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Genetics Exam 3

front 1

What type of aneuploidy is responsible for Turner syndrome in humans?

back 1

monosomy XO

front 2

When nondisjunction occurs early in embryogenesis rather than gametogenesis, what would you expect in the resulting karyotype?

back 2

mosaicism

front 3

In Drosophila and Lepidoptera, sex-chromosome mosaicism produces what type of sexually ambiguous phenotype?

back 3

gynandromorphy

front 4

Why is aneuploidy in animals generally detrimental?

back 4

It results in an imbalance of gene products from affected chromosomes, which alters normal development.

front 5

Which of the following accurately describes a possible meiotic nondisjunction event?
A. Meiosis fails to proceed to completion.
B. Fusion of gametes results in trisomy.
C. Sister chromatids fail to separate during meiosis I.
D. Homologs fail to separate during meiosis I.

back 5

Homologs fail to separate during meiosis I.

front 6

All of the following events occur during normal meiosis except _______.
A. two haploid gametes fuse to form a diploid cell
B. sister chromatids separate from one another during meiosis II
C. one diploid cell produces four haploid cells
D. homologous chromosomes separate from one another during meiosis I

back 6

A. two haploid gametes fuse to form a diploid cell

front 7

True or False:
In order to create the possibility of generating a trisomy, nondisjunction must occur during meiosis II.

back 7

False.

front 8

The phenomenon of ________ in allopolyploids consists of more rapid growth, increased fruit and flower production, and improved disease resistance.

back 8

Hybrid Vigor

front 9

Humans with the condition of familial Down syndrome exhibit which type of chromosomal aberration?

back 9

chromosome 14/21 Robertsonian translocation

front 10

In humans that are XX/XO mosaics, the phenotype is highly variable, ranging from females who have classic Turner syndrome symptoms to females who are essentially normal. Likewise, XY/XO mosaics have phenotypes that range from Turner syndrome females to essentially normal males. How can the wide range of phenotypes be explained for these sex-chromosome mosaics?

back 10

The resulting phenotype depends on the relative percentage of cells with each karyotype at different stages of development.

front 11

A boy with Down syndrome (trisomy 21) has 46 chromosomes. His parents and his two older sisters have a normal phenotype, but each sister has 45 chromosomes.
How many chromosomes do you expect to see in karyotypes of the parents?

back 11

One parent is expected to have 46 chromosomes and the other is expected to have 45 chromosomes.

front 12

A boy with Down syndrome (trisomy 21) has 46 chromosomes. His parents and his two older sisters have a normal phenotype, but each sister has 45 chromosomes.
What term best describes this kind of chromosome abnormality?

back 12

A fusion of two acrocentric chromosomes (Robertsonian translocation).

front 13

A boy with Down syndrome (trisomy 21) has 46 chromosomes. His parents and his two older sisters have a normal phenotype, but each sister has 45 chromosomes.
What is the probability the next child of this couple will have a normal phenotype and have 46 chromosomes?

back 13

Probability 1/3

front 14

The mature mRNA transcribed from the human β-globin gene is considerably longer than the sequence needed to encode the 146-amino acid polypeptide.
Select from the following list the names of three sequences located on the mature β-globin mRNA but not translated.
intron
promoter region
TATA box
3'-UTR
CAAT box
stop codon
5'-UTR

back 14

3' UTR, stop codon, 5' UTR

front 15

Determine which of the following sequences and structures represent part of mature eukaryotic mRNA.
poly-A tail
5'-UTR
termination sequence
start codon
5'-cap
3'-UTR
stop codon
promoter
AAUAAA
intron

back 15

Poly A tail, 5' UTR, start codon, 5' cap, 3' UTR, stop codon, AAUAAA

front 16

A tRNA anticodon is 5' GAA 3'. Answer the following questions.
Which one of the following codons is recognized by this tRNA?

5' TTC 3'
5' CTT 3'
5' UUC 3'
5' CUU 3'

back 16

5' UUC 3'

front 17

A tRNA anticodon is 5' GAA 3'. Answer the following questions.
The first base at the 5' end of the anticodon has relaxed base-pairing rules allowing some tRNAs to bind multiple codon sequences. This is called the third base "wobble" hypothesis. How many codons could a tRNA with the anticodon 5'-GAA-3' recognize?

back 17

2

front 18

A tRNA anticodon is 5' GAA 3'. Answer the following questions. Which amino acid does this tRNA carry?

back 18

Phenylalanine

front 19

Use the following copolymers to decipher the genetic code: UG and UGG.
Which codons are present in the UG copolymer?

back 19

UGU and GUG

front 20

Use the following copolymers to decipher the genetic code: UG and UGG.
Which amino acids are encoded by the UG copolymer?

