Answer:
A. A significant reduction in the numbers of mature CD4 T cells
Explanation:
In the given problem, there is an engineering of the MHC class II protein with the sequence of a single peptide. In addition, it was expressed in the epithelial cells of the thymic cortical. Based on the result obtained from the engineering construction, it is obvious that there would be a large decreases in the CD4 T cells numbers.
You can make a cell extract that is able to perform glycolysis in vitro (in a test tube) if glucose is added. Arsenate is a potent inhibitor of triose phosphate dehydrogenase, the enzyme required for the 6th step in glycolysis. If both arsenate and glucose are added to the cell extract, what happens?
a. ATP levels decrease.
b. Both ATP and pyruvate levels decrease.
c. Both ATP and pyruvate levels increase.
d. ATP levels increase.
Answer: B
Explanation:
If glucose and arsenate are both added to the cell extract, at first glycolysis will start.
In step one of glycolysis, glucose is phosphorylated to glucose-6-phosphate catalyzed by hexokinase which splits the ATP into ADP, and the Pi is added on to the glucose.
In step 3 of glycolysis, fructose-6-phosphate is further phosphorylated to fructose 1,6-bisphosphate. The enzyme is phosphofructokinase. This again involves hydrolysis of another ATP molecule.
A total of two ATP is used.
Step 6 in glycolysis reaction which involves generation of 2 ATP's molecules is inhibited by arsenate. Hence all other glycolytic reaction would not take place. Therefore no ATP is produced and pyruvate is not produced also.
ATP level decreases because ATP is only used up but no ATP is gained from the inhibited pathway. Also the inhibition of the step 6 enzyme cut short the pathway and pyruvate the end product of the pathway is not formed.
A maternal effect can cause the offspring phenotype ratio to depart from that of classic Mendelian inheritance. In a species of snail, the dominant allele N codes for right-handed shell coiling and recessive allele n codes for left-handed shell coiling. If an Nn female with right-handed shell coiling males with an Nn male, what is the shell coiling phenotypic ratio of their offspring?
A. 4.right-handed coil:0, left-handed coil
B. 3. Right-handed coil: 1.left-handed coil
C. 3.bidirectional coil: 1.left-handed coil
D. 0.right-handed coil: 4.left-handed coil
Answer:
B. 3. Right-handed coil: 1.left-handed coil
Explanation:
Phenotype is what you see - the visible or observable expression of the results of genes, combined with the environmental influence on an organism's appearance or behavior.
When Nn is crossed with Nn, they will produce offspring with NN, Nn, Nn and nn genotype.
N - Dominant allele
n - recessive allele
The phenotypic ratio of this offspring is 3 right-handed coil and 1 left-handed coil.
Answer:
B. 3 right-handed coil: 1 left-handed coil
Explanation:
Female genotype : Nn
Male genotype: Nn
N: dominant allele (right handed coil)
n: recessive allele (left handed coil)
When we cross the Male (Nn) x Female (Nn) the phenotypic ratio will be right handed coil 3 : left handed coil 1.
Solution:
Nn x Nn
Four possible genotypes = NN, Nn, Nn, and nn
So, NN = right handed coil
Nn (2) = right handed coil
nn= left handed coil
"As you examine the specimens (slides, whole specimens, etc.) in lab, determine where each species belongs on the phylogenetic tree based on the traits provided. List 5 additional traits you can add to the phylogeny. "
To determine where each species belongs on the phylogenetic tree, examine their shared traits and make inferences about their evolutionary history. Five additional traits can be added to the phylogeny, such as anatomical structures, biochemical reactions, reproductive strategies, behaviors, and genetic sequences.
Explanation:When examining specimens in the lab, you can determine where each species belongs on the phylogenetic tree based on their shared traits. By comparing the traits of different species, you can identify common characteristics and make inferences about their evolutionary history.
For example, if two species share a trait that is not found in any other species, they are likely more closely related and would appear closer on the phylogenetic tree.
To add five additional traits to the phylogeny, you can consider various characteristics such as anatomical structures, biochemical reactions, reproductive strategies, behaviors, and genetic sequences. By including these traits, you can further refine the phylogenetic tree and understand the relationships between different species more comprehensively.
Alleles of the gene that determines seed coat patterns in lentils can be organized in a dominance series: marbled > spotted = dotted (codominant alleles) > clear. A lentil plant homozygous for the marbled seed coat pattern allele was crossed to one homozygous for the spotted pattern allele. In another cross, a homozygous dotted lentil plant was crossed to one homozygous for clear. An F1 plant from the first cross was then mated to an F1 plant from the second cross. a. What phenotypes in what proportions are expected from this mating between the two F1 types? b. What are the expected phenotypes of the F1 plants from the two original parental crosses?
