1. The mixed culture in an experiment was prepared by mixing overnight E. coli and S. aureus cultures in the ratio 2:8. However, there seemed to be more E. coli growth on the resulting streak plate than S. aureus growth. Explain this.
2. It was observed that some colonies of the same bacterial species appeared in different sizes on the streak plates. Give the possible reason for this.
3. Is it better to prepare pure culture in broth or on agar plate? Explain.
In: Biology
Bacteria, Archaea, Chloroplasts, and Mitochondria reproduce by binary fission. Explain why. (paragraph)?
In: Biology
1.Draw and label two cells at anaphase of mitosis and anaphase I of meiosis in a species having 2n = 4 chromosomes (one homologous pair of metacentric chromosomes and a pair of acrocentric chromosomes).
2.Sketch the cell cycle
In: Biology
What are some examples of cell survival signals?
What are some examples of cell death signals?
Stem cells Cloning: read therapeutic (SCNT) vs. reproductive cloning
iPs cells
In: Biology
1. Answer the following completely and in detail.
a) Draw, clearly label, and describe how binding of the trp corepressor to its repressor alters repressor function and transcription in E. coli.
b) Draw, clearly label, and describe the binding of RNA polymerase, repressors, and activators to the lac operon when both lactose and glucose are scarce. What is the effect of these scarcities on transcription of the lac operon?
c) Suppose that a certain mutation in E. coli changes the lac operator so that the active repressor cannot bind. How would this affect the cell's production of B-galactosidase?
THANK YOU!
In: Biology
1/ Could you think of a way to change the parameters in the life table, so that λ will never stabilize to a certain value? (Hint: a simple way to do so involves changing the values of mx).
2/ Among all the assumptions for the exponential growth model, which no longer hold true for the age-structured model?
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In a cross of two flies +/vg Cy/+ +/se +/ab X +/vg +/+ se/se +/ab what proportion of the offspring will be mutant in phenotype for all four markers?
ANSWER: 1/64
Please explain the process
In: Biology
How do Cells Attach to the ECM?
Integrins: Apoptosis: What are triggers? What does this cause in cell? What is importance of location of PS (phosphatidyl serine) in cell membrane? Caspases (role and examples of targets in cell) Apoptosome (look at role of cytochrome c)
In: Biology
In: Biology
Many antibiotics work via selective toxicity. Explain how each of the following antibiotics is toxic to bacterial pathogens, but not* human cells. A) β-lactam drugs (such as penicillin, ampicillin, etc.) B) Sulfa drugs C) Tetracyclines
In: Biology
After the fusion of the viral and endosomal membranes, the negative-stranded genome segments of influenza virus are released into the cytoplasm. Describe what happens to genome segments 4, 5 and 7 during the viral replication cycle (see slides 21 and 28 of lecture 32 Powerpoint presentation). You descriptions should include:
Provide a separate description for each of the genome segments (even though some of the steps are the same for each). Your descriptions should be concise but should include all relevant details.
Genome segment 4:
Genome segment 5:
Genome segment 7
In: Biology
In: Biology
Human Genetics - What will your children be?
The purpose of this exercise is to give you some practice working with the major concepts of genetics using specific example of inherited traits present in humans. During this lab, we will observe patterns of inheritance for three general types of traits: autosomal traits controlled by two alleles (hair color, Rh factor, PTC tasting), an autosomal trait controlled by multiple alleles (blood type), and a sex-linked trait (color-blindness). Before we begin this exercise, we should introduce or review some of the terminology used when discussing genetics.
All of an organism's traits, whether they are visible traits such as hair and eye color or non-visible traits such as blood type, are controlled by genes. A gene is a specific region on a strand of DNA (deoxyribonucleic acid) that controls a specific trait. The 46 long strands of DNA in human cells, each containing many genes, are wound into chromosomes which become visible during mitosis (remember Ex.7). Most human cells are diploid, meaning they contain two sets of genetic information. Diploid human cells contain 23 pairs of chromosomes (46 chromosomes total). A pair of these chromosomes are referred to as homologous chromosomes because both chromosomes have genes that code for the same traits. One of the chromosomes in each pair was inherited from the mother and the other from the father.