back 20

UGG, GUG, GGU

front 21

Use the following copolymers to decipher the genetic code: UG and UGG.
Which amino acids are encoded by the UGG copolymer?

back 21

Valine, Tryptophan, Glycine

front 22

Use the following copolymers to decipher the genetic code: UG and UGG.
Which one of the following amino acids is encoded by both copolymers?

back 22

Valine

front 23

Use the following copolymers to decipher the genetic code: UG and UGG. Based on the results of the copolymer experiment, what triplet code can definitely be assigned to valine?

back 23

GUG

front 24

Below is a partial mRNA sequence from a mutant gene in E. coli. used to answer the following questions.
5'... ACCCGCCACGCC... 3'
Assume the reading frame contains the codon ACC.Which partial amino acid sequence is encoded within the partial mRNA sequence?

back 24

Nterm...T-R-H-A...Cterm

front 25

This gene is a mutant variant that severely slows the growth of the bacteria. The normal version of the encoded protein should contain the partial amino acid sequence (Nterm...T-R-N-A...Cterm). It's known that mutant forms of tRNA synthetases can charge tRNAs with a different amino acid than expected.If a mutant synthetase gene was introduced into the E. coli, which one could produce the normal version of the encoded protein?

back 25

synthetase that charges a histidine tRNA with asparagine

front 26

In the early 1960s, a number of investigators conducted a series of elegant experiments that helped to decipher the genetic code. Which of the following describes a bona fide property of the genetic code?

back 26

The genetic code consists of nonoverlapping codons.

front 27

Khorana used synthetic mRNAs to determine genetic code possibilities. To do so, he translated synthetic mRNA in vitro in the presence of individual 14C-labeled amino acids. Using the repeating dinucleotide Poly-UG, he identified a polypeptide with which two amino acids alternating?

back 27

Cys Val

front 28

Eighteen of the amino acids have two or more synonymous codons. Which two amino acids are the exceptions?

back 28

met, trp

front 29

How many different aminoacyl-tRNA synthetases can be found in a given cell?

back 29

20

front 30

The Shine-Dalgarno sequence in bacteria ________.
A. is a purine-rich consensus sequence found in the 16S rRNA subunit
B. is a pyrimidine-rich consensus sequence found in the 3 UTR of the mRNA
C. is a region of the tRNA molecule involved in formation of charged tRNAs
D. is a purine-rich consensus sequence found in the 5 UTR of the mRNAis a consensus sequence involved in the termination of translation

back 30

is a purine-rich consensus sequence found in the 5 UTR of the mRNA

front 31

A tRNA in the P site of the ribosome will enter the ________ site after translocation of the ribosome.

back 31

E

front 32

Which step in translation initiation is unique to eukaryotes?

back 32

binding of ribosomes to the 5' cap of the mRNA

front 33

A portion of an mRNA attached to a ribosome reads:

5 GACCAUUUUUGA 3

If a tRNA with a histidine amino acid attached is in the P site of the ribosome, a tRNA with which amino acid attached will enter the A site?

back 33

phenylalanine

front 34

During elongation, the charged tRNA is recruited to which location on the ribosome?

back 34

A site

front 35

Binding of what protein initiates translation-termination events that result in polypeptide release and dissociation of ribosomal subunits?

back 35

Release factor

front 36

Once elongation is underway, tRNAs involved in the process occupy a series of sites on the complexed ribosome. The occupation of sites occurs in the following order.

back 36

A Site, P Site, E Site

front 37

When a peptide bond is formed between two amino acids, one is attached to the tRNA occupying the P site and the other _______.
A. is free in the cytoplasm
B. is attached to the tRNA occupying the A site
C. is attached to the tRNA occupying the E site
D. is attached through hydrogen bonds to the mRNA

back 37

is attached to the tRNA occupying the A site

front 38

At which site does the charged initiator tRNA bind during protein synthesis?

back 38

P site

front 39

True or False:
The enzyme EF‑Tu catalyzes the formation of a peptide bond between the amino acid held by the tRNA in the A site and the elongating amino acid chain held by the tRNA in the P site.

back 39

False

front 40

What event occurs during translocation?
A. The polypeptide is cleaved from the terminal tRNA.
B. Amino acids are added to the polypeptide chain.
C. The two ribosomal subunits join to form a complex.
D. mRNA shifts in the 5' direction along the ribosome.

back 40

mRNA shifts in the 5' direction along the ribosome.

front 41

Bacteria group their genes such that they share a single promoter and the mRNA transcript synthesizes several different polypeptides. Collectively, these are referred to as ________ mRNAs, which are part of the operon system.

back 41

polycistronic

front 42

Why do patients with sickle cell disease commonly suffer from chronic anemia?

back 42

Sickled red blood cells are damaged easily and therefore have a shorter life span than normal red blood cells.

