TEST BANK FOR GENETIC ANALYSIS AN
INTEGRATED APPROACH 3RD EDITION
SANDERS / ALL CHAPTERS 1 - 20 /
FULL COMPLETE
GENETICS- Chaviva Korb - ANSWER-
Lecture 1: - ANSWER-
- DNA is packaged as chromatin, chromosomes become visible during
mitosis. - ANSWER-
Homologues- (maternal and paternal forms of same chromosome) -
ANSWER-
Meiosis: G1, S, G2, M. Before replication phase, chromosomes have one
chromatid and after replication, chromosomes have 2 sister chromatids,
held together at the centromere. Ends with 4 daughter cells with 1
chromatid each. - ANSWER-
Key differences from mitosis: Homologs pair (Prophase I), Sister
centromeres act as a single centromere (Metaphase I), Sister chromatids
remain attached (Anaphase I), Meiosis I is a reduction division- meaning
start meiosis I with 46 units and end meiosis I with 23 units ( 2
chromatids in each haploid daughter cells), Meiosis II is an equational
division (identical to mitosis). - ANSWER-
Recombination occurs in Prophase I. Crossing over and recomb is
Exchange of homologous segments between non-sister chromatids. -
ANSWER-
Homologues move apart during anaphase I-disjunction. 223 possible
combinations of chromosomes. In cytokinesis I, Cells divide into two
haploid daughter cells. One cell receives most of the cytoplasm and the
,other becomes the first polar body which doesn't go through Meiosis II.
Meiosis I ends after ovulation. Non dysjunction in meiosis I: can cause
problems like Trisomy 21. - ANSWER-
-Brief interphase between first and second meiotic divisions. No S phase
occurs in Meiosis II. - ANSWER-
-In the second meiotic division a second polar body forms during
oogenesis. In oogenesis, egg is arrested in metaphase II until fertilization.
- ANSWER-
After Meiosis there is a Reduction of chromosome number 2nn. (diploid
vs. haploid cells)- Notation 'n' has to do with amount of DNA (#of
nucleotides), not the necessarily the # of chromosomes. - ANSWER-
Lecture 2: Patterns of Inheritance - ANSWER-
Medelian Inheritance- determined by a single major gene. Based on
independent assortment. Dihybrid cross (2 genotypes) with ry, Ry, rY, Ry
set up on both sidesphenotypic ratio of 9:3:3:1-yellow round, green
round, yellow wrinkled, green wrinkled, If get 9:3:3:1 you know that
segregation is independent. - ANSWER-
Multifactorial inheritance-multiple genetic and non-genetic factors
involved - ANSWER-
Compound heterozygous: has two different mutant alleles for a
character - ANSWER-
Linkage-2 genes physically near each other on a chromosome will not
assort randomly in meiosis. Tightly linked: will get 2 types of gametes ex.
PL and pl. Unlinked: will get 4 types of gametes PL, Pl, pL, pl. - ANSWER-
The frequency of recombination between two genes is proportional to
the distance between the genes. The closer the genes are on the
chromosome the less likely crossing over occurs. Linkage map: 1%
recombination = 1 map unit = 1 centiMorgan (cM), Map distances are
additive. - ANSWER-
,The non-random association between alleles at two locations on a
chromosome is called linkage disequilibrium. If the frequency of
chromosomes with AB=Ab=aB=ab then the genes are in equilibrium. If
frequency of 1 allele is seen more (A more than B for ex) then genes are
in linkage disequilibrium. - ANSWER-
Autosomal dominant inheritance Examples: - ANSWER-
Achondroplasia- FGFR3 mutations, Always full penetrance with
achondroplasia (so normal parents have a child with aplasia then it's a
new mutation). Heterozygous b/c homozygotes usually die in utero -
ANSWER-
Neurofibromatosis- NF1 (neurofibromin) and NF2 (merlin) mutations. -
ANSWER-
Incomplete dominance- mixed phenotype. In cases of disease, Dominant
disorders are more severe in homozygotes then in heterozygotes
(termed also "semidominant") Ie. Familial Hypercholesterolemia. -
ANSWER-
Co-dominance- phenotypic expression of two different alleles for a locus
ie. Blood type. - ANSWER-
Autosomal recessive inheritance: Examples: Cystic fibrosis, Tay-Sachs
disease, Sickle-cell disease - ANSWER-
Pseudodominance: the inheritance of an autosomal recessive trait
mimics an autosomal dominant pattern - ANSWER-
Males are hemizygous with respect to X-linked genes. - ANSWER-
X-linked dominant inheritance: Affected females are twice as common
as affected males but males usually more severely affected or the
disorder may be lethal in males (Rett syndrome). - ANSWER-
X-linked recessive: incidence is much higher in males and affected males
do not usually transmit the disorder unless mother is a carrier.
