young woman in a suspected breast cancer family takes the BRCA1 and BRCA2 genetic tests and receives negative results. That is, she does not test positive for the mutant alleles of BRCA1 or BRCA2. Can she consider herself free of risk for breast cancer??

Answer 128

BRCA1 and BRCA2. These two genes encode tumor

suppressor proteins that are involved in repairing damaged DNA.

woman whose BRCA test results are negative may feel relieved and assume that she is not subject to familial breast cancer. However, her risk of developing breast cancer is still 12 % (the population risk), and she should continue to monitor herself for the disease.

Also, a negative BRCA genetic test does not eliminate the possibility that she carries an inherited mutation in another gene that increases breast cancer risk or that her BRCA1 or BRCA2 gene mutations exist in regions of the genes that are inaccessible to current genetic tests

Women who inherit germ-line mutations in BRCA1 have an approximately 60% chance of developing breast cancer and a 39% chance of developing ovarian cancer.

Those who inherit mutations in BRCA2 have an approximately 45 percent chance of developing breast cancer and a 15 percent chance of developing ovarian cancer.

In men, mutations in these two genes lead to increased risks of both breast and prostate cancers.

Question 129

p53 gene mutation involved ……………??

and the BRCA1 gene, which, when mutated, predisposes women to………………

Answer 129

p53 gene mutation ►in majority of human cancer

BRCA1 gene, which, when mutated,► women breast cancer

Question 135

Sperm DNA Repair by Oocyte

Answer 135

egg attempt Sperm DNA repair only during the first round of DNA replication after fertilization 2pn stage

Question 136

Semenogelin (Sg)

Answer 136

Semenogelin (Sg)

the major protein of the human semen coagulum, is present at high concentrations in seminal vesicle secretions

Question 137

PGCs can identified …………migration close to the Yolk sac an d at the root of the Allantois

• a) day 7 of gestation
• b) 25 days of gestation
• c) 12 week of gestation
• d) 7 week of gestation

Answer 137

b)

about 3 week of gestation = 20-25 days

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