front 43

Which of the following best describes the basic structure of hemoglobin?
A. It is a tetramer composed of either four α-globin molecules or four β-globin proteins.
B. It is a heterodimer composed of one α-globin and one β-globin protein.
C. It is a dimer composed of either two α-globin molecules or two β-globin proteins.
D. It is a heterotetramer composed of two α-globin and two β-globin proteins.

back 43

It is a heterotetramer composed of two α-globin and two β-globin proteins.

front 44

Sickle cell disease is caused by a mutation in the β-globin gene. Although this may be one of the most common globin gene variants, it is not the only one. Approximately how many different alleles of the human α- and β-globin genes have been documented worldwide?

back 44

Several hundred

front 45

The mutation that causes SCD causes what kind of change in the structure of hemoglobin?

back 45

a single amino acid substitution that alters the structure of the tetramer

front 46

Pauling used gel electrophoresis to confirm that homozygous individuals produce a single form of β-globin protein that differs depending on the homozygous genotype as well as that heterozygotes produce proteins of both types in approximately equal concentration. What characteristic(s) of the bands proved this?

back 46

both the positions and the densitometry peaks of each band (both A and B)

front 47

Which of the following techniques describes either the rate of a molecule's migration or its final position in a gel?

back 47

electrophoretic mobility

front 48

Molecular charge is an important characteristic influencing protein migration on a gel. What accounts for the variation in the molecular charge of proteins?

back 48

the amino acid composition of the protein

front 49

The βS and βA alleles can be distinguished on the basis of ________.

back 49

variations in the DNA restriction fragments and different electrophoretic mobilities of the protein products due to a single amino acid substitution. (both A and C)

front 50

How is the DNA visualized on a Southern blot?

back 50

either radioactively labeled probes or chemically labeled probes

front 51

The intercalating agent ________ allows researchers to visualize DNA and RNA on a gel when the gel is exposed to UV light.

back 51

ethidium bromide

front 52

Gel electrophoresis is used to separate biological molecules based on size, charge, and shape. Which of the following statements about electrophoretic mobility is true?
A. Large molecules migrate more quickly than do small molecules.
B. Molecules with greater positive charge migrate toward the positive pole more quickly than do molecules of the same size with less positive charge.
C. Linear molecules will migrate more quickly than will tightly condensed globular molecules with the same mass.
D. Small molecules migrate more quickly than do large molecules.

back 52

Small molecules migrate more quickly than do large molecules.

front 53

What are restriction enzymes?

back 53

endonucleases that cut double-stranded DNA at a precise nucleotide sequence (typically 4-6 nucleotides)

front 54

The experimental technique that involves transfer of DNA from an electrophoresis gel to a membrane, followed by detection with a DNA probe, is known as _______.

back 54

Southern blot

front 55

Globin genes contain ________ segments that ultimately encode the amino acids of the protein and interspersed ________ sequences that do not contribute to encoding the protein product of the gene.

back 55

exon, intron

front 56

Can Southern and northern blotting detect differences between WT and mutant β-globin gene or between WT and mRNA or both?

back 56

Only differences between WT and the gene can be detected.

front 57

True or False:
An RFLP marker that cosegegrates with a disorder exists in most people who have the disorder and is absent in most people who don't have the disorder.

back 57

True

front 58

A portion of the β‑globin allele responsible for sickle‑cell anemia contains the sequence CCTGTGGAG, whereas the same region of the normal β‑globin allele contains the sequence CCTGAGGAG. Which allele would be cut by the restriction enzyme DdeI, which cuts at the sequence CTNAG, where N is any base?

back 58

Only the normal allele will be cut.

front 59

The mutation resulting in the allele for sickle‑cell anemia removes a restriction site from the β‑globin gene such that a specific probe hybridizes to one 1.3‑kb fragment instead of two fragments of 1.1 kb and 0.2 kb. What restriction pattern will be seen in sickle‑cell carriers?

back 59

Three fragments of 1.3, 1.1, and 0.2 kb

front 60

Please briefly explain the role of complementary base pairing in the initiation of translation in prokaryotes. Be specific.

back 60

A portion of the 16S rRNA that is part of the small subunit of the ribosome binds to the Shine-Dalgarno sequence of the mRNA by means of complementary base pairing. This specificity determines where translation will be initiated. The start codon is near the Shine-Dalgarno sequence.

front 61

The initiation of translation in eukaryotes is rather different than in prokaryotes. a) Where is translation initiated in eukaryotes and b) what important implication does this have for the number of polypeptides that can result from a single mature mRNA in eukaryotes?

back 61

a) In eukaryotes, translation is initiated when the initiation complex binds at the 5’ cap. The initiation complex consists of the small subunit of the ribosome, several initiation factors and a charged initiator tRNA. This initiation complex moves along the mRNA until it reaches the correct start codon.
b) because there is only one 5’ cap, only one polypeptide can result from a single mature mRNA.

