BIOL 1090 Lecture Notes - Lecture 9: Phenylketonuria, Polydactyly, Missense Mutation

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13 Feb 2017

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Outside mendel"s garden, things are not so simple : genes may (and usually) have more than 2 alleles. Different mutations in a gene can cause the same disorder. Different mutations in a gene can cause different phenotypes. Recessive allele: involve a loss of gene function: complete loss of function (null allele, partial loss of function (hypomorphic allele) Loss of function mutations are usually recessive because, for most gene, one functional copy is enough for a normal (wild type) phenotype. Dominant allele: if it has the same phenotype in heterozygotes as ain homozygotes. Dominant mutations can involve a loss of protein function or a gain of protein function: loss of protein function mutation: Mutant phenotype of an is between aa and aa (incomplete. Dominance heterozygotes have half the functional gene. Can interfere with function of the wild type protein (dominant dosage) Enhances the function of the wild type protein.

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Homozygous recessive loss of function mutations in any one of 4 different genes in C. elegans worms (genes A, B, C, or D) results in female worms that don't form proper egg-laying structures, called vulvas, as shown in Lines 2-5 of the data table below. In vulvaless females, the fertilized eggs hatch and baby worms develop inside the mother, killing her in the process.

	genotype	phenotype
1	AA BB CC DD (wild type)	normal vulva
2	aa BB CC DD	No vulva
3	AA bb CC DD	no vulva
4	AA BB cc DD	no vulva
5	AA BB CC dd	no vulva

You would like to figure out the order in which these genes act during the creation of the vulva (you cannot assume it will be A--->B--->C--->D as the gene names are arbitrary). Your friend tells you to perform epistasis analysis by making double mutants between the different homozygous recessive mutants and analyzing the phenotype of the double mutant. For example, she asks you to examine the phenotype of aabbCCDD (double mutant of genes A and B) and compare this to the single mutants to figure out whether gene A acts earlier than gene B, or vice versa. You know this won't work. Why not? Pick the ONE BEST choice:

a. The phenotype of the single mutant is already pretty severe. The double mutant will most likely be dead, therefore epistasis analysis won't be possible.

b. Each of the single mutants (aaBBCCDD) and (AAbbCCDD) has the same mutant phenotype i.e., no vulva. Epistasis analysis between any 2 genes is only possible when the mutant phenotypes for each gene is different.

c. Epistasis analysis involves making triple mutants (such as aabbccDD) in order to learn something about how the genes are ordered.

d. Epistasis analysis can never be carried out with null loss of function mutations. The mutations being analyzed all have to be dominant gain of function alleles.

To address your concern, you first generate overactive alleles of either gene A (denoted as A*) or gene B (denoted as B*). As shown in lines 6-7 of the table below, C. elegans that have one of these overactive alleles produce multiple vulvas.

	Genotype	Phenotype
6	A*A BB CC DD	multiple vulvas
7	AA B*B CC DD	multiple vulvas

To find out the order in which these genes act, you combine the overactive alleles with different loss of function alleles and observe the phenotype in double mutants (see Lines 8-11).

	Genotype	Phenotype
8	A*A bb CC DD	multiple vulvas
9	A*A BB cc DD	multiple vulvas
10	A*A BB CC dd	no vulva
11	AA B*B cc DD	multiple vulvas

Based on the data in the table, what is the order in which these 4 genes normally act in wild-type C. elegans in order to produce a wild-type/normal vulva?

a. A----> B----->C----->D ---> Vulva

b. D----> B----->C----->A----> Vulva

c. B----> C----->A----->D----> Vulva

d. C----> B----->A----->D----> Vulva

e. C----> B----->D----->A----> Vulva

f. don't have enough data to make any conclusions

Based on the vulva formation phenotype of the double mutants, which of the following statement(s) accurately describes the genetic interactions between the A* allele and alleles of other genes affecting vulva formation? Pick ALL that apply:

a. The A* allele is epistatic to homozygous recessive loss of function mutations in gene B

b. A homozygous recessive loss of function mutation in gene B is epistatic to the A* allele

c. Homozygous recessive loss of function mutation in gene D is epistatic to the A* allele

d. The A* allele is epistatic to homozygous recessive loss of function mutation in gene D

e. The A* allele enhances the homozygous recessive loss of function mutation in gene D

(tautomerization) The most common form of guanine is the ketoform, not the enol form.

True

False

What term describes the allowed degree of base pair matchingduring annealing, varying from lots of base-pair mismatches allowedto allowing only perfect base-pair matching.

stringency

aneuploidy

polyploidy

quantitative PCR

restriction fragment length polymorphism

I have an individual with a disease we will call SSI-itis,brought on by extreme exposure to cannabis. I do a southern andfind the gene affected. Another of his classmates exhibits theexact same symptoms, but that gene in hiw is wt. A southern in thesecond student finds a mutation in another gene. This is an exampleof

pleitropy

allelic heterogeneity

locus heterogeneity

dominant traits

recessive traits

penetrance

If a person carrying a dominant allele does not demonstrate thephenotype strongly, but still differs from wt, thisdemonstrates?

haploinsufficiency

penetrance

expressivity

monogenic traits

polygenic traits

I find three alleles for a gene in a fruit fly. One allelecreates the ability for the fly to eat graphite. Another allelecreates the ability for the fly to eat coal. The third alleleallows the fly to eat diamond. These alleles are all dominant, andwt cannot eat any of these substances. Which statement below is themost true?

