Designer Genes

Designer Genes (Division C) and Heredity (Division B) are based on genetics and molecular biology (introns/exons, mitosis/meiosis, leading/lagging strand, etc).

You are allowed to bring a note sheet and 2 non-graphing calculators.

DNA
DNA is made up of three things: a phosphate group, a deoxyribose sugar, and heterocyclic rings of carbon and nitrogen (purines have two such rings; pyrimidines, one). The bases found in DNA are adenine and guanine (purines), thymine and cytosine (pyrimidines). The sugar and phosphate groups form the backbone or the sides of the double helix "ladder" and the nitrogenous bases stick out from the chain like "rungs" of the ladder.

Watson, Crick, and Maurice Wilkins are credited with finding the structure of DNA. Rosalind Franklin, however, was the first to use X-ray diffraction to see the DNA helix. Watson and Crick used her work to discover the helical structure in 1953. See The Double Helix by James D. Watson for more information.

Base Pairing
Adenine only bonds with thymine and guanine bonds only with cytosine. This is called base pairing. According to Chargaff's rules, an organism should have equal percentages of adenine and thymine and cytosine and guanine. One way to remember which base pairs with which is to remember the "curvy" letters go together. If Chargaff's rules do not hold, the organism's DNA may be single-stranded rather than double-stranded.

In RNA, uracil replaces thymine and thus, uracil binds with adenine.

DNA Replication
When a cell divides, it makes a duplicate of its DNA in the S-phase of the cell cycle so the daughter cells will have a complete set of chromosomes. This process is DNA replication, also called DNA synthesis. First, topisomerase unwinds the DNA strands, after which the enzyme helicase separates the strands by breaking the hydrogen bonds between nitrogenous bases. This area of separation is called a replication fork.

Before DNA Polymerase can enter the replication fork to make copies of the DNA strands, RNA Primase puts down a "primer" to attract RNA nucleotides, which form hydrogen bonds with DNA bases. The next step is elongation, which creates some difficulties because of how enzymes "read." DNA Polymerase can only copy from 5' to 3'; however, one DNA strand is 3' to 5'. This strand is called the lagging strand, as opposed to the 5'-3' leading strand. While DNA Polymerase can copy the leading strand without a problem, it can only replicate the lagging strand in spurts. The spaces between replicated portions of the lagging strand are called Okazaki fragments.

Once replication is complete, an exonuclease (an enzyme that cleaves nitrogenous bonds) removes the RNA primer. Finally, ligase connects the strands with their complements by catalyzing the phosphodiester bonds with the 3' hydroxyl group and the 5' phosphate.

Karyotypes
A karyotype is a chart that shows each chromosome. Each karyotype displays 23 pairs of chromosomes, including the X/Y chromosomes. Every pair is assigned a number (except for the sex chromosomes; they are always referred to as the X and Y chromosomes). Some genetic disorders can be detected by analyzing the number of chromosomes and/or the sex chromosomes. The gender of the individual can also be deduced from looking at the sex chromosomes. If there is an X and a Y, the individual is a male. A female has two X chromosomes and no Y chromosome. A karyotype is created by stopping cells in cell division and staining the chromosomes, then observing them under a light microscope.

Karyotypes can be used to diagnose genetic diseases. For example, a karyotype can reveal a third chromosome 21, which results in trisomy 21, commonly known as Down Syndrome. It can also reveal Turner syndrome (45, X), a disorder that results in females with one X chromosome, and Klinefelter's syndrome (47, XXY), in which a man has two X chromosomes and one Y chromosome.

Sex determination
In humans, the male and female share 22 of the 23 pairs of chromosomes in each body cell. The 23rd pair is known as the sex chromosomes because it determines the sex of the individual. In the male, the sex chromosome consists of an X and a Y chromosome(XY) while the pair in females consists of two X chromosomes(XX). The male is the one who determines the sex of the child and the female gives an X to all eggs while the male randomly produces about 50% X sperm and 50% Y sperm.

In rare cases, through nondisjunction, a person will have three sex chromosomes. If they have three X (XXX) chromosomes, they are female. If they have even one Y chromosome (XXY), they are male. Although they will show more feminine qualities, any person who has a Y chromosome is considered a male.

