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).

DNA
DNA is made up of three things --> Phosphate group, a five carbon sugar (deoxyribose) and nitrogenous bases. Nitrogenous bases are Adenine and Guanine -- which are purines, and Thymine and Cytosine -- which are pyrimidines. The sugar -- deoxyribose and the 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.

Nitrogenous bases 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.

DNA Replication When a cell divides, it makes a duplicate copy of its DNA so the produced cell will have a set of DNA molecules. This is DNA replication -- it is also called DNA synthesis. Enzymes (DNA polymerase) "unzip" the two strands of the double helix structure and insert appropriate nitrogenous bases.

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.

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.

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.

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


 * 1) $$p^2 + 2pq + q^2 = 1$$
 * 2) $$p + q = 1$$

where

$$p$$ is the frequency of the (homozygous) dominant allele in the population $$q$$ is the frequency of the (homozygous) recessive allele in the population $$p^2$$ is the percentage of the homozygous dominant individuals $$2pq$$ is the percentage of the heterozygous individuals and $$q^2$$ 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 $$q$$. 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 $$p$$. Using the second equation, $$p$$ can be found once $$q$$ has been determined. Step 3: Determine $$p^2$$ and $$q^2$$. Steps 3 and 4 are interchangeable, but finding $$p^2$$ and $$q^2$$ first is generally the common practice. Step 4: Determine $$2pq$$.

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

Step 1: Determine $$q$$. Since aa, or $$q^2$$ 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 $$p$$. Using the second equation, $$ p + q = 1$$. Therefore, $$p$$, or A must be 0.4. (40%) Step 3: Determine $$p^2$$ and $$q^2$$. Now that $$p$$ and $$q$$ (A and a respectively) are both known, $$p^2$$ and $$q^2$$ can be found by squaring each term. In this case, $$p^2 = .16$$ and $$q^2 = .36$$ (16% and 36% respectively). Step 4: Determine $$2pq$$. This can be done two ways. Rearranging the first equation, $$2pq = 1 - p^2 - q^2$$, so $$2pq = .48$$ (48%). Additionally, $$2pq$$ can be found by multiplying $$p$$ and $$q$$ together, then multiplying that by $$2$$.

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).

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

 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. A