Heterozygote females are usually unaffected, but some may express the
condition with variable severity as determined by the pattern of X
, inactivation. A significant proportion of isolated cases are due to new
mutation (Duchenne muscular dystrophy-DMD). - ANSWER-
Pseudoautosomal inheritance- group of genes on the inactive x
chromosome are NOT inactivated. Diseases associated with these genes
are inherited similar to autosomal inheritance. - ANSWER-
Same amount of X-linked gene products between males and females
achieved through dosage compensation. Lyon Hypothesis states that the
inactive X is NOT randomly chosen in each cell Ex. A structurally
abnormal X is preferentially inactivated. Inactivation is NOT complete-
some genes can escape inactivation (ie. Those with a functional homolog
on the Y). Inactivation is NOT permanent- reversed in development of
germ cells (not passed on to gametes). - ANSWER-
The key player is the X-linked gene XIST→ X (inactive) specific transcript.
XIST is transcribed to produce a non-coding RNA that "coats" the X-
chromosome and inactivates it. XIST is only expressed from the inactive
X. The histones on the coated X undergo methylation which causes the
chromosome to condense (heterochromatin), forming a Barr body. -
ANSWER-
-Some genes do not have Y homologue and do not undergo inactivation
(e.g. steroid sulfatase gene) - ANSWER-
-Random/skewed inactivation may result in affected/totally healthy
heterozygotes. - ANSWER-
Variable expression of X-inactivation: On both extremes, a heterozygous
female with recessive x-linked disease could manifest the disease. In a
case with a dominant X-linked trait, in which almost all of a females X
chromosomes with mutation is inactive, might not manifest this disease.
Identical twins could even have diff phenotypes due to skewed X
inactivation. - ANSWER-
Mosaicism- X chromosome inactivation occurs randomly and
inactivation pattern is passed to cell progeny. Result: functional
mosaicism in which female is a mosaic with respect to expression of
genes on X chrom. Ex. Calico Cat, B - dominant orange, b - recessive gene
black, Genes for white: autosomal. - ANSWER-
INTEGRATED APPROACH 3RD EDITION
SANDERS / ALL CHAPTERS 1 - 20 /
FULL COMPLETE
GENETICS- Chaviva Korb - ANSWER-
Lecture 1: - ANSWER-
- DNA is packaged as chromatin, chromosomes become visible during
mitosis. - ANSWER-
Homologues- (maternal and paternal forms of same chromosome) -
ANSWER-
Meiosis: G1, S, G2, M. Before replication phase, chromosomes have one
chromatid and after replication, chromosomes have 2 sister chromatids,
held together at the centromere. Ends with 4 daughter cells with 1
chromatid each. - ANSWER-
Key differences from mitosis: Homologs pair (Prophase I), Sister
centromeres act as a single centromere (Metaphase I), Sister chromatids
remain attached (Anaphase I), Meiosis I is a reduction division- meaning
start meiosis I with 46 units and end meiosis I with 23 units ( 2
chromatids in each haploid daughter cells), Meiosis II is an equational
division (identical to mitosis). - ANSWER-
Recombination occurs in Prophase I. Crossing over and recomb is
Exchange of homologous segments between non-sister chromatids. -
ANSWER-
Homologues move apart during anaphase I-disjunction. 223 possible
combinations of chromosomes. In cytokinesis I, Cells divide into two
haploid daughter cells. One cell receives most of the cytoplasm and the
,other becomes the first polar body which doesn't go through Meiosis II.
Meiosis I ends after ovulation. Non dysjunction in meiosis I: can cause
problems like Trisomy 21. - ANSWER-
-Brief interphase between first and second meiotic divisions. No S phase
occurs in Meiosis II. - ANSWER-
-In the second meiotic division a second polar body forms during
oogenesis. In oogenesis, egg is arrested in metaphase II until fertilization.