Answer:
two types of cross are given,marbled and spottedhere, marbled is dominant (MM) ; spotted is co-dominant (Ss)
in F1 generation → MS (marbled spotted coat)
2. dotted and clear
here, dotted is co-dominant (Dd) ; clear is recessive (dd)
in F1 generation → Dd (clear background with dotted coat)
if MS × Dd , then, it will give,
25% spotted and dotted ( as both are co-dominant {Ss=Dd}) : 50% marbled (as marbled is dominant{ MM}) : 25% spotted ( as spotted {Dd} is co-dominant and clear {dd} is recessive)
a. 25% spotted dotted: 50% marbled: 25% spotted
expected phenotypes from F1 is →MS (marbled)- as marbled is the most dominant among all, it won't let spotted to be expressed.
and Dd - as clear is recessive only dotted will be expressed.
b.marbled and dotted.
Final answer:
The expected phenotypes from the mating of two F1 lentil plants from different crosses are 1:1 marbled to spotted/dotted, as marbled is dominant while spotted and dotted show codominance, and clear is recessive. The F1 offspring from the first cross will have a marbled phenotype, and from the second cross, a dotted phenotype due to codominance.
Explanation:
We are addressing complex inheritance patterns in lentils involving multiple alleles with a dominance series. To predict the offspring's phenotypes, we first need to understand Mendel's laws and how dominance affects the genotype-phenotype relationship. We also employ Punnett squares to visualize genetic crosses.
a. Expected Phenotypes in Offspring from F1 Mating
The F1 plants from the first cross (marbled × spotted) will all exhibit the marbled phenotype, as marbled is dominant over spotted. The F1 plants from the second cross (dotted × clear) will all show the dotted phenotype due to codominance between dotted and spotted alleles, and dominance over clear.
When these two F1 plants are crossed (marbled/dotted × dotted), the expected offspring phenotypes are as follows, assuming each allele has equal chance of passing on:
Marbled: Represents the dominant allele and is expected to show up in half the offspring if the marbled allele is present.
Spotted/Dotted (Codominance): Should appear in the other half of the offspring, as these are the alleles present in both F1 parents. The spotted and dotted patterns are codominant and will be expressed equally if one of each is inherited.
Clear: Will not be seen in the offspring, as it is recessive to all other alleles.
Thus, the phenotypic ratio of the offspring from the cross between the two F1 plants is expected to be 1:1 marbled to spotted/dotted with no clear phenotypes.
b. Expected F1 Phenotypes from Original Parental Crosses
The F1 phenotype resulting from a homozygous marbled crossed with homozygous spotted will be all marbled, following Mendelian inheritance patterns. In the cross of homozygous dotted with homozygous clear, all offspring will display a dotted pattern, indicative of codominance interaction over the clear allele.
Consider a locus with two alleles - B and b. B is dominant, while b is recessive. There is no mutation. B has a selective advantage relative to b, so that the fitnesses of the three genotypes are BB = 1, Bb = 1, and bb = 1-s. In this case, s = 0.50, so that bb homozygotes have 50% fitness of heterozygotes and BB homozygotes. If the population has the following genotypic counts prior to selection of BB = 500, Bb = 250, and bb = 250, what is the expected change in the frequency of B after one generation with selection? Please give your answer to two decimal places.
Answer:
0.09
Explanation:
Before selectionTotal number in population = 1000
Genotype frequencies
Genotype frequency of BB = 500/1000 = 0.5
Genotype frequency of Bb = 250/1000 = 0.25
Genotype frequency of bb = 250/1000 = 0.25
Allele frequencies
Allele frequency of B = BB genotype frequency + half of the Bb genotype frequency = 0.5 + (0.25/2) = 0.625
Allele frequency of b = bb genotype frequency + half of the Bb genotype frequency = 0.25 + (0.25/2) = 0.375
After selectionWe are told that after selection, the genotype frequency of bb is changed as they become 50% less fit. This means the frequency of bb individuals changes from 250 to 125 individuals (50% reduction).
Now the total number of individuals is 500 + 125 + 250 = 875.
Genotype frequencies
Genotype frequency of BB = 500/875 = 0.57
Genotype frequency of Bb = 250/875 = 0.29
Genotype frequency of bb = 125/875 = 0.14
Allele frequencies
Allele frequency of B = BB genotype frequency + half of the Bb genotype frequency = 0.57 + (0.29/2) = 0.715
Allele frequency of b = bb genotype frequency + half of the Bb genotype frequency = 0.14 + (0.29/2) = 0.285
Change in frequency of B after 1 generation0.715 - 0.625 = 0.09
Final answer:
The expected change in the frequency of allele B after one generation of selection, given the provided genotypic counts and selection coefficients, is an increase by approximately 0.09.
Explanation:
First, we can calculate the initial frequencies of alleles B and b. Each BB individual contributes two B alleles, and each Bb individual contributes one B allele. The total number of B alleles in the population is (2 × 500) + (1 × 250) = 1250, and similarly, the total for b alleles is (2 × 250) + (1 × 250) = 750. The total number of alleles is 2000. Therefore, the frequency of B (p) is 1250/2000 = 0.625 and the frequency of b (q) is 750/2000 = 0.375.
