Griffith’s Experiment
Griffith’s Experiment
| Mendelian Genetics |
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| 1862 | 1868 | 1880 |
Genetic Terminology:
Blending Concept of Inheritance:
Gregor Mendel:
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Why peas, Pisum sativum?

GARDEN PEA
Mendel’s Experiments:
a. Seed shape — Round (R) or Wrinkled (r)
b. Seed Color —- Yellow (Y) or Green (y)
c. Pod Shape — Smooth (S) or wrinkled (s)
d. Pod Color — Green (G) or Yellow (g)
e. Seed Coat Color — Gray (G) or White (g)
f. Flower position — Axial (A) or Terminal (a)
g. Plant Height — Tall (T) or Short (t)
h. Flower color — Purple (P) or white (p)


Trait – plant height |
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Alleles – T tall, t short |
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P1 cross TT x tt |
genotype — Tt | |||
| t | t | phenotype — Tall | ||
| T | Tt | Tt | genotypic ratio –all alike | |
| T | Tt | Tt | phenotypic ratio- all alike | |
Trait – plant height |
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Alleles – T tall, t short |
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F1 cross Tt x Tt |
genotype — TT, Tt, tt | |||
| T | t | phenotype — Tall & short | ||
| T | TT | Tt | genotypic ratio —1:2:1 | |
| t | Tt | tt | phenotypic ratio- 3:1 | |
| Trait – Plant Height | |||||||
| Alleles – T tall, t short | |||||||
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F2 cross TT x Tt |
F2 cross tt x Tt |
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| T | t | T | t | ||||
| T | TT | Tt | t | Tt | tt | ||
| T | TT | Tt | t | Tt | tt | ||
| genotype – TT, Tt | genotype – tt, Tt | ||||||
| phenotype – Tall | phenotype – Tall & short | ||||||
| genotypic ratio – 1:1 | genotypic ratio – 1:1 | ||||||
| phenotypic ratio – all alike | phenotypic ratio – 1:1 | ||||||
Problems: Work the P1, F1, and both F2 crosses for all of the other pea plant traits & be sure to include genotypes, phenotypes, genotypic & phenotypic ratios.
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Traits: Seed Shape & Seed Color |
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Alleles: R round Y yellow |
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P1 Cross:
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| ry | Genotype: | RrYy | ||
| RY | RrYy |
Phenotype: | Round yellow seed | |
| Genotypic ratio: | All alike | |||
| Phenotypic ratio: | All Alike | |||
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Traits: Seed Shape & Seed Color |
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Alleles: R round Y yellow |
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| F1 Cross: RrYy x RrYy | ||||
| RY | Ry | rY | ry | |
| RY | RRYY |
RRYy |
RrYY |
RrYy |
| Ry | RRYy |
RRyy |
RrYy |
Rryy |
| rY | RrYY |
RrYy |
r rYY |
r rYy |
| ry | RrYy |
Rryy |
r rYy |
r ryy |
| Genotypes | Genotypic Ratios | Phenotypes | Phenotypic Ratios |
| RRYY | 1 | Round yellow seed |
9 |
| RRYy | 2 | ||
| RrYY | 2 | ||
| RrYy | 4 | ||
| RRyy | 1 | Round green seed |
3 |
| Rryy | 2 | ||
| r rYY | 1 | Wrinkled yellow seed |
3 |
| r rYy | 2 | ||
| r ryy | 1 | Wrinkled green seed |
1 |
Problems: Choose two other pea plant traits and work the P1 and F1 dihybrid crosses. Be sure to show the trait, alleles, genotypes, phenotypes, and all ratios.
Results of Mendel’s Experiments:
| Trait: Pod Color | |
| Genotypes: | Phenotype: |
| GG | Green Pod |
| Gg | Green Pod |
| gg | Yellow Pod |
| Rr | |
| R | r |
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RrYy |
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| RY | Ry | rY | ry |
Other Patterns of Inheritance:

