Category: Curriculum Map

Genetics of Drosophila Melanogaster

 

 

Genetics of Drosophila melanogaster

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

  1. How are the alleles for genes on different chromosomes distributed to gametes? What genetic principle does this illustrate?
    The alleles on different chromosomes are distributed independently of one another, demonstrating Mendel’s Law of Independent Assortment.
  2. Why was it important to have virgin females for the first cross (yielding the F1 generation), but not the second cross (yielding the F2 generation)?
    It was important to have virgin females for the first cross to ensure that the offspring are the result of the desired cross. It was not necessary to isolate virgin females for the second cross because the only male flies to which they had been exposed were also members of the F1 generation.
  3. What did the chi-square test tell you about the validity of your experiment data? What is the importance of such a test?
    The chi-square test showed that the results of our first cross were valid, but that the results of our F1 cross were not normal. It is important to conduct such a test to determine how much your experimental data deviated from what was expected.

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.

Genetics Problems ppt Questions

Genetics Problems
ppt Questions

 

Independent Assortment

1. How many different kinds of gametes could the following individuals produce? Remember the formula 2n where n equals the number of heterozygotes.

     a. aaBb

     b. CCDdee

     c. AABbCcDD

     d. MmNnOoPpQq

     e. UUVVWWXXYYZz

 

P1, F1, and F2 Monohybrid Crosses

2. In dogs, wire-haired is due to a dominant gene (W), smooth-haired is due to its recessive allele (w). Show the results of crossing a homozygous wire-haired dog with a smooth-haired dog.

 

 

 

 

 

3. What kind of cross is this?

4. What was the genotype of all of the puppies? the phenotype?

 

5. The puppies belong to the _________ generation.

6. How would you write the F1 cross for this trait?

7. Show the results of working the F1 cross for this trait.

 

 

 

 

6. What phenotypic ratio did you get from this F1 cross?

7. What genotypic ratio did you get from this F1 cross?

8. Two wire-haired dogs are mated. Among the offspring of their first litter is a smooth-haired pup. If these two dogs mate again, what are the chances of them having another smooth-haired pup?

 

 

 

9. What are the chances that the pup will be wire-haired?

 

10. A Wire-haired male is mated with a smooth-haired female. The mother of the wire-haired male was smooth-haired. What are the phenotypes and genotypes of the pups they could produce? Show how you got your results.

 

 

 

 

 

Incomplete Dominance 

11. In snapdragons, red flower color (R) is incompletely dominant over white flower color (r). The hybrids or heterozygous plants (Rr) are pink in color. Show the genotype for a white flower and for a red flower.

 

12. If a red-flowered plant is crossed with a white-flowered plant, what are the genotypes and phenotypes of the F1 generation plants? Show your work.

 

 

 

 

13. What is the phenotype of the flowers? what is their genotype?

 

14. What genotypes and phenotypes will be produced in the F2 generation? Show your work.

 

 

 

 

 

15. How did the genotypic and phenotypic ratio compare to each other in this incomplete dominance cross?

16. What would the phenotypic ratio have been if this had been complete dominance?

17. What kind of offspring can be produced if a red-flowered plant is crossed with a pink-flowered plant? Show your work.

 

 

 

 

 

18. What kind of offspring is/are produced if a pink-flowered plant is crossed with a white-flowered plant? Show your work.

 

 

 

 

 

Sex-linked Traits

19. What is the genotype for female?  for male?

20. In humans, colorblindness (Xc) is a recessive sex-linked trait. Two people with normal color vision (XC) have a colorblind son. What are the genotypes of the parents?

 

21. What are the genotypes and phenotypes possible among their other children? Show your work.

 

 

 

 

 

22. A couple has a colorblind daughter. What are the possible genotypes and phenotypes of the parents and the daughter?

 

 

Dihybrid Crosses

23. In humans, the presence of freckles is due to a dominant gene (F) and the non-freckled condition is due to its recessive allele (f). Dimpled cheeks (D) are dominant to non-dimpled cheeks (d). Two persons with freckles and dimpled cheeks have two children. One child has freckles but no dimples. The other child has dimples but no freckles. What is the genotypes of the parents? the children?