After selection, the new frequencies of the alleles can be determined. The fitness values are wBB = 1, wBb = 1, and wbb = 0.5. We need to adjust the number of each genotype by its fitness:
Adjusted BB = 500 × 1 = 500Adjusted Bb = 250 × 1 = 250Adjusted bb = 250 × 0.5 = 125The new frequency of B after selection (p') will be the sum of the alleles from BB and Bb genotypes after fitness adjustment over the total adjusted alleles.
New frequency of B (p') = (500 × 2 + 250) / (2 × (500 + 250 + 125))p' = (1250) / (2 × 875)p' = 0.7143 (to four decimal places)The change in the frequency of B after one generation of selection is the difference between the new frequency and the original frequency.
Change in frequency of B = p' - p = 0.7143 - 0.625 = 0.0893 or to two decimal places, 0.09.
Your BFF is a red-headed, left-handed, blue-eyed girl. She finds out that you are taking Human Genetics and wants you to explain to her the genetic basis of her three traits. So, your task is to provide her with the most up-to-date explanation, to the extent that is possible. You have to do so in 400-500 words.
Answer:
Red hair, left handed, and blue eyes are recessive traits.
Any trait has its two allele, dominant and recessive. Dominant trait allele if present with the recessive allele, then the phenotype will be of dominant allele.
Suppose A = dominant allele for brown color hair, a = recessive allele for red color hair.
As mentioned above red color hair is recessive in nature. for the recessive traits to show, the alleles should be recessive only.
Aa (brown hair) (parent 1) is crossed with Aa (parent 2) (brown hair). They both have gametes A and a.
Their children will have genotype AA, aa, and Aa. AA = brown color, aa = red color hair, and Aa= brown color hair.
Same case goes with the left hand and blue eyes traits.
So, the girl parents must be having recessive alleles for all the traits, and she must have inherited all the recessive allele only.
Explanation:
First of all, you should ask your friend if any of her parents also have these characteristics. In this you must explain the concept of heredity, which is the genetic phenomenon that allows us to be able to inherit genetic information from our parents, through DNA. That's because each of us has half the DNA of each of our parents, so we have characteristics in common with them.
After explaining heredity, you should talk about your friend's blue eyes. Blue eyes are characterized by a small amount of melanin, which is the brown pigment that colors our eyes, skin and hair. The more melanin, the darker the eyes. What determines the amount of melanin in the eyes is DNA. If your friend's parents have light eyes, it means they have little melanin and that trait was passed to her through DNA.
Red hair, on the other hand, is the result of a genetic mutation that is passed from parents to children through DNA. People with red hair have a mutation in the gene responsible for the production of melanin. This mutation causes the gene to produce very low levels of melanin for the hair, but it produces large amounts of pheomelanin which is a reddish pigment.
However being left-handed is not related to heredity. Many scientists believe that what determines whether a person is left-handed or not is a set of genes that choose the left or right side of the body as dominant. However, other scientists believe that this is actually linked to a matter of skill and custom.
Which of the following is true of the opioid fentanyl? It is used primarily in conjunction with surgical anesthesia. It is found in the brain tissues and have potent psychoactive effects. It stimulates the brain area controlling nausea and vomiting. It makes the respiratory centers less responsive to carbon dioxide levels in the blood.
Answer: it is used primarily in conjunction with surgical anesthesia
Explanation: it is one of the synthetic opioid , it is used intravenously as anesthesia and to treat pain. It is given with a muscle relaxant and a sedative hypnotic. The effect is quick in the body and central nervous system and can last less than 2 hours.
Answer:
It is used primarily in conjunction with surgical anesthesia.
Explanation:
Fentanyl opioid is an injectable or orally applied medication used to treat pain or used in anesthesia. Its effect occurs very fast providing the quick aliveio of the pain the loss of sensation of the anesthetized place, generally in about an hour or two.
Despite having a great medicinal value, this medication is highly dependent and can leave those who consume it quickly addicted and causing serious damage to your body, even leading to death.
This medication should not be used without a doctor's prescription. It can cause drowsiness, nausea, constipation, pressure drop, addiction, respiratory depression, among others.
The autonomic nervous system can change the rate of the heart by: Group of answer choices beta1 adrenergic receptor activation. Increases in cAMP lead to increased amounts of Na influx (though If channels) and Ca2 influx of the pacemaker cells. This increases the frequency of APs of the pacemaker, and increases the rate of contraction. beta2 adrenergic receptor activation. Increases in cAMP lead to increased amounts of Na influx (though If channels) and Ca2 influx of the pacemaker cells. This increases the frequency of APs of the pacemaker, and increases the rate of contraction. muscarinic ACh receptors activation. Activation leads to reduced activity of Ca2 channels, and increasing activation of K channels, hyperpolarizing cells and reducing the rate of contraction. alpha1 adrenergic receptor activation.
Answer:
The correct answer is: β1 adrenergic receptor activation. Increased in cAMP lead to increased amounts of Na influx (though if channels) and calcium influx of pacemaker cells. This increases the frequency of APs of the pacemaker and increases the rate of contraction.