| Genotype | Phenotype |
| IOIO | Type O |
| IAIO | Type A |
| IAIA | Type A |
| IBIO | Type B |
| IBIB | Type B |
| IAIB | Type AB |
The Hardy-Weinberg formulas allow scientists to determine whether evolution has occurred. Any changes in the gene frequencies in the population over time can be detected. The law essentially states that if no evolution is occurring, then an equilibrium of allele frequencies will remain in effect in each succeeding generation of sexually reproducing individuals. In order for equilibrium to remain in effect (i.e. that no evolution is occurring) then the following five conditions must be met:
Obviously, the Hardy-Weinberg equilibrium cannot exist in real life. Some or all of these types of forces all act on living populations at various times and evolution at some level occurs in all living organisms. The Hardy-Weinberg formulas allow us to detect some allele frequencies that change from generation to generation, thus allowing a simplified method of determining that evolution is occurring. There are two formulas that must be memorized:
p = frequency of the dominant allele in the population
q = frequency of the recessive allele in the population
p2 = percentage of homozygous dominant individuals
q2 = percentage of homozygous recessive individuals
2pq = percentage of heterozygous individuals
Individuals that have aptitude for math find that working with the above formulas is ridiculously easy. However, for individuals who are unfamiliar with algebra, it takes some practice working problems before you get the hang of it. Below I have provided a series of practice problems that you may wish to try out. Note that I have rounded off some of the numbers in some problems to the second decimal place.
PROBLEM #1 You have sampled a population in which you know that the percentage of the homozygous recessive genotype (aa) is 36%. Using that 36%, calculate the following:
PROBLEM #2. Sickle-cell anemia is an interesting genetic disease. Normal homozygous individuals (SS) have normal blood cells that are easily infected with the malarial parasite. Thus, many of these individuals become very ill from the parasite and many die. Individuals homozygous for the sickle-cell trait (ss) have red blood cells that readily collapse when deoxygenated. Although malaria cannot grow in these red blood cells, individuals often die because of the genetic defect. However, individuals with the heterozygous condition (Ss) have some sickling of red blood cells, but generally not enough to cause mortality. In addition, malaria cannot survive well within these “partially defective” red blood cells. Thus, heterozygotes tend to survive better than either of the homozygous conditions. 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?
PROBLEM #3. There are 100 students in a class. Ninety-six did well in the course whereas four blew it totally and received a grade of F. Sorry. In the highly unlikely event that these traits are genetic rather than environmental, if these traits involve dominant and recessive alleles, and if the four (4%) represent the frequency of the homozygous recessive condition, please calculate the following:
PROBLEM #4. Within a population of butterflies, the color brown (B) is dominant over the color white (b). And, 40% of all butterflies are white. Given this simple information, which is something that is very likely to be on an exam, calculate the following:
PROBLEM #5. A rather large population of Biology instructors have 396 red-sided individuals and 557 tan-sided individuals. Assume that red is totally recessive. Please calculate the following:
PROBLEM #6. A very large population of randomly-mating laboratory mice contains 35% white mice. White coloring is caused by the double recessive genotype, “aa”. Calculate allelic and genotypic frequencies for this population.
PROBLEM #7. After graduation, you and 19 of your closest friends (lets say 10 males and 10 females) charter a plane to go on a round-the-world tour. Unfortunately, you all crash land (safely) on a deserted island. No one finds you and you start a new population totally isolated from the rest of the world. Two of your friends carry (i.e. are heterozygous for) the recessive cystic fibrosis allele (c). Assuming that the frequency of this allele does not change as the population grows, what will be the incidence of cystic fibrosis on your island?
PROBLEM #8. You sample 1,000 individuals from a large population for the MN blood group, which can easily be measured since co-dominance is involved (i.e., you can detect the heterozygotes). They are typed accordingly:
| BLOOD TYPE | GENOTYPE | NUMBER OF INDIVIDUALS | RESULTING FREQUENCY |
|---|---|---|---|
| M | MM | 490 | 0.49 |
| MN | MN | 420 | 0.42 |
| N | NN | 90 | 0.09 |
Using the data provide above, calculate the following:
PROBLEM #9. Cystic fibrosis is a recessive condition that affects about 1 in 2,500 babies in the Caucasian population of the United States. Please calculate the following:
PROBLEM #10. In a given population, only the “A” and “B” alleles are present in the ABO system; there are no individuals with type “O” blood or with O alleles in this particular population. If 200 people have type A blood, 75 have type AB blood, and 25 have type B blood, what are the allelic frequencies of this population (i.e., what are p and q)?
PROBLEM #11. The ability to taste PTC is due to a single dominate allele “T”. You sampled 215 individuals in biology, and determined that 150 could detect the bitter taste of PTC and 65 could not. Calculate all of the potential frequencies.
| Genetics of Drosophila melanogaster | ![]() |
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Introduction:
Gregor Mendel revolutionized the study of genetics. By studying genetic inheritance in pea plants, Gregor Mendel established two basic laws of that serve as the cornerstones of modern genetics: Mendel’s Law of Segregation and Law of Independent Assortment. Mendel’s Law of Segregation says that each trait has two alleles, and that each gamete contains one and only one of these alleles. These alleles are a source of genetic variability among offspring. Mendel’s Law of Independent Assortment says that the alleles for one trait separate independently of the alleles for another trait. This also helps ensure genetic variability among offspring.
