Exam 4-Review

Welcome to K101 Recitation

Last week before Thanksgiving!!

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Week 13

​Agenda

• ​Announcements and Events

• ​Short Term Goals?

• ​Genetics Problems

• ​Exam Review

• ​Wrap Up

Genetics Problems!!​

Poll

Hardy Weinberg:

If 9% of an African population is born with a severe form of sickle-cell anemia (ss), what percentage of the population will be more resistant to malaria because they are heterozygous (Ss) for the sickle-cell gene?

a) The frequency of sickle cell disease. (q2)

b) The frequency of the "s" allele. (q)

c) The frequency of the "S" allele. (p)

d) The frequencies of homozygous dominant individuals "SS" (p2)

e) The frequencies of the heterozygote "Ss." (2pq)

• I know how to solve this!!

• Need some hints!!

• Totally confused!!

• Too long to read!!

​Hardy Weinberg... Think about it and try to solve

​If 9% of an African population is born with a severe form of sickle-cell anemia (ss), what percentage of the population will be more resistant to malaria because they are heterozygous (Ss) for the sickle-cell gene?

​

a) The frequency of sickle cell disease. (q2)

b) The frequency of the "s" allele. (q)

c) The frequency of the "S" allele. (p)

d) The frequencies of homozygous dominant individuals "SS" (p2)

e) The frequencies of the heterozygote "Ss." (2pq)

​Hardy Weinberg...Answer

​If 9% of an African population is born with a severe form of sickle-cell anemia (ss), what percentage of the population will be more resistant to malaria because they are heterozygous (Ss) for the sickle-cell gene?

​

a) The frequency of sickle cell disease. (q2) 9% =.09 = ss = q2

b) The frequency of the "s" allele. (q) q = Square root of .09 = .3

c) The frequency of the "S" allele. (p) p = 1 - .3 = .7

d) The frequencies of homozygous dominant individuals "SS" (p2) p2 = 0.49 (49%)

e) The frequencies of the heterozygote "Ss." (2pq) 2pq = 2(.7x.3) = .42 (42%)

Multiple Choice

Hardy-Weinberg: In a population with two alleles for cystic fibrosis, C and c, the frequency the recessive allele is 0.6. What percent of the population would be heterozygous carriers for cystic fibrosis if the population is in Hardy-Weinberg equilibrium?

• 16%

• 48%

• 40%

• 64%

​Pedigree...Think and Try to solve it

​The trait represented by the colored circles and squares below is inherited as a dominant allele. This is not a sex-­linked trait. Shaded individuals show the dominant trait. What is the probably genotype of each individual?

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​a. Are there any homozygous dominant individuals in the pedigree above? How do you know?

b. What is the probability of the trait appearing in offspring if 7 should marry 9?

c. What is the probability of the trait appearing in offspring if 8 should marry 10?

​Pedigree...Answer

​Are there any homozygous dominant individuals in the pedigree above? How do you know? NO!! Some offsprings are recessive.

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What is the probability of the trait appearing in offspring if 7 should marry 9? Trait will NOT appear!!

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What is the probability of the trait appearing in offspring if 8 should marry 10? 50% of the offspring will have the dominant trait.

Multiple Choice

The trait found in the individuals represented by the shaded symbols is rare in the general population. Which of the following patterns of transmission for this gene is/are consistent with this pedigree?

• Autosomal Dominant

• Autosomal Recessive

• X-linked Dominant

• X-linked Recessive

Four genes, B, C, E, and M are located on the same human chromosome. The distances between the genes are below:

 

Genes	Map Unit Distance
C - E	18
C - B	10
B - E	8
C - M	2
E - M	20

​

      Given this information, how would you order the genes on this chromosome?

​a.     B-E-M-C

b.     E-B-C-M

c.     B-E-C-M

d.     C-B-E-M

​Linkage...Think and try to solve it

​Linkage...Answer

​Four genes, B, C, E, and M are located on the same human chromosome. The distances between the genes are below:

 

Genes	Map Unit Distance
C - E	18
C - B	10
B - E	8
C - M	2
E - M	20

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Given this information, how would you order the genes on this chromosome? b. E-B-C-M

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Exam Review!!​

Multiple Choice

Based on your understanding of the sugars, nitrogenous bases, and hydrogen bonding between nucleic acids, what does the diagram below represent?