RNA
RNA (ribonucleic acid) is a single stranded nucleotide chain, not a double helix. While it does not share the same structure as DNA, it has many similar properties. RNA consists of a ribose sugar and hydroxyl group, as opposed to deoxyribose. RNA also consists of Adenine, Guanine and Cytosine, but Thymine is replaced with Uracil.

Types
There are three major types of RNA. While there are many other minor types, these three are heavily involved in translation.
 * Messenger RNA (mRNA): Encodes the sequence of amino acids that becomes a protein.
 * Transfer RNA (tRNA): Transports amino acids to ribosomes during translation. It contains about 80 RNA nucleotides, with an amino acid attached to the 3' end and an a complimentary anticodon attached to the 5' end.
 * Ribosomal RNA (rRNA): Along with ribosomal proteins, rRNA makes up the ribosome which is the organelle that translates mRNA into proteins.

Transcription
Transcription is the process of transcribing DNA into mRNA so that it can be translated into proteins. It is also the first major step of gene expression. Transcription produces a complimentary sequence to the DNA; A bonds with T in DNA, U bonds with A, G bonds with C and C bonds with G. For example:

DNA: GCACGTGTAGCATAGTACTAG mRNA: CGUGCACAUCGUAUCAUGAUC

Transcription occurs in the nucleus during the G1 and G2 phases of the cell cycle. In eukaryotes, it occurs in three distinct stages.

Initiation

 * 1) Activator proteins bind to distal control elements that are located before the DNA sequence known as a promoter. Promoters are located near the start sites of genes, and allow various proteins and enzymes (such as RNA polymerase II) to form an initiation complex that begins transcription.
 * 2) Proteins called transcription factors bind to a specific DNA sequence known as a promoter. At this point in the process, the DNA is still double stranded. RNA polymerase binds to the promoter region shortly after the transcription factors.
 * 3) RNA polymerase unwinds approximately 14 base pairs to form an "open complex" that becomes the transcription bubble. As the RNA polymerase begins creating RNA, it enters the RNA exit channel and leaves behind the initial transcription factors.

Elongation

 * 1) RNA polymerase begins unwinding the double helix and exposes 10-20 nucleotides for transcription at a time. To do this, RNA polymerase uses free-floating RNA nucleotides in the nucleoplasm.
 * 2) RNA polymerase travels from the 3' → 5' direction on the template strand of DNA, producing a mRNA strand in the 5' → 3' direction. This process produces an RNA copy of the 5' → 3' strand of DNA.
 * 3) RNA transcription occurs very quickly, and can involve multiple RNA polymerase working on a single gene. The typical rate of elongation is 10-100 nucleotides/sec.
 * 4) Elongation also involves a proofreading mechanism that can replace incorrect nucleotides. Transcription pauses, allowing RNA editing factors to bind to the new strand of mRNA and edit base order.

Termination

 * 1) The RNA codes for the polyadenylantion (AAUAAA), and the proteins that have been associated with the RNA polymerase stop moving.
 * 2) RNA polymerase continues moving, adding hundreds of adenine nucleotides to the end of the mRNA strand. Spare RNA created like this may be used by enzymes.
 * 3) This termination factor releases the newly created mRNA, which leaves the nucleus and travels to the ribosome where it is translated into a protein.

Interpreting Genetic Code
A sequence of three mRNA nucleotides is called a codon. Each of these codons corresponds with a complimentary anticodon attached to a strand of tRNA. Different tRNA molecules are attached to different amino acids, meaning that each codon corresponds with one amino acid. Since there are only four different RNA nucleotides, there are 64 possible codons. However, there are only 20 standard amino acids. This means that multiple codons can code for the same amino acid.

A chain of amino acids is called a protein. They are responsible for most biological functions in the body such as DNA replication, transcription, transporting molecules, and regulation of gene expression. Proteins are very complex macromolecules and have four different levels of structure. The amino acid sequence created in translation is known as the primary structure.