While diploid cells have two copies of each gene, the copies may
not be identical. For example, each of our cells has two genes that
control hair color, but one gene may code for light hair and one
may code for dark hair. The different versions of a gene are called
alleles. The different alleles of a gene are often not
expressed equally. For each trait, one allele is usually
dominant and the other is recessive. A dominant
allele is expressed whenever it is present, even if only one copy
is present.
A recessive allele must be present in two copies to be
expressed.
When we write out a genetics problem, we often use letters to denote the different alleles for a particular trait. An uppercase letter denotes a dominant allele, and a lowercase letter denotes a recessive allele. As we will see later, other notations may also be used to indicate which alleles are present. If a person has two copies of the same allele (AA or aa), they are considered homozygous for that trait. If they possess two different alleles (Aa), they are considered heterozygous for that trait. When we refer to the actual genetic makeup of an individual, the alleles that are present, this is called the genotype. If we refer to the outward appearance of the individual, or what trait is expressed, then this is referred to as the phenotype of that individual.
Next, we will consider the specific traits that we will be
working with in lab today.
Autosomal traits controlled by two alleles: Genes
for these traits are carried on two homologous autosomal (non-sex)
chromosomes. There are only two alleles or versions for each gene
present in the population.
Ability to taste PTC: PTC (phenylthiocarbimide) is a chemical that tastes bitter to some people and is tasteless to others. The ability to taste PTC is an inherited trait, and the ability to taste PTC is dominant (T) over the non-taster (t) condition.
Hair Color: In this case, we will consider only dark hair or light hair. Dark hair (B) is dominant over light hair (b).
Rh factor: The Rh factor is an antigen found on red blood cells. It is named after the Rhesus monkey where it was discovered. People who possess this antigen on their red blood cells are considered Rh+ (dominant), while those without the antigen are Rh-(recessive). The Rh factor is of concern if an Rh- mother is carrying an Rh+ child, since the mother's body may form antibodies to the child's blood, resulting in numerous health problems.
Autosomal traits controlled by multiple
alleles: Genes for these traits are also carried on
homologous autosomal chromosomes, but there are more than two
alleles (versions) of the gene present in the population. However,
only two of these alleles can be
present in any one individual.
Blood type: A person's blood type or phenotype (A, B, AB, O) is determined by the presence of certain antigens on the surface of their red blood cells. Control for the inheritance of blood types is based on three different alleles: IA, IB, and IO. Again, only two of these alleles will be present in any one individual. The IA allele codes for the "A" antigen, the IB allele codes for the "B" antigen. The presence of the IO allele results in no antigen on the surface of the red blood cell. A person's blood type (phenotype) is the result of the combination of these alleles. The IA and IB alleles are co-dominant and are equally expressed if both are present. The IA and IB are both dominant to the IO allele. The only way for a person to have type O blood is to have both recessive alleles. Fill in the table at the right with the proper phenotype (blood type) for each possible genotype.
Sex chromosomes and sex-linked traits: Sex in humans is determined by the presence or absence of certain whole chromosomes. Females have two X chromosomes and males have an X and a Y chromosome. Genes for sex-linked traits are usually carried on the X chromosome and are absent on the homologous Y chromosome. As a result, males only have a single copy of any gene located on the X chromosome and they will express whichever allele is present, even if it is normally a recessive allele.
Color-blindness: The gene that controls color vision is located on the X chromosome and has two alleles: normal (X) and color-blind (X°). The normal condition is dominant over the color-blind condition. Females that are phenotypically normal, but heterozygous for color-blindness (X X°) are often referred to as carriers
Now that you have become familiar with the basic terms of genetics and the specific traits that we will observe today, you are ready to put this knowledge to use. To do this, we will play a "family" game. During the course of the lab, each of you will determine your genetic makeup, find a mate, and determine the combination of traits that would be possible in your offspring. Here is how it works:
Step 1: Names - These will be used to determine your mates. Men will draw a last name from the beaker marked males and women will draw a last name from the beaker marked female. If we do not have equal numbers of men and women, some people may have multiple mates.