- ANSWER-
After Meiosis there is a Reduction of chromosome number 2nn. (diploid
vs. haploid cells)- Notation 'n' has to do with amount of DNA (#of
nucleotides), not the necessarily the # of chromosomes. - ANSWER-
Lecture 2: Patterns of Inheritance - ANSWER-
Medelian Inheritance- determined by a single major gene. Based on
independent assortment. Dihybrid cross (2 genotypes) with ry, Ry, rY, Ry
set up on both sidesphenotypic ratio of 9:3:3:1-yellow round, green
round, yellow wrinkled, green wrinkled, If get 9:3:3:1 you know that
segregation is independent. - ANSWER-
Multifactorial inheritance-multiple genetic and non-genetic factors
involved - ANSWER-
Compound heterozygous: has two different mutant alleles for a
character - ANSWER-
Linkage-2 genes physically near each other on a chromosome will not
assort randomly in meiosis. Tightly linked: will get 2 types of gametes ex.
PL and pl. Unlinked: will get 4 types of gametes PL, Pl, pL, pl. - ANSWER-
The frequency of recombination between two genes is proportional to
the distance between the genes. The closer the genes are on the
chromosome the less likely crossing over occurs. Linkage map: 1%
recombination = 1 map unit = 1 centiMorgan (cM), Map distances are
additive. - ANSWER-
,The non-random association between alleles at two locations on a
chromosome is called linkage disequilibrium. If the frequency of
chromosomes with AB=Ab=aB=ab then the genes are in equilibrium. If
frequency of 1 allele is seen more (A more than B for ex) then genes are
in linkage disequilibrium. - ANSWER-
Autosomal dominant inheritance Examples: - ANSWER-
Achondroplasia- FGFR3 mutations, Always full penetrance with
achondroplasia (so normal parents have a child with aplasia then it's a
new mutation). Heterozygous b/c homozygotes usually die in utero -
ANSWER-
Neurofibromatosis- NF1 (neurofibromin) and NF2 (merlin) mutations. -
ANSWER-
Incomplete dominance- mixed phenotype. In cases of disease, Dominant
disorders are more severe in homozygotes then in heterozygotes
(termed also "semidominant") Ie. Familial Hypercholesterolemia. -
ANSWER-
Co-dominance- phenotypic expression of two different alleles for a locus
ie. Blood type. - ANSWER-
Autosomal recessive inheritance: Examples: Cystic fibrosis, Tay-Sachs
disease, Sickle-cell disease - ANSWER-
Pseudodominance: the inheritance of an autosomal recessive trait
mimics an autosomal dominant pattern - ANSWER-
Males are hemizygous with respect to X-linked genes. - ANSWER-
X-linked dominant inheritance: Affected females are twice as common
as affected males but males usually more severely affected or the
disorder may be lethal in males (Rett syndrome). - ANSWER-
X-linked recessive: incidence is much higher in males and affected males
do not usually transmit the disorder unless mother is a carrier.
Heterozygote females are usually unaffected, but some may express the
condition with variable severity as determined by the pattern of X
, inactivation. A significant proportion of isolated cases are due to new
mutation (Duchenne muscular dystrophy-DMD). - ANSWER-
Pseudoautosomal inheritance- group of genes on the inactive x
chromosome are NOT inactivated. Diseases associated with these genes
are inherited similar to autosomal inheritance. - ANSWER-
Same amount of X-linked gene products between males and females
achieved through dosage compensation. Lyon Hypothesis states that the
inactive X is NOT randomly chosen in each cell Ex. A structurally
abnormal X is preferentially inactivated. Inactivation is NOT complete-
some genes can escape inactivation (ie. Those with a functional homolog
on the Y). Inactivation is NOT permanent- reversed in development of
germ cells (not passed on to gametes). - ANSWER-
The key player is the X-linked gene XIST→ X (inactive) specific transcript.
XIST is transcribed to produce a non-coding RNA that "coats" the X-
chromosome and inactivates it. XIST is only expressed from the inactive
X. The histones on the coated X undergo methylation which causes the
chromosome to condense (heterochromatin), forming a Barr body. -
ANSWER-
-Some genes do not have Y homologue and do not undergo inactivation
(e.g. steroid sulfatase gene) - ANSWER-
-Random/skewed inactivation may result in affected/totally healthy
heterozygotes. - ANSWER-
Variable expression of X-inactivation: On both extremes, a heterozygous
female with recessive x-linked disease could manifest the disease. In a
case with a dominant X-linked trait, in which almost all of a females X
chromosomes with mutation is inactive, might not manifest this disease.
Identical twins could even have diff phenotypes due to skewed X
inactivation. - ANSWER-
Mosaicism- X chromosome inactivation occurs randomly and
inactivation pattern is passed to cell progeny. Result: functional
mosaicism in which female is a mosaic with respect to expression of
genes on X chrom. Ex. Calico Cat, B - dominant orange, b - recessive gene
black, Genes for white: autosomal. - ANSWER-