It is possible to interpret a DNA sequence into an amino acid sequence by using a chart like the one shown below.
 * Find the RNA nucleotides that would pair with the DNA nucleotides.
 * DNA: TAC AGG TAG CTA GTT ATT
 * RNA: AUG UCC AUC GAU CAA UAA
 * Follow the sequence of nucleotides on the chart from the inside out. For example, the RNA sequence AUG is found at the beginning of every protein and codes for Methionine. In the center of the circle, start with the A quadrant, then follow the U quadrant and then the G quadrant.
 * RNA: AUG UCC AUC GAU CAA UAA
 * Amino Acids: Methionine Serine Isoleucine Aspartic Acid Glutamine Stop

Translation
Translation is the process of translating the mRNA created during transcription into a protein. These proteins are responsible for different genetic traits such as hair/eye color, blood type, or hereditary conditions such as color blindness. It takes place in the ribosome, an organelle with three chambers and two subunits that consists of rRNA and other proteins. The three chambers are the A site (Aminoacyl-tRNA binding site), the P site (Peptidyl-tRNA binding site) and the E site (Exit site). All of these chambers are located in the large subunit. Like transcription, it occurs in three steps.

Initiation

 * 1) The small subunit attaches to the mRNA, holding it in place throughout translation.
 * 2) The Methionine tRNA bonds to the start codon AUG.
 * 3) The large subunit arrives and completes the translation initiation complex.

Elongation

 * 1) Amino acids are brought to the ribosome by tRNA molecules and are added to the polypeptide chain one by one.
 * 2) The anticodon on a tRNA molecule binds to the mRNA codon at the A site.
 * 3) An rRNA molecule in the large subunit catalyzes the formation of a peptide bond between the amino acid on the tRNA and the polypeptide chain.
 * 4) The ribosome moves the mRNA to from the P site to the E site, where the tRNA is released.

Termination

 * 1) The stop codon on the mRNA reaches the A site.
 * 2) Release factors bind to the stop codon at the A site.
 * 3) A water molecule is added to the end of the polypeptide instead of an amino acid, and hydrolysis releases the chain so it can be folded into its final structure.

Epigenetics
Epigenetics is the study of changes in organisms not caused by the alteration of genetic code. Epigenetics revolves around gene expression, not the DNA itself. It affects how genes are read by cells, and how they produce proteins. Think of the human genome as a filing cabinet, and the genes as folders that contain the instructions to make a protein. Certain folders might be marked as important, or others could be marked as less important. These epigenetic marks control the expression of genes. It's the reason that even though every cell in your body has the same DNA at its core, different cells have different functions. A liver cell would open different folders in the filing cabinet than a brain cell would, because it would need to make different proteins.

Epigenetic marks take the form of molecular tags that are placed in different places on the histone, and each one has a different effect. They can make DNA more accessible to proteins, or purposefully make it less accessible so that a specific gene isn't transcribed or translated. Some epigenetic marks are very long and cover large stretches of DNA, or others are gathered at the start of genes. Epigenetic marks can also change over time. These changes can be caused by anything from chemical additives in plastics to DNA errors during replication.

Some epigenetic marks can also be inherited through generations. This is how environmental factors are passed down through generations. Addictive behavior is inherited in this way, and the effects nutrient deprivation can be passed down in this way too. However, passing down epigenetic tags is different than passing down genes. Reproductive cells undergo a process called reprogramming, and this process is supposed to erase all epigenetic tags. However, on some genes it fails and leaves these tags in place to be passed down to another generation. In mammals, about 1% of genes escape epigenetic reprogramming.

Transcriptional
Gene expression can be controlled during or after transcription. The rate of gene transcription is often controlled by allowing or denying RNA polymerase access to the gene. Termination can also occur early, preventing the gene from being transcribed properly. Transcriptional regulation can also occur when RNA polymerase is attempting to escape the promoter complex to start transcribing DNA. Protein factors can also alter the rate of transcription.

Lac and Trp Operons
Lac and Trp Operons are examples in prokaryotic gene regulation. Most prokaryotic genes such as in E.coli are always turned "on", but others are active only when products are needed by the cell, so their expression must be regulated.

