Genetics and Molecular Biology

Question 1

man and woman are both affected by an autosomal dominant disorder with 80% penetrance in all affected individuals. They are both heterozygotes for the disease causing mutation. which of the following is the probability that they will produce a phenotypically normal offspring?

Answer 1

probability that a child receives the dominant trait multiplied by the penetrance of the trait=risk of the child expressing the abnormal phenotype.
expression of normal phenotype=probability of receiving dominant trait-number from 1st calculation:

example:
75% chance of receiving disease causing gene (RR and 2Rr)
80% penetrance
0.75 X 0.80=0.6 or 60%
75%-60%=15%

also 25% will be geneotypically normal (rr)

So, probability of phenotypically normal child= 25%+15%=40%

Question 2

for a autosomal recessive condition, if a phenotypically normal child is born to parents that are both heterozygous for the condition, the probability that the child is heterozygous for the condition is

Answer 2

2/3. (Rr, Rr, and RR)

not 1/2 b/c we already know that the child is not (rr) so it’s not included in the overall probability calculation.

Question 3

Fragile X syndrome- inhereitance, gene effected presentation, male vs female

Answer 3

X-linked

• trinucleotide repeat (CGG)
• affects methylation of FMR1 gene
• macro-orchidism, long face with large jaw, large everted ears, autism, mitral valve prolaspe
• more severe in males b/c they only have 1 X chromosome
• think Xtra large-testes, jaw, and ears

Question 4

molecule for responsible for adding RNA primer to initiate DNA eukaryotic replication, not polymerase III because

Answer 4

• polymerase III is found in prokaryotics only

- primase is a complex of proteins that synthesize a short fragment of RNA. required before DNA can be synthesized

Question 5

capping vs splicing vs polyadenylation at 3’ end of growing RNA transcript

Answer 5

capping- occurs immediately after synthesis of first 30 nucleotides. triphosphate of GTP condenses with 5’diphosphate to make a cap that is recongnized during protein synthesis and protects RNA
splicing-splice acceptor site on RNA at 3’ end is AG (note it’s GU on 5’end) these high conserved sequences are needed for correct spicing of introns and exons.
polyadenylation-occurs near the 3’ end. an AAUAAA sequence is recognized cleaved and then poly-A polyerase add hundreds of adenylate residues

Question 6

findings in osteogenesis imperfecta are best explained due to what (genetic term) and not multiple mutation because?

Answer 6

osteogenesis imperfecta is an example of a disease in which a single mutation leads to multplie features (pleiotrophy).
it is cause by single mutation

Question 7

allelic heterogeneity vs locus heterogeneity vs pleiotrophy

Answer 7

• different mutations in single locus alleles leads to one feature (albinism)
• different mutations in different locus leads to one feature (b-thalasemmia)
• a single mutation leads to multiple features (osetogenesis imperfecta)

Question 8

purpose of

• Southern,
• Northern
• Western
• Southwestern blot
• Dot (slot)

Answer 8

-to determine which restriction fragments of DNA are associated with a particular gene, genetic relatedness, indirect mutation analysis within families
-to measure sizes and amts of mRNA to determine gene expression
-measure amt of antigen or proteins, determine translation
-detect and characterize DNA-bind proteins and where on DNA they bind
-detect specific DNA,RNA, protein or antibody
all above techniques need electrophoresis except Dot (Slot)

Question 9

which biochemical techniques can be used to fine disease causing alleles?

Answer 9

PCR- amplify DNA sequences 
or Dot (slot)- use probes to find disease causing DNA, RNA, protein, or antibody

Question 10

disease associated with DNA repair defect

• base excision
• direct pyrimidine dimer repair
• global genomic nucleotide excision repair (GGNR)
• transcription-coupled nucleotide repair
• mismatch repair
• non-homologous end joining

Answer 10

• none
• none (mediated by photolyase)
• xeroderma pigmentosum* (loss of ability to remove UV-radiation-induces pyrimidine dimers)-can lead to disease although photolyase can’t
• developmental defects (loss of ability to repair sequences removed by GGNR
• HNPCC or Lynch syndrome*
• ataxia telangiectasia*

Question 11

mutation in hMSH2 of hMLH1 leads to

Answer 11

defective mismatch repair- Lynch syndrome HNPCC

Question 12

• inherited
• vs sporadic rentiblastoma
• how PCR is used to detect between sporadic and heritable forms of a disease? i.e retinoblastoma

Answer 12

-many tumors in both eyes, body, and autosomal dominant inheritance
-one tumor in one eye, no increased risk of other tumors compared to general population, not passed to offspring
PCR
-sporadic-cells other than tumor will show normal two alleles
-heritable- 1 chormosome will show deletion or mutation in all cells of body. tumor will have 2 mutations

Question 13

def of Biochem technique called DNA footprinting

Answer 13

tech used to detect DNA binding proteins by comparing the fragmentation patterns (after digestion with DNase I) of DNA bound to proteins and DNA not bound to protein. Assume binding protein protects DNA sequences from being digested with DNase I

Question 14

Fluorescence-activated cell sorting (FACS)- definition, purpose

Answer 14

• antibodies coupled to fluorescent markers to detemrine surface markers on whole cells
• can determine stage of maturity or activation of cells (i.e. WBCs)

Question 15

how to solve this problem the using gametes instead of mating calculations:
autosomal recessive disease has a general population frequency of 1%. a heterozygous mans marries a woman from the general population. what is the probability that he and his partner will have an child affected with the disorder.

Answer 15

for the man there’s a 50% chance (since he’s a know heterozygous) that one of his sperm carrying the defective allele will fertilize the egg

for the woman her chance (since she’s from the general population) of her eggs carrying the defective allele is equal to the allele frequency q=0.10.

chance of child effected will disease is 0.5 X 0.1= 0.05

Question 16

two day year old baby develops tetany and has dsymorphic facies, with low-set ears and micrognathia, and decreased serum Ca2+ levels. child has what? not patau’s because?