Upon mating a bunch of the flies carrying these alleles, Ishould not find a fly that can eat graphite, coal, and diamond.

I would consider the ability to eat these substances an exampleof pleitropy because they are all carbon.

I would expect the ability to eat diamond to be dominant to theability to eat graphite.

These flies are demonstrating locusheterogeneity.

Expression vectors in prokaryotes do not make functionaleukaryotic gene products in bacteria very well because

the codon sequence for prokaryotes is different than the codonsequence in eukaryotes

there are no disulfide bridges formed in proteins normally madein prokaryotes

prokaryotic expression vectors cannot translate eukaryticsequences

eukaryotic genes have introns, and prokaryotes don't

eukaryotic genes have exons and prokaryotes don't

Which of the following is not used in PCR amplification ofDNA?

Double-stranded DNA template

Oligonucleotide primers

DNA polymerase

DNA ligase

ddNTPs

I create a knockout mouse using the agouti/black fur systemdiscussed in the online lecture and in the book. When I cross theagouti offspring of the originally engineered mouse, I find a ratioof 2 agouti mice to 1 black furred mouse. What is the bestexplanation?

The gene knocked out was recessive.

The gene knocked out was dominant.

The gene knocked out was a lethal gene.

The knockput was integrated into a random spot, and did notknockput the original gene.

A loss of function mutation is usually recessive, but canactually be a dominant mutation because of what effect?

haploinsufficiency

dominant negative mutations

allelic heterogeneity

locus heterogeneity

pleitropy

A genetic carrier of a disease can have a form of the diseasethat only shows intermittent symptoms, while the full homozygousrecessiuve individual would show all of the symptoms of thedisease.

True

False

Wild type, or wt, refers to the genotype that is the typicalgenotype found in nature. Now read that again. Once more. Nowanswer.

Answer

True

False

mauvekangaroo793

___1. Nationally, the most common cause of Down Syndrome is
a. nondisjunction of Chromosome 21 during meiosis
b. translocation of Chromosome 21 onto another chromosome
c. inversion of a large section of Chromosome 21
d. deletion of the Down Syndrome gene
___2. In humans, sex is determined by
a. The presence or absence of the Y chromosome
b. The ratio of X chromosomes to sets of autosomes
c. The temperature during fetal development
d. Which parent is stronger at the time of conception
___3. A translocation is
a. a chromosomal abnormality in which a piece of one chromosome breaks off and attaches to a different chromosome
b. an abnormality in which the chromosomes are translocated out of the nucleus
c. a condition in which the chromosome has broken in two places and come back together with the middle segment backward
d. an abnormality in which there is an extra copy of one chromosome
___ 4. In snapdragons, a red-flowered plant crossed with a white-flowered plant produces all pink-flowered offspring. But crossing two pink-flowered plants produces red, pink, and white in a 1:2:1 ratio. This is a good example of
a. Incomplete dominance
b. Dominance of the red allele over the white allele
c. Lethal effects of the pink allele
d. Sex-linkage of the flower-color alleles
___ 5. If a person with type AB blood marries a person with Type O blood, what blood types are possible in their children? (Assume this trait is a single-gene trait, not complicated by interaction with other genes).
a. A, B, AB, or O
b. A or B only
c. AB or O only
d. AB only 2
___6. In what phase of meiosis are haploid nuclei first formed? a. Anaphase I b. Metaphase II c. Telophase II d. Metaphase I e. Telophase I
___7. What is the phenotype of a heterozygous tall pea plant? a. tall b. Tt c. TT d. tt e. dominant
___ 8. If genes A and B are far apart on a chromosome, and A and C are close together, which of the following observations is expected?
a. Crossing over will occur more often between A and C than between A and B
b. Crossing over will occur at equal frequencies between A and C, and between A and B
c. Crossing over will occur more often between A and B than between A and C
___9. Sometimes an organismâs genotype for one gene masks the effect of a different gene. This phenomenon is called
a. Dominance
b. Incomplete dominance
c. Co-dominance
d. Epistasis
e. Variable expressivity
___10. The dominant allele of a gene is
a. The one that is expressed most often in a population
b. The one that is expressed most often in any particular cross
c. The one that is expressed in an individual that has two different alleles
d. The one that is most healthy to have, that gives the greatest fitness
___11. In human cells, DNA replication occurs
a. In S phase of interphase
b. In G1 phase of interphase
c. In G2 phase of interphase
d. In prophase of mitosis
e. Whenever the cell needs to read the DNA to make proteins
___ 12. A process in which recipient cells take up genes from DNA molecules floating free in the surrounding medium is called
a. conjugation
b. specialized transduction
c. generalized transduction
d. transformation 3
___ 13. Most calico cats are female. However, a few rare males are observed. What is the genotype of a male calico cat?
a. X BX 0
b. X OY
c. X BY
d. X BYY
e. X BX OY
___ 14. A type of red-green color-blindness is a sex-linked (X-linked) recessive trait in humans. Can a woman have this type of color-blindness?
a. Yes, a woman could be color-blind if her mother and father were both heterozygous.
b. Yes, a woman could be color-blind but only if she has an abnormal number of sex chromosomes.
c. No, women can't show sex-linked traits because they lack a Y chromosome.
d. No, women can't show sex-linked traits because they have two X chromosomes.
e. Yes, a woman could be color-blind if her father was color-blind and her mother was a carrier (heterozygous).

BIOL 1090 Lecture Notes - Lecture 9: Phenylketonuria, Polydactyly, Missense Mutation

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