An operon is a group of genes transcribed together by a single promoter. The lac operon was the first to be discovered. In the model bacterium E. coli, this operon is transcribed in the presence of lactose to give the bacterium the ability to digest this source of energy. It has three parts: lacA, lacY, and lacZ, as well as a promoter, a regulator, a terminator, and an operator. To activate lactose digestion abilities, an isomer of lactose (allolactose) binds to the gene's repressor, allowing the operon to be transcribed

Whereas the lac operon gives E. coli the ability to digest lactose, the trp operon shuts off the bacterium's capability to metabolize tryptophan. As such, it is an example of a repressible operon. In the presence of lactose, its five structural genes (trpA, trpB, trpC, trpD, and trpE), which code for tryptophan synthase, will be repressed so E. coli can metabolize lactose instead. Lac operons are inductible operons due to the fact that genes are expressed in the presence of a substance (lactose).

Single-Factor Crosses (Monohybrid)


The images to the right are examples of Punnett squares, named after the geneticist Reginald C. Punnett. Punnett squares show the cross between alleles and the genotype of the resulting offspring. Since both of the Punnet squares in the diagram only cross one trait (one pair of alleles), it is called a monohybrid or single-factor cross. Likewise, when two traits (two pairs of alleles) are crossed, it is called a dihybrid or two-factor cross.

The first Punnett square shows a cross between two heterozygous plants. The second Punnett square shows a cross between a homozygous tall plant and a homozygous short plant. The letters inside the boxes represent the genotype of each offspring. For example, in the first square, the genotypes of the offspring will be TT, Tt, and tt (2 of the 4 offspring will have the same genotype-Tt).

It is helpful to memorize the genotypic and phenotypic ratios of a heterozygous monohybrid cross. If two heterozygotes are crossed (like the first Punnet Square in the image to the right) then the genotypic ratio will always be:

1 D/D: 2 H: 1 R/R

and the phenotypic ratio will be:

3 D: 1 R

where D=homozygous dominant, R=homozygous recessive, and H=heterozygous.

Memorizing other simple crosses (such as a single-factor homozygous dominant x homozygous recessive cross) is useful and saves time on tests. Here are some simple monohybrid crosses with their respective genotypic and phenotypic ratios.

AA x aa (Homozygous dominant x Homozygous recessive)

Genotypic ratio: 0 D/D: 4 H: 0 R/R

Phenotypic ratio: 4 D: 0 R

AA x Aa (Homozygous dominant x Heterozygous)

Genotypic ratio: 2 D/D: 2 H: 0 R/R

Phenotypic ratio: 4 D: 0 R

Aa x aa (Heterozygous x homozygous recessive)

Genotypic ratio: 0 D/D: 2 H: 2 R/R

Phenotypic ratio: 2 D: 2 R

Some important Punnett Square terms are defined below. On tests, be extra careful when you spot these terms as they are easily confused with each other.


 * Genotype: The different combinations of the alleles.
 * Phenotype: The physical appearance of the offspring.


 * Genotypic ratio: The ratio of the combination of alleles.
 * Phenotypic ratio: The ratio of the physical appearance.

Two-Factor Crosses (Dihybrid)
Two factor crosses, or dihybrid crosses, are similar to single-factor crosses except that in a two-factor cross, two traits are crossed rather than one trait in a single-factor cross. An example of a two-factor cross is pictured to the left.



Here, two heterozygotes are crossed (RrYy x RrYy). The "R" allele represents the shape of the seed and the "Y" allele represents the color. It is important to note the genotypic and phenotypic ratios for a heterozygous dihybrid cross. Regardless of the alleles, if two dihybrid heterzygotes are crossed, then the resulting phenotypic ratio will be:

9 D/D: 3 D/R: 3 R/D: 1 R/R (D = dominant, R = recessive).

and the genotypic ratio will be:

1 D/D: 2 D/H: 1 D/R: 4 H/H: 4 H/D: 1 R/D: 2 R/H: 1 R/R (H = heterozygous).

So, the phenotypic ratio for the pictured dihybrid cross is:

9 round/yellow:3 round/green: 3 wrinkled/yellow: 1 wrinkled/green.