Answer 16

• DiGeorge Syndrome* (22q11), absence of thymus and parathyroid leads (improper development of 3rd and 4th pharyngeal pouches) to manifestations
• Patau’s presents with intellectual disability, physical abnormalities (heart defect, hypotonia, micropthalmia, cleft lip and palate, and extra digits and toes)*

Question 17

genetic inheritance of Marfan’s

Answer 17

autosomal dominant

Question 18

first occurrence of autosomal dominant condition in a family is usually caused by

Answer 18

new mutation that occurred in one of the gametes transmitted by one parent

Question 19

a defect in enzyme that has a normal mRNA expression is due to nonsense mutation and not RNA splicing because

Answer 19

nonsense mutation can lead to truncated product that can be abnormal in shape leading to decreased function or rapid degradation
-RNA-splicing defect would affect mRNA expression levels and size

Question 20

if there is male to male transmission of the disease then what do you know about the inheritance of a disease? what don’t you know? how do you determine the what you don’t know?

Answer 20

-know that is autosoma*
-don’t know recessive vs dominant
-dominant will be seen in multiple generations* (50% chance) unless incomplete penetrance
recessive will skip generations (25% chance)*

Question 21

Barr body are seen in individuals with what genotype?

Answer 21

must have at least two or more X chromosomes.

i.e. normal females and Klinefelter (47,XXY)

Question 22

trinucleotide expansion in Huntington’s leads to loss of function or gain of function mutation?

Answer 22

thought to produce a toxic gain of function

Question 23

for an autosomal recessive disease what is the carrier frequency approximately equal to?

Answer 23

=2pq

or approximately 2q (because value of p is close to 1 b/c q is so small) p+q=1

Question 24

-name of technique used to determine when and in which parent gamete non-disjunction occurs?
parents:
-result if occurs in meiosis I
-result if occurs in meiosis II
child:
-result if nondisjunction occurs during 1st mitotic division

Answer 24

• restriction fragment length polymorphism
• child receives two DIFFERENT alleles from parent
• child receives two SAME alleles (failed splitting of sister chromatids) from parent
• mosaic genotype (some cells have one genotype while other cells have another)

Question 25

reason for difference in severity of Becker’s and Duchenne muscular dsytrophy

Answer 25

in-frame deletion or insertions in the becker form and frameshift deletions or insertions in the Duchenne form

Question 26

Robertsonian translocation-def and name two types, which one leads to disease? give 2 examples

Answer 26

chromosomal translocation. (usually 13,14,15,21,22)
two types
non-reciprocal-two large arms join and two small arms are lost- associated with Down’s and Patau’s
reciprocal-DNA is exchanged- usually no abnormal phenotype

Question 27

presentation of Klinefelter syndrome

Answer 27

• most common congenital abnormality causing primary hypogonadism
• most common genotype is 47,XXY resulting form nondisjunction of sex chromosomes
• abnormal male secondary characteristics (small testes, small penis, and gynecomastia) and mental retardation

Question 28

how do you use monozygotic twin concordance and dizygotic twin concordance to determine inheritance of a trait, (autosomal recessive, dominant, or multifactorial)

Answer 28

• if DZ is less than 50% of MZ then can’t be solely autosomal dominant
• if DZ is less than 25% of MZ then can’t be solely recessive
• so must be multifactorial (some genetic and environmental component)

Question 29

with a multifactorial inheritance what is risk of infecting 2nd sibling if 1st sibling has disease compared to

• general population
• monozygotic concordance rate
• approximately equal to

Answer 29

-higher risk than population (b/c there’s some genetic effect)
-lower risk than MZ (b/c there’s some environmental effect)
approximately equal to DZ or square root of general population risk

Question 30

trisomy vs triploidy

Answer 30

• extra copy of a single chromosome (i.e. Down’s trisomy 21)

- 3 extra copies of all chromosomes

Question 31

define chiasma

Answer 31

cross like structure produced when a homologous pair of chromosomes aligns to exchange genetic material

Question 32

define synapsis

Answer 32

formal name for pairing of homologous chromosomes with which meiosis begins

Question 33

reasons for muscular dystrophy in males females?

Answer 33

males-recombination event that gives rise of a frameshift mutation
females-new mutation (unequal crossing over in females at meiosis I)

Question 34

given phenotypically normal parents what is the probability that a female sibling of a male affected by an X-linked recessive disease will give birth to an affected child

Answer 34

1/8=p1 p2p3
p1-sister is carrier=1/2 (mother not father is carrier b/c neither parent has disease
p2- sister (if carrier) will pass disease to offspring. =1/2 (only passes on 1 of 2 X chromosomes)
p3-offspring of sister is male (only male will have disease). =1/2 (50% chance she will have a male)

Question 35

given carrier frequencies of an autosomal recessive disease what is probability that parents will give birth to child with disease?

Answer 35

must calculate probability will be a carrier (carrier frequency) and multiply by chance that EACH parent will pass along mutated gene (1/2).

So, (carrier frequency 1of 1st parent x 1/2) X (carrier frequency of 2nd parent x 1/2)= carrier frequency X carrier frequency X 1/4

Question 36

given carrier frequencies of an autosomal recessive disease what is probability that parents will give birth to child with disease?

Answer 36

must calculate probability will be a carrier (carrier frequency) and multiply by chance that EACH parent will pass along mutated gene (1/2).

So, (carrier frequency 1of 1st parent x 1/2) X (carrier frequency of 2nd parent x 1/2)= carrier frequency X carrier frequency X 1/4