 

 

24. What are the possible phenotypes and genotypes of the children that they could produce? Show all your work.

 

 

 

 

 

 

 

 

 

 

25. What phenotypic ratio did you get?

26. What genotypic ratio did you get?

27. What are the chances that they would have a child whom lacks both freckles and dimples?  What would be the child’s genotype?

 

28. A person with freckles and dimples whose mother lacked both freckles and dimples marries a person with freckles but no dimples whose father did not have freckles or dimples. What are the chances that they would have a child whom lacks both freckles and dimples? Show the genotypes of the parents and all the offspring.

 

 

 

 

 

 

 

 

29. In dogs, the inheritance of hair color involves a gene (B) for black hair and a gene (b) for brown hair. A dominant (C) is also involved. It must be present for the color to be synthesized (made). If this gene is NOT present, a blond condition results. Complete the following table:

 

Genotype Phenotype Color Deposition gene
BB or Bb CC or Cc
bb CC or Cc
BB or Bb cc
bb cc

 

 

30. A brown-haired male, whose father was a blond, is mated with a black-haired female ,whose mother was brown-haired and her father was blond. What is the genotype of the man and woman? Show the genotypes and phenotypes of all of their offspring.

 

 

 

 

 

 

 

Population Genetics or Hardy-Weinberg Law

Sixteen percent (16%) of the human population is known to be able to wiggle their ears. This trait is determined to be a recessive gene. Use the following equations to answer this population genetics problem:

1 = p2 + 2pq + q2                                      then use 1 = p + q

p2 – frequency of homozygous dominants

2pq – frequency of heterozygotes

q2 – frequency of homozygous recessives

p – frequency of dominant allele

q – frequency of recessive allele

31. What percent of the population is homozygous dominant for this trait? Show your work.

 

 

 

 

 

32. What percent of the population is heterozygous for this trait? Show your work.

 

 

 

 

 

Multiple Alleles – ABO Blood Type 

33. Henry Anonymous, a film star, was involved in a paternity case. The woman bringing the suit had two children. One child had blood type A and the other child had blood type B. Her blood type was O, the same as Henry’s. The judge in the case awarded damages to the woman, saying that Henry had to be the father of at least one of her children. was the judge correct in his decision? Show how you got your answer.

 

 

 

Genetics Study Guide

Genetics Study Guide 

The two genes or alleles that combine to determine a trait would be the organism’s _______________.
Type AB blood, having two genes dominant for a trait, is an example of ________.
State Mendel’s law of segregation.
Rr x Rr is an example of what type of cross —– P1, F1, or F2?
If both alleles are the same in a genotype, is the genotype homozygous or heterozygous?
Which cross is a cross between two hybrids —– P1, F1, or F2?
__________ dominance results in the blending of genes in the hybrid. Give an example using flower color.
What is another term for a heterozygous genotype?
The _____________ is the physical feature such as round peas that results from a genotype.
How many traits are involved in a monohybrid cross?
What type of organism was used in the first genetic studies done by Gregor Mendel?
What is a karyotype?
The two genes for a trait represented by capital & lower case letters are called __________.
How many traits are involved in a dihybrid cross?
Which of Mendel’s laws states that the dominant gene in a pair will be expressed?
If both alleles are the same, is the genotype homozygous or heterozygous? Write an example.
Write an example of a hybrid or heterozygous genotype.
The genes for sex-linked traits are only carried on which chromosome?
Who is considered to be the “father of genetics”?
A second filial or F2 cross is also called a ____________ cross.
The failure of chromosomes to separate during meiosis (egg & sperm formation) is known as _________________.
A cross between two pure or homozygous organisms is called what type of cross —– P1, F1, or F2?
What genetic disorder results from a sex-linked trait that affects color vision?
The genetic disorder called _______________ is known as the “free bleeders” disease.
Having three 21st chromosomes causes the genetic disorder known as _________.
A person suffering from the genetic disorder called ______________ can not digest fats.
_____________________ disease is a genetic disorder where red blood cells carry less oxygen.
Work a P1 cross for plant height in peas.
Work an F1 cross for plant height in peas.
BACK

 

Hardy-Weinberg Problems

 

POPULATION GENETICS AND THE HARDY-WEINBERG LAW

 

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:

  1. No mutations must occur so that new alleles do not enter the population.
  2. No gene flow can occur (i.e. no migration of individuals into, or out of, the population).
  3. Random mating must occur (i.e. individuals must pair by chance)
  4. The population must be large so that no genetic drift (random chance) can cause the allele frequencies to change.
  5. No selection can occur so that certain alleles are not selected for, or against.