Explanation:
cAMP generated in response to β1 adrenergic receptors,results in excitation contraction coupling by initiating PKA and causing phosphorylation of L type Ca2+ channels and ryanodine receptors. This increases the concentration of intracellular Ca2+ ions in atrial cells, ventricular cells and AV pacemaker cells. This increase in firing rate. As a result rate of contraction increases.
In one experiment, scientists raised mice in germ-free conditions so the mice lacked intestinal microbes. The mice were fed a low-fat diet rich in complex plant polysaccharides, such as cellulose, that are often called fiber.
When the mice were 12 weeks old, the scientists transplanted the microbial community from the intestine of a single "donor" mouse into all of the germ-free mice. Then they divided the mice randomly into two groups and fed each group a different diet.
Group 1 (the control group) continued to eat a low-fat, high-fiber diet.
Group 2 (the experimental group) ate a high-fat, high-sugar diet.
Identify the components of this experiment by matching the terms with the appropriate category below:
Factors being tested Factors controlled Factors to be measured
(independent variable) (Kept consistent) (dependent variable)
1. initial composition of the microbial community
2. age of the mice
3. change in body fat
4. diet
5. weight gain
6. final composition of the microbial community
Answer:
The answers are:
1. initial composition of microbial community (controlled variable)
2. age of the mice (controlled variable)
3. change in body fat (dependent variable)
4. diet (independent variable)
5. weight gain (dependent variable)
6. final composition of the microbial community (controlled variable)
Explanation:
Before I go ahead to explain each of the choices of classification, let us define each of the different types of variable.
1. Independent variable: this is a variable that is under the control of the experimenter. It is made up of the factors that are being tested. In our example, the independent variables is; diet. The aim of the experiment is to study the effect of the type of diet on the mice. The diet can also be varied and assigned at will by the experiment.
2. Dependent variable: a dependent variable is one whose outcome is reliant on the effect of the changes in the independent variable. It is the factor to be measured. In our example, the factors that obey these definitions are; change in body fat and weight gain. these two factors depend on the types of diet taken by the mice, hence they are the factors to be measured.
3. Controlled variable: these are variables that are kept constant throughout the experiment to rule out biases, because they produce the same effect on the dependent variable as the independent variable and must be keep constant or uniform in the test groups. in our example, the factors that are controlled include; initial and final compositions of the microbial communities and the age of the mice, because the microbial composition if not uniform among the groups can cause biases in how the diets are processed, and the age of the mice also must be uniform to avoid age-related differences in the processing of the diets
Bacteria are grown in N15 over time, then switched to growth in N14 medium. After three divisions (60 min) what is the percentage of bacterial cells that have N15 labeled DNA? (A) 100% (B) 50% (C) 25% (D) 10% (E) 5%
Answer:
C. 25%
Explanation:
A bacterial cell gives rise to two daughter cells by one division. If a single bacterial cell enters into cell division, it will form a total of 2^3= 8 cells after three rounds of cell division. The two DNA strands of the parent DNA duplex having N15 will be present in two bacterial cells out of the total 8 cells. This would occur since the process of DNA replication forms a new DNA duplex with one parental strand and one new strand. So, after three round of cell division 25% cells will have DNA with N15 (2/8 x 100= 25%)
Final answer:
After three divisions in 14N medium starting with 15N-labeled DNA, due to semi-conservative replication, 25% of bacterial cells will contain DNA with 15N label per Meselson and Stahl's experiment.
Explanation:
The experiment by Meselson and Stahl with E. coli growing in heavy nitrogen (15N) and then switching to light nitrogen (14N) demonstrates the semi-conservative model of DNA replication. After the first division in 14N medium, each DNA molecule contains one strand with 15N and one with 14N, making it fifty percent 14N. By the principle of semi-conservative replication, after three divisions, there will be one more division than there are strains that have any 15N isotopes. Thus, one quarter of the bacterial cells should contain 15N since those are the ones that did not undergo the last division and still contain one original heavy strand. This makes the answer (C) 25%.
Biennials produce only a small leaf crown in the first year of growth.
True
False
The answer is true because biennialsare mostly two.
True, biennials like carrots and beets produce a small crown of leaves in the first year of growth as they focus on vegetative development and storage of resources, preparing for flowering and seeding in their second year.
True. Biennials are plants that have a two-year lifecycle. In the first year, they focus on vegetative growth, developing a small crown of leaves and storing resources in their roots and other structures. This vegetative phase includes crops such as carrots, beets, and cabbage, where typically only a rosette of basal leaves is produced. During the second year, these plants utilize the stored resources to flower, produce fruit, and set seeds, followed by the end of their lifecycle. The cold period, or winter, plays a crucial role in vernalizing these plants, preparing them for the reproductive phase. Due to the significance of this vegetative growth in the first year, these plants do not need to be re-planted annually like annuals, but they do not survive beyond the second year or season like perennials.