front 62

Suppose you have isolated RNA from the cells of flower petals and also from leaf cells of the California Poppy for RT-PCR.
a) which enzyme does RT-PCR use?
b) which primer would you use to study all the genes expressed in each tissue?
c) what is the term for the results of this RT-PCR (one word)?

back 62

a) reverse transcriptase
b) You choose a poly-dT primer to bind with the poly A tail of all mRNA.
c) This will produce cDNAfor all mRNA in the tissue.

front 63

Now you use the resulting RT-PCR products and primers specific for gene R for standard PCR for the sample from the petals and the sample from the leaves. Gene R is expressed in the leaf cells, but not in the petals; the entire gene is 10 kb in length with 5 exons that have a total length of 1.2 kb. Please draw a gel that shows the final PCR products from each sample. Label your gel carefully and briefly explain your results.

back 63

Your gel should show three lanes each with its own well. One lane is for the molecular weight markers, one is for the PCR products that used cDNA from petals as a template and another for the sample that used cDNA from leaf cells as a template.
The one band of 1.2 kb will be in the lane with the PCR products from the leaf cells. The other lane from the PCR products from the petals will have no bands. Gene R is not expressed in those cells and thus there would have been no RNA from that gene in petal cells. The lane with the molecular weight marker should show bands of varying sizes that will help determine the size of the 1.2 kb band.

front 64

a) Ribosomes are composed of twosubunits that are each composed of proteins and rRNA.
b) How is translation terminated?

back 64

Translation is terminated when the ribosome comes to the stop codon and no charged tRNA move into the A site, because no tRNA have an anticodon for any of the three stop codons. Instead a protein called a release factor moves into the A site and the polypeptide is released and the two subunits of the ribosome dissociate

front 65

What does the term quaternary refer to with regard to protein structure? What is a familiar example (one word)?

back 65

A quaternary structure is formed when two or more polypeptides form a functional protein. Hemoglobin is a familiar example.

front 66

Deciphering the genetic code: a) What is found in a cell-free translation system that is necessary for in-vitro translation? b) Why is DNase added?

back 66

a) The cell-free translation system contains tRNA’s, functional ribosomes and other proteins. Thus translation can occur in vitro.
b) DNase is added to remove any DNA present in the cell-free extract to prevent transcription from occurring and thus the presence of RNA molecules besides the synthetic ones that the scientist adds. If those indigenous RNA molecules were to be translated this would severely affect the ability of the researcher to interpret the polypeptides found after translation.

front 67

Sickle cell disease: a)at the molecular level how do the hemoglobin molecules that contain the sickle cell disease causing beta-globins behave that causes the red blood cells to form their characteristic sickle shape? b) How does the fitness of the heterozygotes with genotype HbA HbS vary depending on the altitude?

back 67

a) the hemoglobin molecules will aggregate together and form long chains or fibers.
b) heterozygotes have a similar fitness as homozygote for the dominant allele at lower elevations, but at higher elevations and with increased exertion individuals often suffer serious physical symptoms.

front 68

a) What is the process by which members of a gene family such as the globin gene family originally arose within an individual (two words)? b) different genes in the globin gene family are expressed in different tissues. Please briefly explain.

back 68

a) gene duplication
b) myoglobin is expressed in muscle tissue, while aand bglobin genes are expressed in blood cells

front 69

Briefly, what is a pseudogene?

back 69

A pseudogene does not code for a functional product due to the presence of a premature stop codon or a deletion that results in the loss of important amino acids. Insertions or deletions can also result in frameshift mutations.

front 70

What is the basic structure of an amino acid? What is the portion of the molecule that varies?

back 70

An amino acid has a central carbon, an amino group, and a carboxyl group and an R group. The R group varies.

front 71

What are the four main groups based on characteristics of the side chain that amino acids can be placed into? These groups vary in their ability to form particular types of bonds. What are two examples?

back 71

The R group or side chain of an amino acid can be 1) non-polar, 2) polar, 3) basic or 4) acidic. For example, serine has a polar side chain, while glycine has a non-polar side chain.

front 72

What are the four levels of structure of a protein? What are the two types of secondary structure?

back 72

primary – the sequence of amino acids

secondary - the formation of either an alpha helix or beta sheet with hydrogen bonds

tertiary – additional folding based on hydrogen, ionic or disulphide bonds

quaternary – the bonding of separate polypeptides into a functional protein (not all proteins have a quaternary structure

front 73

Why does it matter which amino acid is at a particular site in a polypeptide (in general and specifically)?

