Genetics Unit 2

Select filters that were used in the bioinformatics analysis that identified mutations in the XIAP gene as being responsible for Nic Volker's symptoms.

- filters for novel variants - filters for recessive inheritance - filters to exclude silent mutations

how might a non-actionable outcome of whole genome sequencing to identify a disease gene still provide useful information?

- it may provide information that helps identify a future treatment for the disease - it may help patients to guide their reproductive decisions

Reasons a mutation may remain unidentified even when a whole-genome sequence is available

- the mutation may affect splicing of the disease gene's transcript - the mutation may affect transcription of the disease gene

In a rare dominant condition it is likely that:

- the patient will be heterozygous for the causative allele. - related patients will have the same rare mutant allele - unrelated patients might have a different mutation in the same gene

molecules that can undergo nucleic acid hybridization

- two complementary single-stranded DNAs - two complementary single-stranded RNAs - a single-stranded DNA and a complementary single-stranded RNA

In humans, the genes for red-green color blindness (R = normal, r= color blind) and hemophilia A (H = normal, h = hemophilia) are both X-linked and only 3 map units apart. A woman whose mother is color blind and whose father has hemophilia A is pregnant with a boy. If the alleles for color blindness and hemophilia A are rare in the population, what is the probability that the baby will have normal vision and normal blood clotting?

0.015

Genes A and B are on different (nonhomologous) chromosomes. If an individual's genotype is Aa Bb, what is the genotypic ratio of gametes produced

1 AB: 1Ab: 1aB: 1ab

Step of nucleotide excision repair

1) UvrA + UvrB complex tracks along the DNA in search of a thymine dimer 2) a complex of UvrB + UvrC makes cuts on both sides of the thymine dimer 3) DNA polymerase fills the gap 4) DNA ligase seals the nick

Steps of Base Excision Repair

1) glycosylase recognizes an abnormal base and cleaves the bond between the base and the sugar 2) AP endonuclease recognizes the missing base and cleaves the DNA backbone 3) exonuclease remove nucleotides creating a gap 4) DNA polymerase fills the gap using the undamaged strand as a template 5) DNA ligase seals the nick in the DNA backbone.

Steps of methyl-directed mismatch repair system

1) parental strands are marked with methyl groups 2) Mut S and MutL recognize mismatch in replicated DNA 3) MutL recruits MutH and GATC; Much makes a short nick in strand opposite methyl tag 4) DNA exonuclease excise DNA from unmethylated new strand 5) repair and methylation of newly synthesized DNA strand

how many SSR loci per person are examined and catalogued in the CODIS database?

13

Suppose the L and M genes are on the same chromosome but separated by 100 map units. What fraction of the progeny from the cross L M / l m × l m / l m would be L m / l m?

25 %

DNA polymerases use their ________ activity to remove a mismatched base.

3' -> 5' exonuclease

The map of a chromosome interval is: A——10 m.u.——B——40 m.u.——C From the cross A b c / a B C × a b c / a b c, how many double crossovers would be expected out of 1000 progeny?

40

In Drosophila, the genes y (yellow body) and car (carnation eyes) are located at opposite ends of the X chromosome. In doubly heterozygous females (y+car+ / y car), a single chiasma is observed somewhere along the X chromosome in 90% of the examined oocytes. No X chromosomes with multiple chiasmata are observed. What percentage of the male progeny from such a female would be recombinant for y and car?

45%

Suppose the map for a particular human chromosome interval is: a——1 m.u.——b——1 m.u.——c——1 m.u.——d——1 m.u.——e——1 m.u.——f In a man heterozygous for all six genes, what fraction of his sperm would be recombinant in the a-f interval?

5%

Assume that the mutation rate for a given gene is 5 × 10−6 mutations per generation. For that gene, how many mutations would be expected if 10 million sperm are examined?

50

In Drosophila, the genes b, c, and sp are linked and arranged as shown below: b——30 m.u.——c——20 m.u.——sp This region exhibits 90% interference. How many double crossovers would be recovered in a three-point cross involving b, c, and sp out of 1000 progeny?

6

An actionable outcome came from the bioinformatics analysis behind variations in these genes

CFTR; XIAP

In humans, the genes for red-green color blindness (R = normal, r= color blind) and hemophilia A (H = normal, h = hemophilia) are both X-linked and only 3 map units apart. Suppose a woman has four sons, and two are color blind but have normal blood clotting and two have hemophilia but normal color vision. What is the probable genotype of the woman?

Hr/hR

trinucleotide repeat expansion disorders

Huntington disease and fragile x syndrome

In total, what gametes will be produced by an A B / a b individual?

More AB and ab; fewer Ab and aB.