Step 2: Determine sex & sex-linked traits - Men will draw only ONE X chromosome and ONE Y chromosome. Women will draw TWO X chromosomes. The X chromosome will have an allele for normal vision or color blindness.
Step 3: Determine autosomal traits - In order to find out your traits, you will draw TWO chromosomes with alleles for each of the following traits: hair color, ability to taste PTC, Rh factor, and blood type.
Step 4: Take a social break and find your mate. When you find someone with the same last name, have a seat beside your new mate.
Step 5: Once you have become acquainted with your mate;-) arrange your alleles in front of you in the order shown below for Jane Smith.
Genotype XX
Bb
IA IO Rh+Rh- TT
Phenotype
Female with normal vision Dark hair
Blood type A
Rh+
PTC Taster
Step 6: Fill in the table below with the information for you and your mate. If needed, review the information on pages 1-2
MY RESULTS
my phenotypes: SEX/VISION = xBxb (female normal vision), BLOOD TYPE=IAIA (A), +/- BLOOD TYPE = Rh-Rh- (-), PTC TASTING = Tt (taster), HAIR COLOR = BB (dark)
Mates phenotypes: SEX/VISION = xbY (colorblind male), BLOOD TYPE= IAIO (A), +/- BLOOD TYPE = Rh- Rh- (-), PTC TASTING = tt (non taster), HAIR COLOR= Bb (dark)
Mates Phenotypes:
Step 7: You are now ready to determine all the possible gametes that each of you could contribute to your offspring. Remember, gametes are formed by meiosis, which is reduction division. Each gamete will only get one allele from each pair of homologous chromosomes.
The number of possible gametes is equal to 2 (because we are dealing with 2 alleles) raised to the ?n?
power, where ?n? equals the number of pairs of heterozygous alleles. For example, looking at Jane Smith=s genotype we can see that she is heterozygous for three traits: hair color (Bb), blood type (IAIO), and Rh factor (+-). According to our formula, she could have (23) = 8 different gametes. How many possible gametes could you have? _____ How many possible gametes could your mate have?_____
Step 8: Each person should now determines their own gametes and writes them below. The method illustrated below using Jane Smith=s genotype will allow you to systematically find all possible gametes for you and your mate. Each column represents one possible gamete. Jane could have eight different gametes (23 = 8), so there should be eight columns shown below when we are finished. You may have more or less than eight columns depending on the number of heterozygous pairs of alleles that you have. Choose one trait that Jane (or you) is heterozygous for (ex., Rh factor) and give the possible gametes. (Hint: there should be 2, one for each allele).
Use the space below to write out all of the possible gamete combinations for you and your partner.
The total number of different looking children that you and your mate could produce can be determined by multiplying your number of possible gametes times your mate's number of possible gametes. Considering only these traits, how many different children could you and your mate produce?_________
Look at your gamete possibilities and genotypes and those of your mate to answer the following questions. Use simple Punnett squares to determine the possible outcomes of the crosses where needed.
1. Will any of your children be color-blind?
2. What are the possible hair colors of your children?
3. Will any of your children be non-tasters?
4. What are the possible blood types of your children?
5. Will any of your children be blood type O-?
6. Will any of your children be AB+?
7. Could you possibly produce a child who would have dark hair, could taste PTC, and have AB- blood?
8. Would you bet on the sex of your next child? Why or why not?
9. If you were a betting person, are there any traits, concerning your child, on which you would bet? Explain.
10. Today we can determine the genetic makeup of individuals and determine whether they carry traits that are potentially harmful such as Huntington=s disease. Knowing this, would you participate in a genetic typing for certain various characteristics before you get married? Why or why not?
In: Biology
To test the effectiveness of a new drug (Phasomaxtrin) designed to treat Quarantinitis, researchers conducted a clinical trial. The subjects were 512 adult American volunteers with advanced Quarantinitis. Half were randomly assigned to take the drug and the other half were randomly assigned to take a placebo. Neither the subjects nor the doctors who evaluated them knew who was in which group. After three years, 32 percent of those who got Phasomaxtrin were no longer afraid to keep staying inside, compared with only 23 percent of those who got the placebo. (The difference is statistically significant.)
In: Biology