Mendel’s laws have their limitations. For example, if two genes are on the same chromosome, the assortment of their alleles will not be independent. Also, for genes found on the X chromosome, expression of the trait can be linked to the sex of the offspring. Our knowledge of genetics and the tools we use in its study have advanced a great deal since Mendel’s time, but his basic concepts still stand true.
Drosophila melanogaster, the common fruit fly, has been used for genetic experiments since T.H. Morgan started his experiments in1907. Drosophila make good genetic specimens because they are small, produce many offspring, have easily discernable mutations, have only four pairs of chromosomes, and complete their entire life cycle in about 12 days. They also have very simple food requirements. Chromosomes 1 (the X chromosome), 2, and 3 are very large, and the Y chromosome – number 4 – is extremely small. These four chromosomes have thousands of genes, many of which can be found in most eukaryotes, including humans.
Drosophila embryos develop in the egg membrane. The egg hatches and produces a larva that feeds by burrowing through the medium. The larval period consists of three stages, or instars, the end of each stage marked by a molt. Near the end of the larval period, the third instar will crawl up the side of the vial, attach themselves to a dry surface, and form a pupae. After a while the adults emerge.
Differences in body features help distinguish between male and female flies. Females are slightly larger and have a light-colored, pointed abdomen. The abdomen of males will be dark and blunt. The male flies also have dark bristles, sex combs, on the upper portion of the forelegs.
Hypothesis:
After performing a dihybrid cross between males with normal wings and sepia eyes and females with vestigial wings and red eyes, we expect to see only hybrids with normal wings and red eyes in the first filial generation. Then we expect to observe a 9:3:3:1 ratio of phenotypes in the second filial generation.
Materials and Methods:
The materials used for this lab were: culture vial of dihybrid cross, isopropyl alcohol 10%, camel’s hair brush, thermo-anesthetizer, petri dish, 2 Drosophila vials and labels, Drosophila medium, fly morgue.
A vial of wild-type Drosophila was thermally immobilized and the flies were placed in a petri dish. Traits were observed. A vial of prepared Drosophila was immobilized and then observed under a dissecting microscope. Males and females were separated and mutations were observed and recorded. The parental generation was placed in the morgue. The vial was placed in an incubator to allow the F1 generation to mature.
The F1 generation was immobilized and examined under a dissecting microscope. The sex and mutations of each fly were recorded. Five mating pairs of the F1 generation were placed into a fresh culture vial, and the vial was placed in an incubator. The remaining F1 flies were placed in the morgue. The F1 flies were left in the vial for about a week to mate and lay eggs. Then the adults were removed and placed in the morgue. The vial was placed back in the incubator to allow the F2 generation to mature. The F2 generation was immobilized and examined under a dissecting microscope. The sex and mutations of each fly were recorded.
Results:
Table 1 Phenotypes of the Parental Generation
| Phenotypes | Number of Males | Number of Females |
| Normal wings/red eyes | 0 | 0 |
| Normal wings/sepia eyes | 3 | 0 |
| vestigial wings/red eyes | 0 | 4 |
| vestigial wings/sepia eyes | 0 | 0 |
Table 2 Phenotypes of the F1 Generation
| Phenotype | Number of Males | Number of Females |
| Normal wings/red eyes | 78 | 95 |
| Normal wings/sepia eyes | 0 | 0 |
| vestigial wings/red eyes | 0 | 0 |
| vestigial wings/sepia eyes | 0 | 0 |
Table 3 Phenotypes of the F2 Generation
| Phenotypes | Number of Males | Number of Females |
| Normal wings/red eyes | 4 | 7 |
| Normal wings/sepia eyes | 4 | 5 |
| vestigial wings/red eyes | 0 | 1 |
| vestigial wings/sepia eyes | 0 | 0 |
| normal red/mutated body shape | 2 | 0 |
| normal sepia/mutated body shape | 1 | 0 |
Questions
Discussion and Conclusion:
The results of our parental cross turned out just as expected, but our F2 generation was not normal. Some sort of mutation must have occurred that caused the strange body shape seen in several individuals of our F2 generation.
| Heart Dissection |
Introduction
Mammals have four-chambered hearts and double circulation. The heart of a bird or mammal has two atria and two completely separated ventricles. The double-loop circulation is similar to amphibians and reptiles, but the oxygen-rich blood is completely separated from oxygen-poor blood. The left side of the heart handles only oxygenated blood, and the right side receives and pumps only deoxygenated blood. With no mixing of the two kinds of blood, and with a double circulation that restores pressure after blood has passed through the lung capillaries, delivery of oxygen to all parts of the body for cellular respiration is enhanced. As endotherms, which use heat released from metabolism to warm the body, mammals require more oxygen per gram of body weight than other vertebrates of equal size. Birds and mammals descended from different reptilian ancestors, and their four-chambered hearts evolved independently – an example of convergent evolution.
Objective
Using a pig heart, students will observe the major chambers, valves, and vessels of the heart and be able to describe the circulation of blood through the heart to the lungs and back and out to the rest of the body. (The pig heart is used because it is very similar to the human heart in structure, size, & function.)
Materials
Dissecting pan, dissecting kit, safety glasses, lab apron, pig heart, & gloves
Procedure – External Structure

Front or Ventral Side of the Heart

Procedure – Internal Anatomy:

Tricuspid Valve
When you have finished dissecting the heart, dispose of the heart as your teacher advises and clean, dry, and return all dissecting equipment to the lab cart. Wash your hands thoroughly with soap.
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