back 73

The function of a protein often depends on its secondary, tertiary and quaternary structure. That structure depends on hydrogen, ionic and disulphide bonds between different regions of the polypeptide.

front 74

What are the functions of a ribosome?

back 74

The ribosome aids in initiation by binding to the mRNA and helps identify the start codon. During elongation the ribosome helps the anticodon of the tRNA and codon of the mRNA be in position so that complementary base pairing can occur between them. Importantly, the ribosome also catalyzes the peptide bond between amino acids (the last one that was added to the polypeptide chain and the one on the tRNA in the A site).

front 75

What are three important regions of a ribosome?

back 75

Aminoacyl site (A site) – binds a new tRNA molecule containing an amino acid to be added to growing polypeptide chain

Peptidyl site (P site) – holds tRNA to which polypeptide is attached

Exit site (E site) – provides avenue for exit of tRNA after its amino acid has been added to chain

front 76

What are the three stages of translation? Very briefly, and in general, what happens in each of them?

back 76

Initiation – the small subunit of the ribosome attaches at or near the 5’ end of the mRNA, the initiator RNA binds to the start codon and the large subunit of the ribosome joins the small subunit. Elongation is the addition of amino acids to a growing polypeptide chain. Note that the ribosome moves not the mRNA. Termination occurs when the ribosome reaches the stop codon and the components separate.

front 77

Please describe the basic structure of a ribosome. In general, how is the ribosome of a bacterium different from that of a eukaryote?

back 77

There is a large subunit and a small subunit. In bacteria the large subunit is 50S and the small subunit is 30 S, while the assembled ribosome is 70S. In eukaryotes the large subunit is 60S and the small subunit is 40 S, while the assembled ribosome is 80S.

The eukaryotic ribosome is a bit larger than the prokaryotic ribosome.

front 78

Bacteria: How is translation initiated in prokaryotes? What is the Shine-Dalgarno sequence? Where does the initiator tRNA enter the ribosome?

back 78

A small portion of the 16S rRNA has complementary base pairing with the Shine-Dalgarno region of the mRNA. Once these two have bound, the initiator tRNA enters the P site and the large subunit binds to the smaller subunit to form a functional ribosome.

front 79

What is the role of the 5’ cap in the initiation of translation in eukaryotes?

back 79

The eIF4 (eukaryotic initiation factor 4) binds to the 5’ cap. If the 5’ cap is not present translation does not occur.

front 80

What is the role of the eukaryotic initiation factor 4 (eIF4)?

back 80

It binds to the 5’ cap of the mRNA.

front 81

When does the large subunit enter the picture?

back 81

When the initiation complex is bound to the correct start codon the 60S subunit binds to the complex to form the functional ribosome.

front 82

During elongation, where do the charged tRNA’s enter the ribosome and where do they leave?

back 82

The charged tRNA’s enter A site (except for the initiator tRNA which enters at the P site).

front 83

How does translation terminate? What is the role of release factors?

back 83

When the ribosome reaches a stop codon a release factor enters the A site causing the dissociation of the ribosome, mRNA and polypeptide.

front 84

Why can’t transcription and translation be coupled in eukaryotes, but can be coupled in prokaryotes?

back 84

In eukaryotes, transcription and RNA processing occur in the nucleus whereas translation occurs in the cytoplasm. In prokaryotes both processes can occur simultaneously as both processes occur in the cytoplasm.

front 85

What is the functional difference between a monocistronic and a polycistronic mRNA? Do eukaryotes have polycistronic mRNA? Why not?

back 85

A polycistronic mRNA codes for more than one polypeptide, while a monocistronic mRNA only codes for one polypeptide. All eukaryotic mRNA are monocistronic, but both types are present in bacteria.

front 86

What determines the reading frame?

back 86

The start codon

front 87

What does redundant mean with regard to the genetic code?

back 87

More than one codon codes for the same amino acid.

front 88

Which codon position is the wobble position? What does this mean?

back 88

It is the third codon position. Complementary base pairing is not as strict in this position.

front 89

What is inosine and where is it found? Which bases does it have complementary base pairing with?

back 89

Inosine is a modified base and can be found at the 5’end of the anticodon of the tRNA. It can pair with U, C, or A.

front 90

Does the genetic code overlap? What would a consequence of an overlapping genetic code?

back 90

The genetic code does not overlap. If it did a single base pair substitution would result in more than one amino acid substitution.

front 91

What is a frameshift mutation and causes them? Why are they a serious type of mutation?

back 91

A frameshift mutation shifts the reading frame of a mRNA. It is caused by either insertions of deletions of bases (not in multiples of three).

front 92

Is the genetic code read continuously or are there spacers in it?

back 92

The genetic code is read continuously. Otherwise a insertion or deletion or bases not in a multiple of three would not change the reading frame.

front 93

What is needed for in-vitro translation? Why is DNase added?