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:

 

p2 + 2pq + q2 = 1 and p + q = 1

 

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:

  1. The frequency of the “aa” genotype.
  2. The frequency of the “a” allele.
  3. The frequency of the “A” allele.
  4. The frequencies of the genotypes “AA” and “Aa.”
  5. The frequencies of the two possible phenotypes if “A” is completely dominant over “a.”

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:

  1. The frequency of the recessive allele.
  2. The frequency of the dominant allele.
  3. The frequency of heterozygous individuals.

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:

  1. The percentage of butterflies in the population that are heterozygous.
  2. The frequency of homozygous dominant individuals.

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:

  1. The allele frequencies of each allele.
  2. The expected genotype frequencies.
  3. The number of heterozygous individuals that you would predict to be in this population.
  4. The expected phenotype frequencies.
  5. Conditions happen to be really good this year for breeding and next year there are 1,245 young “potential” Biology instructors. Assuming that all of the Hardy-Weinberg conditions are met, how many of these would you expect to be red-sided and how many tan-sided?

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:

  1. The frequency of each allele in the population.
  2. Supposing the matings are random, the frequencies of the matings.
  3. The probability of each genotype resulting from each potential cross.

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:

  1. The frequency of the recessive allele in the population.
  2. The frequency of the dominant allele in the population.
  3. The percentage of heterozygous individuals (carriers) in the population.

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.

ANSWERS

Fungi

 

Fungi
All Materials © Cmassengale

Characteristics

  • Eukaryotic 
  • Do not contain chlorophyll
  • Nonphotosynthetic
  • Absorptive heterotrophs – digest food first & then absorb it into their bodies
  • Release digestive enzymes to break down organic material or their host
  • Store food energy as glycogen
  • Most are saprobes – live on other dead organisms
  • Important decomposers & recyclers of nutrients in the environment
  • Most are multicellular, but some unicellular like yeast
  • Some are internal or external parasites; a few are predators that capture prey
  • Nonmotile
  • Lack true roots, stems, & leaves
  • Cell walls are made of chitin (a complex polysaccharide)
  • Grow as microscopic tubes or filaments called hyphae that contain cytoplasm & nuclei
  • Hyphal networks are called mycelium
  • Some are edible
  • Reproduce by sexual & asexual spores
  • Antibiotic penicillin comes from Penicillium mold
  • Classified by their sexual reproductive structures
  • Grow best in warm, moist environments preferring shade
  • Mycology – study of fungi
  • Fungicide – chemicals used to kill fungi
  • Includes yeasts, molds, mushrooms, ringworm, puffballs, rusts, smuts, etc.
  • Fungi may have evolved from prokaryotes by endosymbiosis

Vegetative (nonreproductive) Structures of Fungi

  • Body of a fungus made of tiny filaments or tubes called hyphae
  • Hyphae contain cytoplasm & nuclei and has a cell wall of chitin


HYPHAE

  • Each hyphae is one continuous cell
  •  Hyphae continually grow & branch
  • Septum (septa-plural) are cross walls with pores to allow the movement of cytoplasm in hyphae
  • Hyphae with septa are called septate hyphae
  • Hyphae without septa are called coenocytic hyphae

  • Tangled mats of hyphae are known as mycelium
  • All hyphae within a mycelium share the same cytoplasm so materials move quickly
  • Hyphae grow rapidly from the tips by cell division
  • Stolon is a horizontal hyphae that connects groups of hyphae to each other
  • Rhizoids are rootlike parts of hyphae that anchor the fungus

Reproductive Structures

  • Most fungi reproduce asexually & sexually
  • Asexual reproduction produces genetically identical organisms & is the most common method used
  •  Sexual reproduction in fungi occurs when nutrients or water are scarce
  • Fruiting bodies are modified hyphae that make asexual spores
  • Fruiting bodies consist of an upright stalk or sporangiophore with a sac containing spores called the sporangium