Three-Factor Crosses (Trihybrid)


Like single- and double-factor crosses, three-factor crosses (trihybrid) show three different traits that are crossed (see the image to the right for an example). Trihybrid crosses are rarely seen on tests, so don't spend too much time practicing them until the later stages of competition.

Incomplete dominance
In some unusual cases such as 4 o'clock flowers, gene pairs for a given trait fail to establish dominance and the heterozygous condition is expressed as an intermediate between the two alleles. Often, to draw attention to this situation, the letter 'I' is used to designate the gene allele.

Example: In 4 o'clock flowers, the genotype RR (homozygous dominant) appears red, rr (homozygous recessive)appears white, and Rr (heterozygous)appears pink. In all cases of incomplete dominance, the number of genotypes equals the number of phenotypes.

Epistasis
Epistatis is where one set of genes stops or inhibits the action of another genes. Epistasis genes can either be recessive or dominant. The gene for no pigment (p) in the skin(albinism) is recessive to normal pigmentation(P). For any pigment to appear at all, at least one gene for enzyme S must also be present. That's like even if there is a pigment, but enzyme S is not present, the person is albino. PpSs? is normal, PPss? is albino, ppSS is albino, and so on. To not be albino, there needs to be at least one P and one S.

Sex-linked traits
Sex-linked traits are features that are associated with the genes on the sex chromosomes, usually X. Examples of those are recessive genes for color-blindness and hemophilia.

Sex-influenced traits
Sex influenced traits are traits that show up more in one sex than they do in the other as a definite phenotype. Usually influenced more by hormones in the male or female.

Multiple genes
Most phenotypic features are controlled by more than one set of non-allelic genes acting on them, such as height, skin color, intelligence, and hair and eye color. Usually this type of problem is seen as a typical two or three, etc factor cross with the more dominants, the more expression of the trait in question.

Multiple alleles
There may be more than the usual two alleles for any given gene. Especially, this appears in fur or pelt conditions of domestic animals. The problem usually uses 'I' (for incomplete dominance) and some prearranged superscript. The most common example found on tests is the ABO blood type system found in humans.

Lethal alleles
While lethal alleles do not affect the way you set up your Punnett square, they can appear to alter Mendelian ratios. A lethal genotype is one that causes death before the individual can reproduce and pass their genes on to the next generation. As such, they remove an expected progeny class after a specific cross. For example, in Mexican hairless dogs, the genotype hh means that te dog is hairy, Hh means that the dog is hairless, but HH means that they die as embryos--thus the term "lethal".

Conditions
The Hardy-Weinberg Law states that a population will maintain the exact allele and genotype frequencies over each generation unless five specific influences are introduced into the population. These are:


 * 1) Mutations
 * 2) Gene flow (migration in/out of the population)
 * 3) Small population
 * 4) Natural selection
 * 5) Non-random mating

For a population to be in Hardy-Weinberg equilibrium, it must not have any of the 5 conditions listed above. Here are the explanations for each condition:


 * 1) Mutations: Mutations introduce new alleles into the population.
 * 2) Gene flow: Like mutations, migration can introduce new alleles (or diminish another allele)
 * 3) Small population: Genetic drift is likely to occur in a smaller population.
 * 4) Natural selection: If some traits are discriminated for/against, the genotype frequencies will not be in equilibrium over the generations.
 * 5) Non-random mating: Like natural selection, non-random mating could discriminate for/against traits.

An example of Hardy Weinberg: Let's say we were in a world where everyone had either purple or blue skin. S is purple skin, and s is blue skin. The probability of either one of these genes occurring is constant, and both probabilities have to add to 1. Given the probabilities of both blue and purple skin, lets say p for purple and b for blue, the probability of having two purple skin alleles (SS) would be pp, and having a blue and a purple (Ss) would be pb, and so on.