back 93

A cell-free translation system contains ribosomes, enzymes, tRNA and a synthetic mRNA added by the scientist.

front 94

What synthetic mRNA’s did the first experiments use?

back 94

Poly U that resulted in a polypeptide with only phenylalaine followed by poly C (only proline), poly A (only lysine) and poly G (only glycine).

front 95

What synthetic mRNA’s did the second experiments use?

back 95

The second group of experiments used copolymers of repeating units such as CACACACACACACACA. This latter synthetic mRNA gives a polypeptide with alternating histidine and threonine.

front 96

What did the final definitive experiments use?

back 96

The final conclusive experiments used all 64 triplets individually to resolve some ambiguities and confirm identities of some codons.

front 97

The genetic code is mostly universal. What does that mean? How many times did the genetic code evolve?

back 97

All organisms use the same genetic code. However, few organisms have a few codons that code for something different. The mitochondrial genetic code is also slightly different.

front 98

What is the difference between a charged and an uncharged tRNA?

back 98

A charged tRNA has an amino acid attached to it, whereas an uncharged tRNA does not have an amino acid attached to it.

front 99

Which enzymes charge the tRNA? Does this process require energy? How many of those enzymes are there?

back 99

Aminoacyltransferases charge the tRNA molecules.

These enzymes attach the correct amino acid to the right tRNA molecule, that is the one with the anticodon that matches the codon for the amino acid.

front 100

Why study gene families?

back 100

A gene family arises when an ancestral gene is duplicated usually more than once where all genes in the family arose from a common ancestral gene. We have concentrated on the globin gene family, but most eukaryotic protein-coding genes belong to gene families.

front 101

What are homologous genes?

back 101

Homologous genes have arisen from a common acnstral gene. There are two main types orthologous and parologous genes. Orthologous genes are any of the same gene in both species and have arisen from a speciation event. Paralogous genes have arisen by means of gene duplication and represent a comparison of genes within a species or among species. Paralogs may have similar, but slightly different functions – they may be expressed at different stages of the organisms life cycle or may show tissue specific expression.

front 102

Using a chimp and gorilla as the organisms, please give an example of two paralogs and two orthologs.

back 102

The a-globin gene in the chimp and the a-globin gene in the gorilla are orthologs; the a-globin gene in the chimp and the b-globin gene in the chimp or the gorilla are paralogs.

The myoglobin gene in the chimp and the myoglobin gene in the gorilla are orthologs;
the z-globin gene in the chimp and the g-globin gene in the chimp or the gorilla are paralogs.

front 103

Different members of the globin gene family have tissue specific expression. Please give an example.

back 103

In a fetus the hemoglobin molecule contains two molecules of a-globin and two molecules of d-globin, but in an adult the hemoglobin molecule contains two molecules of a-globin and two molecules of b-globin. Thus different globin genes are expressed at different stages in the life of the organism.

front 104

Different members of the globin gene family have life stage specific expression. Please give an example.

back 104

There is also tissue specific expression – myoglobin is expressed in the muscles and different globin genes are expressed in immature red blood cells and form hemoglobin proteins.

front 105

What is a pseudogene?

back 105

A pseudogene does not code for a functional product due to the presence of a premature stop codon or an insertion that adds unnecessary amino acids or deletion that results in the loss of important amino acids. Insertions or deletions can also result in frameshift mutations. Some of these may result in premature stop codons.

front 106

What is the basic gene structure of the alpha and beta globin genes?

back 106

Both types have three exons of similar size and two introns.

front 107

What is sickle cell disease and what is its genetic basis? What do the mutant hemoglobin molecules do that causes trouble?

back 107

Sickle cell disease is a serious disease that results in severe anemia, muscle pain and tissue damage that is caused by a mutation in the b-globin gene that substitutes valine for glutamic acid. Hemoglobin molecules that include mutant b-globin proteins will aggregate together and form long chains or fibers.

front 108

What causes the anemia symptoms of sickle cell disease?

back 108

The oxygen carrying capacity of the hemoglobin molecules is diminished.

front 109

In most environments heterozygotes have the same phenotype as individuals with the homozygous dominant genotype. In which two environments do the heterozygotes have decreased or increased fitness?

back 109

Heterozygotes have increased fitness in an environment with malaria as they are able to clear the parasite from the blood more easily. However, at high elevations they are more susceptible to anemia and other ill effects of the presence of the sickle cell trait due to their lower oxygen carrying capacity.

front 110

What is a gel made of and why is that a useful material?

back 110

Gels are made of agarose or polyacrylamide. Neither interacts with nucleic acids or proteins and allow both molecules to move through pores within them.

front 111

On what basis are nucleic acids separated? On what basis are proteins separated?