• 1 is ribose, 2 is Uracil (U), 3 is Adenine (A) 

• 1 is deoxyribose, 2 is Guanine (G), 3 is Cytosine (C) 

• 1 is ribose, 2 is Cytosine (C), 3 is Guanine (G)

• a. 1 is deoxyribose, 2 is Adenine (A), 3 is Uracil (U) 

Multiple Choice

Which of the following DNA replication enzymes is incorrectly described?

• Topoisomerase: relieves strain in the DNA helix ahead of the replication fork

• Ligase:  Makes a covalent phosphodiester bond to join Okasaki fragments

• Primase: Removes RNA and adds the equivalent DNA to the 3’ end of the Okasaki fragment

• DNA Polymerase III: Adds new nucleotides to the free 3’ OH end of a nucleotide chain

Multiple Choice

Cytosine makes up 38% of the nucleotides in a sample of DNA from an organism. Approximately, what percentage of the nucleotides in this sample will be thymine?

• 12%

• 24%

• 31%

• 28%

Multiple Choice

The strands that make up DNA are antiparallel. This means that

• the twisting nature of DNA creates nonparallel strands.

• one strand contains only purines and the other contains only pyrimidines.

• one strand is positively charged and the other is negatively charged.

• the 5' to 3' direction of one strand runs counter to the 5' to 3' direction of the other strand.

Multiple Choice

Which of the following scientists is incorrectly matched with their discoveries?

• Rosalind Franklin - X-ray crystallography of Photo 51

• Maurice Wilkins –  The bases in DNA follow the role of A=T, G=C

• James Watson and Francis Crick – Double Helical Structure of DNA

• Linus Pauling:  DNA is a triple helix with the bases on the outside

Multiple Choice

What is the function of topoisomerase?

• relieving strain in the DNA ahead of the replication fork

• stabilizing single-stranded DNA at the replication fork

• elongation of new DNA at a replication fork by addition of nucleotides to the existing chain

• unwinding of the double helix

Multiple Choice

Which of the following is the attachment site for RNA polymerase?

• the Upstream Promoter Elements

• the AUG codon

• the TATAAA box

• the small subunit of the ribosome

Multiple Choice

In virtually all organisms, the first AUG codon codes for:

• the termination of chain elongation.

•   the initiatior amino acid of translation, f-Met.

• the termination of transcription.

• a termination tRNA molecule.

Multiple Choice

The portion of the molecule labeled 5 is:

• the attached amino acid.

• a single-stranded region.

• the codon.

• a double-stranded region.

Multiple Choice

Use the codon chart below to determine the answers to the next 3 questions:

What amino acid sequence will be generated from the following mRNA sequence?   5’ AUGGAUCGUUUAUCCUUG 3’

• Met-Arg-Glu-Arg-Glu-Phe

•   Met-Asp-Leu-Ser-Leu-Arg

• Met-Leu-Phe-Arg-Glu-Glu

• Met-Asp-Arg-Leu-Ser-Leu

Multiple Choice

Frameshift mutations result from:

• the insertion or deletion of one or more base pairs.

• the substitution of a stop codon for an amino acid-specifying codon.

• the substitution of one base pair for another.

•   the substitution of more than one base pair.

Multiple Choice

One of the mRNA codons specifying the amino acid leucine is 5´-CUA-3´.  Its corresponding anticodon is:

• 3´-GAU-5´.

•     5´-GAU-3´.

•    5´-GAT-3´.

• 3´-GAT-5´.

Multiple Choice

Achondroplasia is caused by a change of mRNA codon from GGA to AGA. Achondroplasia is thus an example of a

• Silent mutation

• Missense Mutation

• Nonsense M

• Frameshift Mutation

Multiple Choice

When translating secretory or membrane proteins, ribosomes are directed to the ER membrane by

•    a signal sequence of RNA that precedes the start codon of the message.

•    a chemical signal given off by the ER.

• a signal-recognition particle that brings ribosomes to a receptor protein in the ER membrane.

• a specific characteristic that identifies it as a “bound” ribosome itself.