Equation
There are two equations used in the Hardy-Weinberg Law:


 * 1) [math]p^2 + 2pq + q^2 = 1[/math]
 * 2) [math]p + q = 1[/math]

where

[math]p[/math] is the frequency of the (homozygous) dominant allele in the population [math]q[/math] is the frequency of the (homozygous) recessive allele in the population [math]p^2[/math] is the percentage of the homozygous dominant individuals [math]2pq[/math] is the percentage of the heterozygous individuals and [math]q^2[/math] is the percentage of the homozygous recessive individuals.

Remember, the equations only apply if the population is in Hardy-Weinberg equilibrium.

Solving a Hardy-Weinberg Problem
A typical Hardy-Weinberg problem will resemble the sample problem below:

IMPORTANT: Before attempting to solve the problem, it is critical to analyze all of the given information and approach it in the correct manner. Make sure to check your work after finishing! One mistake will throw off the entire problem. When solving a problem, make sure to work in the order as follows:

Step 1: Determine [math]q[/math]. Since a dominant phenotype can have either a homozygous or heterozygous genotype, it is easier to find the recessive allele first (unless an exact homozygous/heterozygous dominant value is given). Step 2: Determine [math]p[/math]. Using the second equation, [math]p[/math] can be found once [math]q[/math] has been determined. Step 3: Determine [math]p^2[/math] and [math]q^2[/math]. Steps 3 and 4 are interchangeable, but finding [math]p^2[/math] and [math]q^2[/math] first is generally the common practice. Step 4: Determine [math]2pq[/math].

The answers and work (using the four steps) for the sample problem are shown below:

Step 1: Determine [math]q[/math]. Since aa, or [math]q^2[/math] is 36%, then a (the frequency of the recessive allele-this is q in Hardy-Weinberg terms) must be 60%, or 0.6. Step 2: Determine [math]p[/math]. Using the second equation, [math] p + q = 1[/math]. Therefore, [math]p[/math], or A must be 0.4. (40%) Step 3: Determine [math]p^2[/math] and [math]q^2[/math]. Now that [math]p[/math] and [math]q[/math] (A and a respectively) are both known, [math]p^2[/math] and [math]q^2[/math] can be found by squaring each term. In this case, [math]p^2 = .16[/math] and [math]q^2 = .36[/math] (16% and 36% respectively). Step 4: Determine [math]2pq[/math]. This can be done two ways. Rearranging the first equation, [math]2pq = 1 - p^2 - q^2[/math], so [math]2pq = .48[/math] (48%). Additionally, [math]2pq[/math] can be found by multiplying [math]p[/math] and [math]q[/math] together, then multiplying that by [math]2[/math].

So, the answers to the sample questions are:


 * 1) .36 (this was given to us in the problem)
 * .6
 * .4
 * 1) 48%

Note: Frequency is always expressed as a decimal (and percentages are expressed as percents).

RFLP Analysis
DNA sample is broken into pieces (and digested) by restriction enzymes and the resulting restriction fragments are separated according to their lengths by gel electrophoresis. Though now largely obsolete due to the rise of inexpensive DNA sequencing technologies, RFLP analysis was previously used for DNA profiling.

Polymerase Chain Reaction
Polymerase Chain Reaction, abbreviated as PCR, is a method of quickly making billions of copies of a desired section of DNA. For a virtual lab that clearly explains the process, visit http://learn.genetics.utah.edu/content/labs/pcr/.

The Polymerase Chain Reaction is another way of creating large numbers of a specific piece of DNA, other than cloning DNA.

In PCR, DNA primers are employed on opposite ends of the DNA sequence. They are necessary of the initiation of DNA replication. Then, a single strand of DNA is used as the template to produce double stranded DNA through polymerization.

Individual strands of DNA are unwinded from double stranded DNA using heat. Thus, PCR consists of heat treatment to unwind the DNA, then the binding of primers to the DNA, then polymerization to form another strand. This repeats, and will quickly and exponentially multiply the amount of DNA available.

The key step in the development of PCR was the isolation and use of a heat resistant DNA polymerase (TacDNA Polymerase).

PCR is better than the conventional cloning of DNA due to the fact that PCR can be used with only very small and impure samples of DNA.