18. Which of the following is NOT a factor in the movement of water into tissues?
A. active transport
B. salinaity gradient
C. fluid pressure
D. diffusion
E. osmotic gradient
Answer:D
Explanation:diffusion involved movement of molecules ( gas,) from high concentration to low concentration without a semipermeable membrane. Whereas movement of water require active transport through specialized xylem tissues.
If Sylvia is described as trustworthy, altruistic and tender-minded, then she probably is fairly high on the Big Five superfactor known as extraversion neuroticism openness agreeableness conscientiousness
Answer:
Agreeableness
Explanation: cares about others and contributes to the happiness of others.
Answer:
non argumentive ad will most likely agree
Explanation:
In poodles, black fur is dominant to white fur. A black poodle is crossed with a white poodle. In a litter of four, all of the puppies are black. What is the BEST conclusion?
Answer:
The black poodle is true breeding (homozygous) for the black fur colour.
Explanation:
Let the allele for fur colour be represented by B.
Black fur (B) is dominant over white fur (b).
Black poodle (B _) is crossed with white poodle (bb):
B _ x bb
Progeny (all black) = Bb, Bb, _b and _b.
Since all the progeny are black, _b = Bb (remember, B is dominant over b)
Hence, B_ is also BB.
The best conclusion is that the black poodle is true breeding (homozygous) for the black fur colour.
Answer:
Explanation: all the puppies in the F1 generation are black because homozygous black (BB) is crossed with a homozygous white(bb) and since the black fur is dormant all the puppies will be black ie heterozygous black (Bb).
The recent hominin fossil finds from Ileret, Kenya, negate the conventional view held since 1960 that H. habilisand H. erectusevolved one after the other. Instead, they lived side by side in eastern Africa for perhaps half a million years.A. TrueB. False
Explanation:
The recent finding of the fossils which showed that Homo erectus and Homo habilis lived side by side in eastern Africa for perhaps half a million years challenged the conventional way that these two species evolved one after the other(H.habilis 1.44 million years old and H.erectus 1.55 million years old)
The fossils were found in Kenya and took years to prepare the specimens for study and to be sure of the identification of the species, the scientists saidUniversity of Utah geologists determined the dates of the fossils from volcanic ash depositsThe most recent Homo habilis that had been known was about the same age as the earliest Homo erectus, said Daniel Lieberman, a professor of biological anthropology at Harvard University, “Now we have extended the duration of the habilis species, and there’s no doubt that it overlaps considerably with erectus”The fact that the two hominid species lived together in the same lake basin for so long and remained separate species, Meave Leakey said in a statement from Nairobi, “suggests that they had their own ecological niche, thus avoiding direct competition”Fungi and plants form mutually beneficial symbiotic relationships called mycorrhizae. Classify the resources as a. supplied by the fungus or the plant. b. Supplied by fungus. c. Supplied by fungus d. Supplied by plant.
Answer:
Mycorrhiza is a symbiotic association in between the green plants and fungus.
The micorrhiza are located in the roots of the vascular plants. The Green plants performs the process of photosynthesis and the sugars produced are supplied to the Micorrhizas.
The fungus supplies the nutrients, water such as phosphorus that is absorbed from the soil to the plant body.
Answer:Answer:
Mycorrhiza is a symbiotic association in between the green plants and fungus.
The micorrhiza are located in the roots of the vascular plants. The Green plants performs the process of photosynthesis and the sugars produced are supplied to the Micorrhizas.
The fungus supplies the nutrients, water such as phosphorus that is absorbed from the soil to the plant body.
Explanation:
Answer:
Mycorrhiza is a symbiotic association in between the green plants and fungus.
The micorrhiza are located in the roots of the vascular plants. The Green plants performs the process of photosynthesis and the sugars produced are supplied to the Micorrhizas.
The fungus supplies the nutrients, water such as phosphorus that is absorbed from the soil to the plant body.
1. List the cellular structures over which an action potential travels, starting at the dendrites and traveling to where neurotransmitter molecules are released.
Answer:
A neuron or a nerve cell is a unique cell, which performs an essential function of transmitting the nerve impulse in the form of action potential from one nerve cell to another. The point of communication between two neurons is known as a synaptic junction from where the transmission of a signal between presynaptic axon end and the postsynaptic dendrite occurs.
The synapse present in the dendrite receives a signal in the form of neurotransmitter from the presynaptic axon. This results in the formation of an action potential that gets transmitted towards the cell body of a neuron. From the cell body, the conduction of impulse takes place via a long tubule composition known as an axon, which constitutes the nodes of Ranvier and reaches the nerve endings or the axon terminals.
In the terminals, the change in potential results in the opening of the ions channels that discharge neurotransmitters in the form of acetylcholine, which further combines with the ligand-receptor situated on the next dendrite and thus repeats the process. Thus, the action potential travels from the dendrite to the cell body, and from there it travels to the terminals of the axon and eventually towards the ion channels and the ligand-receptor.
b) WRITE a brief (between 150-200 words) email to a scientific colleague that explains (IN YOUR OWN WORDS) how the cholera bacterium causes diarrhea AT THE CELLULAR LEVEL and what role OSMOSIS plays in this.