Multiple Choice

Enhancers  (Distal Control Elements)

• increase the rate of eukaryotic gene transcription

• may be located upstream or downstream from the genes they regulate

• may be located thousands of base pairs away from the promoter

• all of the above

Multiple Choice

Ubiquitination of proteins:

• Targets them for export from the cell

• Targets them for degradation in the proteosome

• Uses an enzyme called Dicer to chop the protein up into little pieces

• Is the most common way to regulate the expression of genes in eukaryotes

Multiple Choice

In human females, one X chromosome per cell is randomly inactivated during early embryonic development, forming a Barr Body.  Much of the DNA in this chromosome is in the form of

• Introns

• Heterochromatin

• Euchr

• Tetrads

Multiple Choice

MicroRNAs (miRNAs) and small interfering RNAs (siRNAs) both function to "silence" genes, by

• Binding to complementary RNA sequences to prevent translation

• Causing chromatin de-acetylation and tight winding of the DNA

• Binding to complementary DNA sequences to prevent transcription

• Activating the Dicer enzyme to rapidly degrade the mRNA

Multiple Choice

Alternative RNA splicing

• increases the rate of transcription.

• can allow the production of proteins of different sizes from a single mRNA.

• can allow the production of similar proteins from different RNAs.

• is a mechanism for increasing the rate of transcription.

Multiple Choice

In eukaryotes, histone acetylation __________.

•   inactivates genes

•   reduces the size of the genome

•    imprints genes from one parent

• promotes transcription

Multiple Choice

In the bacteriophage shown below, the functions of the structures labeled 2 and 3 are

• 2= protection of the genetic material, 3= locomotion.

•   2= protection of the genetic material, 3= replication.

•   2= protection of the genetic material, 3= attachment to a host cell

• 2= to take over host cell mechanisms, 3= attachment to a host cell.

Multiple Choice

All of the following are DNA viruses except:

• The Herpes virus that causes cold sores

•   The Influenza virus that causes the ‘flu.

• The Human Papilloma virus that causes cervical cancer

• The Parvoviruses that cause the common cold

Multiple Choice

Retroviruses like the HIV virus differ from other viruses by:

• the sugar coating on their capsids.

• the shape of their capsid.

• Using the enzyme reverse transcriptase to copy their RNA into DNA.

• lytically destroying their hosts.

Multiple Select

Prion diseases or Taupoathies of humans include (select all that apply)

• Kuru.

• Alzheimer’s disease

• COVID

• CTE (Chronic traumatic encephalitis)

• Scrapie.

Multiple Choice

Why does a 500-bp piece of DNA travel faster than a 1500-bp piece of DNA during agarose gel electrophoresis?

• A 500 bp piece is less likely to be tangled up in the agarose strands than the 1,500 bp piece

• A 500 bp piece has fewer (-) charges than a larger piece of DNA

• A larger, 1500 bp piece is strongly attracted to the (-) pole and prefers to stay near the wells

• A 500 bp piece has fewer (+) charges than a larger piece of DNA

Multiple Choice

The migration of DNA within an agarose gel during electrophoresis is primarily based on:

• The negative charge of the DNA

• The volume of the DNA sample loaded

• The size of the DNA fragments

• The number of DNA fragments

Multiple Choice

In the pGLO lab, how could you confirm that a transformed cell was expressing the bla gene?

• It would fluoresce green

•   It would grow on Amp plates

•   If would not grow on ‘LB only’ plates

• If would only grow on ‘LB only’ plates

Multiple Choice

E. coli can absorb the sugar Arabinose from its surroundings. When Arabinose is present with pGLO:

•  a. It binds to a repressor protein at the promoter of the Arabinose Operon

• b. It changes the AraC repressor protein so RNA polymerase can transcribe the Ara operon.

• c. It binds to DNA and functions as an enhancer to turn on the Green Fluorescent Protein

• Both  a and b

Multiple Choice

After transformation of HB101 by pGLO, why are the cells grown in LB broth for 20 minutes before plating on Amp plates?

• There is a phenotypic lag of 20 minutes before the b- gene for Amp resistance can be expressed

• The Green Fluorescent Protein takes 20 minutes to be transcribed and translated

•   The HB101 bacteria are weak after the heat shock treatment, and need 20 minutes to recover

• The pGLO plasmid does not fully enter the HB101 cells until 20 minutes after the heat shock

Questions??​

​Wrap Up

No HW due next week!!​

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​On the white board near you, write a topic that you might need more review on!!