Gel Electrophoresis
Gel Electrophoresis is one of the most useful techniques to study macromolecules, especially proteins or nucleic acids. In gel electrophoresis, charged molecules are pulled through a gel (usually purified agar known as agarose) and this separates the molecules. Larger molecules move more slowly through the gel since they get caught in the gel matrix, and molecules with greater charges move faster since the electric field is what's pulling the molecules.

Molecules will have a negative charge, and thus will move towards the positive poles.

Blotting
Blotting is a method used for isolating some certain molecule from a sample. In the case of DNA, it is first cut by restriction enzymes and sorted by size with gel electrophoresis. A blotting membrane is placed over the gel, and a paper towel is used to absorb buffer through the membrane. The buffer moves through the membrane and flows upward, leaving the DNA behind on the other side of the membrane.

Southern
Southern Blotting - getting DNA from a film

Northern
Northern Blotting - getting RNA from a film

Western
Western Blotting - getting proteins from a film

Eastern
Eastern Blotting - detecting post-translational modifications in proteins

Practice Test
This practice test focuses on DNA. For full-length tests see the Designer Genes Test Exchange.

1. Which of the following nucleotide pair bonds would be found in a DNA molecule? a. adenine-guanine b. guanine-cytosine c. adenine-cytosine d. cytosine-uracil

2. The backbone of a DNA molecule is made of which two components? a. phosphate molecules and ribose sugars b. deoxyphosphate molecules and ribose sugars c. phosphate molecules and deoxyribose sugars d. deoxyphosphate molecules and deoxyribose sugars

3. Ribosomes are made of: a. rRNA and protein b. tRNA and mRNA c. rRNA and mRNA d. protein and mRNA

4. Watson and Crick were the first to suggest that DNA is: a. a short molecule b. the shape of a double helix c. a protein molecule d. protein and tRNA

5. The chromosome abnormality that occurs when part of one chromosome breaks off and is added to a different chromosome is: a. deletion b. nondisjunction c. translocation d. inversion

6. Which of the following would be least likely to happen as a result of a mutation in a person's skin cells? a. skin cancer b. reduced functioning of the skin cell c. no change in the functioning of the skin cell d. the person's offspring have mutated skin

7. The process by which a DNA molecule is copied is called: a. binary fission b. mitosis c. replication d. translation

8. A DNA nucleotide may be made up of a phosphate group along with: a. a deoxyribose sugar and uracil b. ribose sugar and adenine c. deoxyribose sugar and thymine d. ribose sugar and cytosine

9. Which series is arranged in order from largest to smallest in size? a. chromosome, nucleus, cell, DNA, nucleotide b. cell, nucleus, chromosome, DNA, nucleotide c. nucleotide, chromosome, cell, DNA, nucleus d. cell, nucleotide, nucleus, DNA, chromosome

10. Messenger RNA is formed in the process of: a. transcription b. translation c. replication d. mutation

11. X rays, ultraviolet light, and radioactive substances that can change the chemical nature of DNA are classified as: a. growth regulators b. metamorphic molecules c. hydrolytic enzymes d. mutagens

12. After DNA replication, the two DNA strands that are produced are: a. are complimentary b. are identical c. must replicate again d. cannot replicate again

13. Bacteriophages are: a. tiny bacteria b. bacteria of the same type c. lipids and ribonucleic acids d. viruses

14. In translation, the order in which codons bond to mRNA is determined by: a. rRNA b. tRNA c. Base pairing rules d. Lagging strand

15. In RNA, the code word AUG that specifies methionine can also serve as a(n): a. anticodon b. stop codon c. initiator codon d. all are correct

16. The two strands of a DNA double helix are: a. identical b. purines c. pyrimidines d. complementary

17. Both DNA and RNA: a. contain ribose b. are single stranded c. contain nucleotides d. contain uracil

Answers to the Review Questions

1. B 2. C 3. A 4. B 5. C 6. D 7. C 8. C 9. B 10. A 11. D 12. A 13. D 14. C 15. D 16. D 17. C

Resources
[[Media:Gangsta DG Notes.pdf|gangsta_duck's Designer Genes Notes]]

[[Media:GFNowhere_designer_notes.pdf|GuyFromNowhere's Designer Genes Notes]] Molecular Biology of the Cell notes