Answer:
The bacterium Vibrio cholerae is the primary cause of cholera disease that mainly infects the small intestine and primarily leads to the dehydration of the body.
Explanation:
The genetic analysis reveals that the aforementioned causative bacteria surpass the acidic conditions of the stomach and eventually reaches the intestinal wall and attaches to it. This is followed by the production of toxic protein by the bacterium. This protein is taken inside the cell via receptor mediated endocytosis followed by its binding to the host protein Arf6. This binding leads to the production of cAMP that results in the dehydration process. This mechanism leads to excessive accumulation of chloride ion in the intestine preventing the entry of sodium ion.
These two ions are associated with the creation of water-salt environment in the intestine that leads to tremendous diarrhea via the process of osmosis.
Hence, we can say that cholera bacterium affects the individuals at the cellular level and osmosis plays a vital role in the process.
You are observing a specimen under the microscope. You have located a cell and brought it into sharp focus using the 10x objective (below left). However, when you switch to the 40x objective, the cell is no longer visible (below right)! What has happened and how would you correct this? [2 pt; G5.2]
Answer:The cell was no longer visible because the image wasn't centered before switching to higher objective of 40X.
Explanation:
While using microscopes, the image will approximately remain in focus if you adjust the focus during change of magnification. Therefore, there is need to always adjust the focus during change in magnification.
If object is not centered before switching to a higher power objective, object will not be visible.
To avoid this, center the object before switching to a higher power objective. This will help you find the object after switching the objective. And it is also important to use fine focus and not coarse focus.
Answer:
When you shift from low power(10x) to high power(40x), the 40X objective lense moves directly over the specimen and this in turns results in the following changes:
•increase in the magnification of a specimen
•decrease in the light intensity which makes the image appears dim
•decrease in the area of field of view, i.e the specimen will appear larger
•decrease in depth of field, i.e. you won't be able to view an entire surface of the specimen.
And the only way to correct this is to shift back to 10X,centre the specimen and then again switch back to 40X
1. True/False: DNA replication is considered "semi-conservative" because only half of the chromosomes are copied during S phase of interphase.
Group of answer choices
True
False
2.The monomers of DNA consist of
Group of answer choices
a. deoxyribose, phosphate, and a nitrogenous base
b. deoxyribose, phosphate, and hydrogen bonds
c. ribose, phosphate, and nitrogenous bases
d. ribose, phosphate, and hydrogen bonds
e. none of the above are correct
Answer:
1) The statement is false
2) Option A) deoxyribose, phosphate, and a nitrogenous base
Explanation:
1) In semi-conservative DNA replication:
- a parent double-stranded DNA splits in two.
- Each strand is then read by the enzyme, DNA polymerase, to ensure accurate synthesis of a new daughter strand
- the newly synthesized strand contains nucleotides that are complimentary to free nucleotides present in the parent strand.
Thus, because the parent strand is retained in the newly synthesized DNA, DNA replication is described as semi-conservative
2) DNA consists of several repeating units of polydeoxynucleotides where each is made up of a nitrogenous base (Adenine, thymine, cytosine or guanine) linked to
deoxyribose sugar by an N-glycosidic linkage, and then the sugar linked to a phosphate group by phosphoester bond.
Answer: 1. True.
Option A. deoxyribose, phosphate and nitrogenous base.
Explanation:
DNA replicates in semi conservative because when a parent helix replicates, it produce two daughter helixes which contain one of the two helical strand of the parent. During S phase in interphase, DNA is unwind by enzyme helicase and it is duplicated. The two single strands is use as a template to form identical strands.
2. The monomers of DNA are also called nucleotides. They include 5 carbon sugar(deoxyribose, Nitogenous base and phosphate group.
Which of the actions below is preformed by all cells to maintain homeostasis
A.They divide rapidly into many specialized cells
B.They use light energy to produce food molecules
C.They obtain energy from nutrients and remove waste products
D.They combine with another cell to increase in size
Answer: Option C.They obtain energy from nutrients and remove waste products
Explanation:
Homeostasis is the maintenance of a stable internal environment. Hence, cells allow processes within its cytoplasm that produces energy by degrading macromolecules like carbohydrates or lipids while expelling by-products such as gases, salts etc out of its cell membrane.
Answer:C
Explanation: they maintain a balance between the nutrients obtained from food and waste product generated during digestion of food.
Carlita was hired today for a job as a laboratory assistant. She will work for professor of molecular biology who wants to make new discoveries about genes, DNA , RNA, and their functions. Carlitas work will most likely contribute to which of these specific research goals? PLEASE HELPPP HURRYYY
Answer: Option C) Determining the link between genes and the traits they influence.
Explanation:
First, the field of molecular biology seeks to understand the relationship between genes and proteins. Now, since genes are specific sequences on the DNA that codes for functional proteins, and these proteins make up the various traits physically expressed in living organisms such as skin color, blood group etc.