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Proteins are separated on the basis of charge and shape, while nucleic acids are separated primarily on the basis of molecular weight and to a smaller extent on shape.

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How are nucleic acids visualized in a gel? How are proteins visualized in a gel?

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Nucleic acids are often visualized with ethidium bromide, although many other newer dyes exist. Proteins can be visualized with a general protein stain such as Coomassie blue.

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Why do some proteins have different electrophoretic mobilities? Do all proteins that have had amino substitutions have different electrophoretic mobilities? Why or why not?

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They have different amino acid composition – recall that amino acids may have different charges that will affect the overall charge of the protein. No, some amino acid substitutions do not affect electrophoretic mobility.

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Linus Pauling’s work – what were the samples he ran out on the gel? What was their conclusion? Which important concept did he come up with as a result of this work?

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They ran out b-globin proteins from individuals of three different genotypes on a gel. They were able to demonstrate that the individuals with the bSbS genotype had a single protein that migrated more slowly than did the protein from individuals with a bAbA genotype and that individuals with a bAbS genotype had two bands, one of each size.

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What is a SNP? What are the three possible genotypes with regard to a SNP?

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Single nucleotide polymorphism. Homozygous for one form of the SNP, homozygous for the other form of the SNP and heterozygous for the two forms of the SNP.

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In general, what are two ways that SNP’s can be found?

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They can be easily found by aligning sequences of DNA and looking for differences. They can also be detected with RFLP analyses if the SNP affects a restriction site.

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What is a restriction endonuclease? and what does it do? What is a restriction sequence (an equivalent term is restriction site)?

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They are enzymes that cut DNA at specific locations called restriction sequences (or restriction sites) that are usually six bp in length and may be palindromes.

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If genomic DNA is cut with restriction endonuclease such as EcoRI how many different fragments will result? How will those fragments look when run out on a gel?

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A large number of fragments will result; they will look like a smear or blur.

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How can mutation change the presence or absence of a restriction sequence?

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A base substitution can change a base such that the restriction sequence is no longer recognized by the restriction endonuclease (Figure 10.2). A base substitution can also change a base in a sequence that is already similar to a restriction sequence but differs by one base – after the substitution the sequence now matches a restriction sequence and can be recognized by a restriction endonuclease.

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What is the overall goal using molecular probes? What type of probe can be used to find a particular nucleic acid sequence?

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Molecular probes can be used to find a single molecule in a heterogeneous mixture of molecules.

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What does the term hybridization mean in the context of Southern and Northern blot techniques?

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Hybridization refers to complementary base pairing between two different molecules, the probe and the target molecule.

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What are the basic steps of a Southern blot technique?

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Cut DNA (usually genomic DNA) with a restriction endonuclease; run out resulting fragments on a gel; blot (transfer) fragments onto a nylon membrane, probe with a radioactive DNA probe; expose X-ray film by placing membrane next to X-ray film; the X-ray film will be exposed by the radioactivity from the probe if it has bound to any fragments on the membrane. This allows one to detect one or more fragments that the radioactive probe hybridized with.

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What are the basic steps of a Northern blot technique?

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Extract RNA; run out on a gel; blot (transfer) onto a nylon membrane; probe with a specific radioactive DNA probe; expose x-ray film by placing next to membrane; the X-ray film will be exposed by the radioactivity from the probe if it has bound to any molecules on the membrane. This allows one to detect one or more RNA molecules that the radioactive probe hybridized with and thus detect presence and size of the RNA molecule.

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You want to determine if an individual is a heterozygote for the sickle cell trait. Can you use results of a Northern blot experiment to determine that? Why or why not?

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No, the wild-type and mutant SCD alleles of b-globin gene produce a mRNA of the same size. Thus northern blot analysis does not detect variation in this example.

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How are proteins detected in a Western blot?

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Proteins are visualized with primary antibodies coupled to secondary antibodies coupled to an enzymatic reaction that produces light or a colored band.

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Why are we interested in chromosome number?

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Chromosome number changes can be important in evolution and species diversification, but chromosome number changes can also be important in human (and plant and animal) disease.

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Why are mules sterile?

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A donkey is 2n=62 and a horse is 2n=64. A mule is 2n=63 and meiosis rarely results in viable gametes.

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Meiosis – what separates during meiosis I? What separates during meiosis II?

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Homologous chromosomes (homologs) separate during meiosis I and sister chromatids separate during meiosis II.

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One species is 2n=14 and a related species is 2n=28. Is this difference euploidy or aneuploidy? How do you know?

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This is euploidy as it represents a difference of two entire sets of chromosomes. 2n=28 is twice that of 2n=14.

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Meiotic non-disjunction – what is this and when can it occur?