If the arrangement of alleles is altered, so the individual's genotype is A b / a B, what gametes will be produced overall?

More Ab and aB; fewer AB and ab

Genes A and B are so close together on the same chromosome that crossing-over occurs between A and B in only 1 in a million meioses. An AA BB individual mates with an aa bb individual to produce an F1 that is A B / a b. If 100 progeny are observed, what genotypic ratio is expected in the gametes produced by the F1?

Only parental 1 AB:1ab

point mutation

a change in a single base pair in the genetic material

high-throughput sequencing

ability to rapidly sequence large amounts of DNA

Alkylating agents, such as ethylmethane sulfate (EMS), are mutagens because they

add ethyl or methyl groups to bases

Hydroxylamine

adds -OH to cytosine, allowing it to pair with adenine

ethylmethane sulfonate

adds an ethyl group to guanine, which then pair incorrectly

After a tautomeric shift in adenine,

adenine bonds with cytosine

breast cancer

caused by failed double-strand break repair

hereditary Colorectal cancer

caused by failed mismatch repair

xeroderma pigmentosum

caused by failed nucleotide excision repair

A nucleotide deletion during DNA replication

causes the amino acids encoded after the deletion to be incorrect.

forward mutation

changes a wild-type allele to a different allele

What structure is responsible for maintaining the connection between homologous chromosomes until anaphase of meiosis I?

cohesion complexes that are located distal to the crossover and hold sister chromatids together.

UV light and other ionizing radiations damage DNA molecules by

creating thymine dimers between adjacent thymines in the DNA chain.

in order to repair damage via homologous recombination, cells use many of the enzymes responsible for

crossing over

types of mutations that can change the number of nucleotides in a gene

deletion; insertion

nitrous acid

dominates cytosine and adenine, changing their pairing properties

DNA damage caused by x-rays

double-strand breaks

In a frameshift mutation all of the amino acids before the shift are changed. true or false

false

Notation of A+ -> a

forward mutation when the mutation is recessive to the wild-type allele

nonanonymous DNA polymorphisms

genetic variants that affect phenotypes through changes in protein function

Double strand breaks are typically repaired by:

homologous recombination and non homologous end-joining

In a tautomeric shift,

hydrogen atoms move to form a base with altered hydrogen properties.

proflavin

intercalates into the double helix, causing deletions and insertions.

5-bromo-uracil

is incorporated into DNA instead of thymine, but can pair with guanine

identification of many disease-causing mutations depends on:

knowledge of variants from many other people's genomes

Another name for a chromosome is a _______, because it contains alleles that are often inherited together.

linkage group

During DNA replication, a base that does not obey the normal AT/GC base pairing rule is added. After replication, this type of mutation will most likely be corrected by

methyl-directed mismatch repair

An agent that can change DNA structure and cause mutations is known as a

mutagen

When DNA is damaged by UV light and is not repaired,

neither DNA replication nor transcription can occur and the organism will probably die

What is the consequence to a bacterial cell of a mutation that inactivates the enzyme responsible for methylating the A in the DNA sequence 5′ GATC?

parental and new DNA strands cannot be distinguished during mismatch repair.

transversion mutations

purine changes to pyrimidine or vice versa

The first step in base excision repair is

recognizing and excising the incorrect base from a nucleotide

deamination

removal of an amino group from a base.

depurination

removes and adenine or guanine from DNA

unstable trinucleotide repeats

repeated sequences of three bases which can increase or decrease in number generation after generation.

Proofreading by DNA polymerase involves the removal of

several bases on the newly-synthesized strand of DNA

A bacterial mismatch repair system is able to correct replication errors that insert an incorrect nucleotide. How is this repair system able to determine which mismatched base is incorrect?

some bases of the parental strand are methylated while non on the new strand are

DNA microarray

technology that allows researchers to genotype many SNP loci simultaneously.

In light repair,

the covalent bonds between the thymine dimers are broken

The diploid garden pea plant has 14 chromosomes. The haploid fungus Neurospora crassa has 7 chromosomes. Neither organism has separate male and female individuals. Therefore, the number of linkage groups in these two organisms is

the garden pea has 7 linkage groups and neurospora has 7

strand that it repaired in a mismatch repair system

the nonmethylated strand

Exposure to UV light from the sun or tanning beds causes

thymine dimers

Replacing an adenine nucleotide with a guanine is an example of a

transition

Replacing a thymine nucleotide with a guanine is an example of a

transversion

The addition of a single base pair to a gene's DNA sequence can cause a frameshift mutation. true or false

true

DNA repair via homologous recombination

typically occurs between sister chromatids during the G2 phase of the cell cycle.

cause of thymine dimers

uv light