SPORANGIOPHORES

  • Types of fruiting bodies include basidia, sporangia, & ascus
  • Spores – haploid cells with dehydrated cytoplasm & a protective coat capable of developing into new individuals
  • Wind, animals, water, & insects spread spores
  • When spore lands on moist surface, new hyphae form

Asexual Reproduction in Fungi

  • Fungi reproduce asexually when environmental conditions are favorable
  • Some unicellular fungi reproduce by mitosis
  • Yeast cells reproduce by budding where a part of the cell pinches off to produce more yeast cells

  • Athlete’s foot fungus reproduce by fragmentation from a small piece of mycelium
  • Most fungi reproduce asexually by spores
  • Penicillium mold produces spores called conidia without a protective sac on the top of a stalk called the conidiophore

Sexual Reproduction in Fungi

  • Fungi reproduce sexually when environmental conditions are unfavorable
  • No male or female fungi
  •  Two mating types — plus (+) and minus (-)
  • Fertilization occurs when (+) hyphae fuse with (-) hyphae to form a 2N or diploid zygote
  • Some fungi show dimorphism (ability to change their form in response to their environmental conditions)

Classification of Fungi

  • Fungi are classified by their reproductive structures
  • The 4 phyla of fungi are Basidiomycota, Zygomycota, Ascomycota, & Deuteromycota

Zygomycota

  • Called sporangium fungi or common molds
  • Includes molds & blights such as Rhizopus stolonifer (bread mold)

  • No septa in hyphae (coenocytic)
  • Asexual reproductive structure called sporangium & produces sporangiospores
  • Rhizoids anchor the mold, release digestive enzymes, & absorb food
  • Asexual reproductive structure called sporangium & produces sporangiospores
  • Sexual spore produced by conjugation when (+) hyphae & (-) fuse is called zygospore
  • Zygospores can endure harsh environments until conditions improve & new sporangium

 

 

Basidiomycota

  • Called club fungi
  •  Includes mushrooms, toadstools, puffballs, bracket fungi, shelf fungi, stinkhorns, rusts, & smuts
  • Some are used as food (mushroom) & others cause crop damage (rusts & smuts)
  • Seldom reproduce asexually
  • Basdiocarp made up of stalk called the stipe & a flattened cap
  • Stipe may have a skirt like ring below cap called the annulus
  • Gills are found on the underside of the cap & are lined with basidia
  • Basidium – sexual reproductive structure that make basidiospores
  • Basidiospores are released from the gills & germinate to form new hyphae & mycelia
  • Vegetative structures found below ground & include rhizoids (anchor & absorb nutrients), hyphae, & mycelia

Ascomycota

  • Called sac fungi
  • Includes yeast, cup fungi, truffles, powdery mildew, & morels

  • Some are parasites causing Dutch elm disease & chestnut blight
  • Sac Fungi can reproduce both sexually and asexually
  •  Yeast reproduce asexually by budding (form small, bud-like cells that break off & make more yeasts)
  • Asexual spores called conidia form on the tips of specialized hyphae called condiophores
  • Ascocarp – specialized hyphae formed by parent fungi during sexual reproduction
  • Ascus – sacs within the ascocarp that form spores called ascospores

Lichens

  • Symbiotic association between a sac fungus & a photosynthetic green algae or cyanobacteria
  • Both organisms benefit (algae makes food & fungus supplies moisture, shelter, & anchorage)
  • Grow on rocks, trees, buildings, etc. & help form soil
  • Crustose lichens grow on rocks & trees; fructose lichens grow shrub-like; foliose lichens grow mat-like on the soil

Mycorrhizae

  • Symbiotic association of a fungus living on plant roots
  • Most plants have mycorrhizae on their roots
  • Fungus absorbs sugars made by plant
  • Plants absorb more water & minerals with aid of the fungus

Importance of Fungi

  • Fungal spores cause allergies
  • Molds, mildew, rusts, & smuts damage crops
  • Yeasts are used to make beer & bread
  •  Antibiotic penicillin
  • Decomposers & recyclers of nutrients
  • Mushrooms eaten as food
  • Help form blue cheeses
  • Aspergillus is used to make soy sauce
  • Cause athlete’s foot & ringworm
  • Amanita is poisonous mushroom
  • Can cause yeast infections