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Chromosomes or sister chromatids do not separate properly during either meiosis I or meiosis II respectively

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What is the difference between an organism that is trisomic for chromosome 2 and one that is monosomic for chromosome 2?

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An individual that is trisomic has three copies of chromosome 2, but an individual that is monosomic for chromosome 2 has only one copy of that chromosome.

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Gene dosage – what is it? And, in general, how can it be harmful?

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Gene dosage is the number of copies of a gene. If gene dosage differs among genes it can be harmful, thus an individual with three copies of some genes but only two copies of most genes may have phenotypic effects especially in animals.

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What are the differences among individuals of Datura stramonium that are trisomic for different chromosomes and how do they differ from the typical diploid?

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The text shows differences in fruit morphology among individuals that are trisomic for each of the 12 chromosomes. The typical diploid has a different morphology than any of the aneuploids.

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Why aren’t there more humans that are trisomic for different chromosomes besides 13, 18, and 21?

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They are not viable due to serious gene dosage effects on development.

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What are some examples of human chromosome numbers for sex-chromosome trisomics?

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XXY – Klinfelter’s syndrome; XXX – triple X syndrome; XYY – Jacob syndrome

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Trisomy 21 – what is another common name for this syndrome? Genes in which region may be responsible for the symptoms of the syndrome?

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Trisomy 21 may also be called Down’s syndrome. Genes in the Down’s syndrome critical region of chromosome 21 seem to be responsible for most of the phenotypic effects.

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Why are trisomics semisterile?

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Meiosis does not go well because the three homologous chromosomes can either form a trivalent or a bivalent with an univalent. Both of those situations will result in only half of the gametes being viable.

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What is a genetic mosaic and how is this relevant to Turner’s syndrome?

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A genetic mosaic has some cells with a particular chromosome number, while other cells have another chromosome number. A woman with Turner’s syndrome may have some cells with only one X chromosome, some cells with two X chromosomes, and even some cells with three X chromosomes (see Figure 13.5). This may affect the severity of the individuals symptoms.

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Suppose a man has Klinefelter’s syndrome, that is, he is XXY. Neither of his parents are colorblind, but he is. Please explain carefully how this could happen. Be specific.

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If his mother is a heterozygote for red-green color-blindness and non-disjunction during meiosis II occurred she could produce a gamete that is Xcb Xcb. This gamete could fuse with a gamete with a Y chromosome and give a zygote with two X chromosomes and one Y chromosome. Xcb Xcb Y. Thus he is colorblind. If non-disjunction occurred during meiosis I then he would have one copy of each type of X chromosome and would not be colorblind.

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What is a gyandromorph and in which organisms are they found?

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This is an organism that is half one sex and half the other sex (see Figure 13.6). This can occur in fruit flies and moths.

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Down’s syndrome: a) if an individual with Down’s syndrome has a total of 47 chromosomes, what was the cause of his disease? Be specific. b) if meiotic disjunction is part of your answer to part a – did non-disjunction occur during meiosis I or meiosis II? Please explain.

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  1. a) meiotic non-disjunction in one parent, usually the mother (about 90 or more per cent of the time). b) in this case it is irrelevant whether it occurred in Meiosis I or Meiosis II, although in fact, with Down’s syndrome non-disjunction occurs more often during Meiosis I (about 70% of the time when non-disjunction was maternal).

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  1. a) What is a Robertsonian translocation? b) how is it different from a reciprocal translocation?

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  1. a) A Robertsonian Translocation occurs when two acrocentric chromosomes fuse together and thus a translocation carrier has one less chromosome than the normal diploid number, for example, 2n=45 versus 2n=46. b) a reciprocal translocation involves the exchange of genetic material from one non-homologous chromosome to another. There is no change in chromosome number.

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A family has a child with Down’s syndrome and she has a total of 46 chromosomes. What is the likely cause of this child’s Down’s syndrome?

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The child has trisomy 21 with one chromosome the result of a Robertsonian translocation. One parent have the following karyotype 45 (rob 14:21). Then during meiosis the parent may produce gametes with the Robertsonian translocation 14:21, and another copy of 21; when this gamete fuses with a normal gamete with one copy each of chromosomes 14 and 21, the zygote will have trisomy 21, but only has 46 chromosomes).

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This child and the child in question eighteen both have Down’s syndrome and both have sisters with normal phenotypes. When the two sisters grow up are either of them more likely or less likely to have a child with Down’s syndrome? Please explain carefully.

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The girl whose sister with Down’s syndrome arose from meiotic non-disjunction is not any more likely to have a child with Down’s syndrome than usual. On the other hand the girl whose sister with Down’s syndrome arose from the presence of a Robersonian translocation is more likely herself to have a child with Down’s syndrome as she has a significant probability of being a carrier for the Robertsonian translocation. This type of Down’s syndrome is termed familial Down’s syndrome.