Then, the specific research goal of Carlita would be determining how genes control the expression of individual traits at the molecular level
Answer: C
Explanation:
If Harry's negligent act injures Sally, and Susan, while attempting to come to Sally's aid breaks her arm in the process then, Harry is liable for the harm to __.
Answer: Sally
If Harry's negligent act injures Sally, and Susan, while attempting to come to Sally's aid breaks her arm in the process then, Harry is liable for the harm to Sally.
Explanation:
As a mathematical expression, we can say the action of Harry H, is directly proportional to the injury to Sally S; while the mistake of Susan N is directly proportional to injury to Sally S.
So, if H = S and N = S, it is safe to say
H = N = S.
Thus, Harry negligence create room for Susan mistake, which eventually harmed Sally.
In a population of 3000 fruit flies, 270 of them contain white eyes. White eye color is a recessive trait. What are the allelic frequencies for the red eye allele and white eye allele
Answer:Red is 91%, white is 9%
Explanation:3000-270 =2730 red in percentage is 2730/3000 *100/1 =91%
White is 270/3000*100/1= 9%.
Allele frequency or gene frequency is the relative frequency of an allele expressed as a fraction or percentage.
Explain why the underproduction of albumin by a cirrhotic liver contributes to excessive filtration, which leads to ascites. Refer again to the forces that determine net filtration pressure in the hepatic capillaries, and to the function of albumins
Answer:
Albumin is produced mainly by the liver. Albumin is the major protein that is acts as the carrier protein for steroids, thyroid hormones and fatty acids in the blood.
Albumin stays in blood vessels as they are too large to move across the capillary walls, in blood vessels by contributing to osmotic pressure or colloid osmotic pressure that maintain water volume. So, the albumin is the protein that is involve primarily on the water reabsorption.
Suppose you want to know the proportion of A_B_ among the total progeny, you may first estimate which genotypes represent A_ and which genotypes represent B_. Clearly AA and Aa for the first genotype A_.
Answer and Explanation:
Complete question:
"What is the expected proportion of progeny with the following broad genotypes each having a distinct phenotype in a typical Mendelian dihybrid cross:A_B_ : A_bb : aaB_ : aabb. Be able to derive the above ratios directly by studying alleles for one gene at a time (individual Punnett) and applying the product rule.
Example: (1 AA: 2Aa: 1aa) (1BB : 2Bb: 1bb) Suppose you want to know the proportion of A_B_ among the total progeny, you may first estimate which genotypes represent A_ and which genotypes represent B_. Clearly AA and Aa for the first genotype A_. And BB and Bb for the second genotype".
Answer:
In a typical Mendelian dihybrid cross, we have:
Parental) AaBb x AaBb
Gametes) AB Ab aB ab
AB Ab aB ab
Punnet Square) AB Ab aB ab
AB AABB AABb AaBB AaBb
Ab AABb AAbb AaBb Aabb
aB AaBB AaBB aaBB aaBb
ab AaBb Aabb aaBb aabb
F1 phenotypic frequencies)
9/16 A-B-
3/16 aaB-
3/16 A-bb
1/16 aabb
But if you do not want to do the dihybrid cross, you might just perform the cross for each gene and then apply the product rule. This is:
Hybrid Cross for gene A)Parental) Aa x Aa
Gametes) A a A a
Punnet Square) A a
A AA Aa
a Aa aa
F1 genotypic frequencies)
1/4 AA
2/4 Aa
1/4 aa
F1 phenotypic frequencies)
3/4 A-
1/4 aa
Hybrid Cross for gene B)Parental) Bb x Bb
Gametes) B b B b
Punnet Square) B b
B BB Bb
b Bb bb
F1 genotypic frequencies)
1/4 BB
2/4 Bb
1/4 bb
F1 genotypic frequencies)
3/4 B-
1/4 bb
To estimate the progeny phenotypic proportions (A-B-, aaB-, AAB-, aabb), from the genotypic proportions of progeny, you can sum the genotypic frequency of homozygote plus heterozygote, to get the proportions A- and B-.
AA=1/4Aa=2/4
A- = AA+Aa= 1/4 + 2/4 = 3/4
BB= 1/4Bb= 2/4
B- = BB + Bb = 1/4 + 2/4 = 3/4
Then multiply these proportions to get the progeny proportions, like this:
A-B-= (AA+Aa) x (BB + Bb) = 3/4 x 3/4 = 9/16
A-bb = (AA + Aa) x bb = 3/4 x 1/4 = 3/16
aaB- = aa x (BB + Bb) = 1/4 x 3/4 = 3/16
aabb = aa x bb = 1/4 x 1/4 = 1/16
Note: You can get the same results using directly the phenotypic proportions of each cross.
