This webpage is created for EDUC 5324 - Integrating Technology Into Education Course.

Designer Genes

esigner 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 - heterocyclic rings of carbon and nitrogen (purines have two such rings; pyramidines, one). The bases found in DNA are adenine and guanine (purines), thymine and cytosine (pyrimidines). The sugar 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.

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.

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.

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).

Punnett squares

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

Karyotype of a male.

2 Punnet squares.

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.

Special Punnet Squares

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.

Incomplete dominance in the F1 generation. When the F1's are selfed, all 3 phenotypes will be expressed.

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".

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

Southern

Southern Blotting- getting DNA from a film

Northern

Northern Blotting- getting RNA from a film

Western

Western Blotting- Getting proteins from a film

Hardy-Weinberg Equilibrium

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:

Mutations
Gene flow (migration in/out of the population)
Small population
Natural selection
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:

Mutations: Mutations introduce new alleles into the population.
Gene flow: Like mutations, migration can introduce new alleles (or diminish another allele)
Small population: Genetic drift is likely to occur in a smaller population.
Natural selection: If some traits are discriminated for/against, the genotype frequencies will not be in equilibrium over the generations.
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:

where

is the frequency of the (homozygous) dominant allele in the population

is the frequency of the (homozygous) recessive allele in the population

is the percentage of the homozygous dominant individuals is the percentage of the heterozygous individualsand

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:

In a certain population, the percentage of the homozygous recessive genotype (aa) is 36%. Using only that information, find:

The frequency of the recessive genotype.
The frequency of the recessive allele.
The frequency of the dominant allele.

The percent of the heterozygous individuals.

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

. 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

. Using the second equation, can be found once has been determined.

Step 3: Determine and . Steps 3 and 4 are interchangeable, but finding and first is generally the common practice.Step 4: Determine .

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

Step 1: Determine

. Since aa, or 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

. Using the second equation, . Therefore, , or A must be 0.4. (40%)Step 3: Determine and . Now that and (A and a respectively) are both known, and can be found by squaring each term. In this case, and (16% and 36% respectively).

Step 4: Determine

. This can be done two ways. Rearranging the first equation, , so (48%). Additionally, can be found by multiplying and together, then multiplying that by

So, the answers to the sample questions are:

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

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

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 molecules that are made 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

Resources

gangsta_duck's Designer Genes Notes

GuyFromNowhere's Designer Genes Notes

Molecular Biology of the Cell notes

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