Starting with a population of genetically identical mice, you discover two new independent mutant strains in which all of the animals have epileptic seizures. In both strains, you know that the epileptic seizures are due to a single DNA mutation. You cross a mutant mouse from one strain to a mutant mouse from the second strain and find that none of their offspring undergo spontaneous seizures. From this experiment you would conclude that the two mutant strains of mice most likely have mutations in:
a-the same DNA base position within a particular gene
b-the same gene, but not necessarily the same DNA base position.
c-two different genes.
Answer:
B - the same gene, but not necessarily the same DNA base position. They do not show seizures when they are crossed, but show seizures in independent populations.
When they are not in the same position and hence they do not come together to express the disease.
Thus, the answer is B Explanation
a. If they are in the same DNA base positions,, their offspring might have expressed epileptic seizures.
c. If they are from different genes it would have interfered with the trait.
Final answer:
The two mutant strains of mice most likely have mutations in two different genes, as inferred from the observation that their offspring did not exhibit epileptic seizures, suggesting compensatory heterozygosity.
Explanation:
When a mutant mouse from one strain is crossed with a mutant mouse from another strain and the offspring do not exhibit epileptic seizures, it suggests that the mutations are likely in two different genes. This conclusion comes from the understanding that if the mutations were in the same gene (whether at the same DNA base position or different ones), the offspring would more likely show the dominant phenotype, in this case, seizures - if the seizure trait was a result of a dominant allele.
Since none of the offspring have seizures, this indicates that the mice are likely heterozygous for a recessive seizure allele from each parent, and thus each mutation is compensating for the other in a complementary fashion, which is consistent with the mutations being in two different genes.
n corn long ears (L) is dominant to short ears (l); glossy kernels (G) is dominant to opaque kernels (g) and high starch (S) is dominant to normal starch (s). A heterozygous plant with long ears, glossy kernels and high starch is crossed with a short eared plant with opaque kernels and normal starch. The following phenotypes are observed. Which genes are linked? (Gene name is based on the dominant trait.)
The question is incomplete. I have attached the complete question
Answer:
Ears and starch are linked, glossy isn't linked to either
Explanation:
The deviation from expected frequencies indicates that ears and starch genes are linked
You are about to give your first speech in front of your speech class and are nervous about getting up in front of everyone. As you anticipate getting up in front of the class, you begin to feel butterflies in your stomach and a pounding in your chest. Using this example, explain how the sympathetic and parasympathetic divisions of the autonomic nervous system work together to help your body react to this situation.
Answer:
Sympathetic division is responsible for the nervousness, pounding in the chest and the butterfly movement in the stomach while parasympathetic division overest a and digest the nervousness.
Explanation:
The nervousness experienced when someone is about to speak in front of a class is caused by the sympathetic system which have divergent effects as many different effector organs are activated together for that same purpose. I.e More oxygen are inhaled and delivered to skeletal muscle. The respiratory, cardiovascular, and musculoskeletal systems are all activated at the same time causing an unstable reaction to the central nervous system...
On the other hand, the parasymthetic division causes the central nervous system to rest and digest at a very slow rate..
At a point in time the central nervous system triggered an Homeostasis action which posses a balance between the two divisions.I.e symthetic and parasymthetic and then brings about a balance to the body reaction at that same point.
The sympathetic and parasympathetic divisions of the autonomic nervous system work together to help your body react to the stress of giving a speech. The sympathetic division triggers the fight-or-flight response, causing the pounding in your chest and butterflies in your stomach. The parasympathetic division helps to calm your body down after the event.
Explanation:When facing a nerve-wracking situation like giving a speech, the sympathetic and parasympathetic divisions of the autonomic nervous system work together to help your body react. The sympathetic division is responsible for your body's fight-or-flight response, which increases heart rate and causes the release of adrenaline. This leads to the pounding in your chest and the feeling of butterflies in your stomach.
The parasympathetic division then helps to calm your body down after the stressful event, bringing heart rate and breathing back to normal.
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Based on your results in Part A, which of the following statements most appropriately reflects the relative effectiveness of the vancomycin treatment, teixobactin treatment, and the control?
Teixobactin reduced the number of MRSA colonies about as effectively as did vancomycin relative to the control.
Neither teixobactin or vancomycin were effective at reducing the number of MRSA colonies relative to the control.
Teixobactin was significantly more effective than the vancomycin at reducing the number of MRSA colonies relative to the control.
Teixobactin was significantly less effective than the vancomycin at reducing the number of MRSA colonies relative to the control.
Answer:
Option C is correct,which states that Teixobactin was significantly more effective than the vancomycin at reducing the number of MRSA colonies relative to the control.
Teixobactin reduced the number of MRSA colonies about as effectively as vancomycin relative to the control. Statement 1
Explanation:The most appropriate statement reflecting the relative effectiveness of vancomycin treatment, teixobactin treatment, and the control is:
Teixobactin reduced the number of MRSA colonies about as effectively as did vancomycin relative to the control.
This means that both vancomycin and teixobactin were similarly effective in reducing the number of MRSA colonies compared to the control group. Therefore, teixobactin was not significantly more or less effective than vancomycin.
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