Answer:
POSITIVE CONTROL
c) Active Activator
d) Active Repressor
NEGATIVE CONTROL
f) Inactive repressor
a) Inactive Activator
b) cAMP
Explanation:
The positive controls are those groups in the experiment whose treatments are expected to confirm previously known results, thereby enabling the comparison of these results with the target group. In this case, both active activators and repressors are able to give results that can be compared with the test group
The negative controls are those experimental groups where no response is expected. In this case, both inactive activators and repressors, and cAMP (which is a secondary messenger in diverse biological processes but is not involved in this pathway), are not expected to produce any measurable response.
Where do sperm mature?
O A. Epididymis
O B. Urethra
O c. Seminiferous tubules
O D. Vas deferens
Answer:
A.
Explanation:
Sperm migrate from the seminiferous tubules to the epididymis.
Consider this plant cell.
The organelles in a plant cell are labeled. Part E is a complex of vesicles and folded membranes within the cytoplasm.
Which organelle is labeled E?
Golgi apparatus
chloroplast
ribosome
nucleus
Mark this and return
The correct answer is A. Golgi apparatus
Explanation:
In a plant cell, the cytoplasm contains organelles such as the chloroplast, the ribosome, the Golgi apparatus, among others. From these organelles, the Golgi apparatus is the one that is composed of vesicles and folded membranes, which makes it look like a system of flattened, long, and folded sacs or membranes connected. Additionally, this organelle is in charge of process proteins and lipids by packaging and modifying these before they go to other sections or organelles in the cell. According to this, the organelles labeled E is likely the Golgi apparatus because only this includes a set of vesicles and folded membranes.
Answer:
A. Golgi apparatus
Explanation:
Species of bacteria can evolve more quickly than
species of mammals because bacteria have
Answer:
540MIL0IAER
P
Explanation:
F
VFVVEV
Answer:
Bacteria have smaller populations, grow faster, and also share more genes compared to mammals.
Explanation:
Because bacteria reproduce quicker and are also in smaller populations, they can evolve quicker because more genes are shared which can be helpful. This is due to natural selection, which will affect bacteria more than mammals because their small populations will make natural selection more significant to be harmed by mutations and allow them to evolve.
A study from the National Institutes of Health states that the human body contains trillions of microorganisms that make up 1% to 3% of the body's mass. Use this information to estimate the average mass of a microorganism.
Answer:
If the mass of the human body is 60 kg so the average mass of microorganism is 1.2 kg.
Explanation:
If we suppose the mass of human body is 60 kg and we know the percentage that is 1 to 3% so we will take the average microorganism mass of which is 2%. So by doing multiplication of 60 with 2 and then divided by 100, we get 1.2 kg. The reason for dividing by 100 is that the average mass present in percentage form so for converting the percentage into standard form we have to divide it by 100. So we conclude that in every human with a mass of 60 kg have 1.2 kg microbes present in their body.
_________16. Which one of the following should NOT be associated with electron transport chain?
A. Absorption of solar energy C. Movement of H+
B. Formation of ATP D. Cytochromes
which organ of the body doesn't rest?
Answer:
Brain
Explanation:
It is always active, and regulates everything going through your body. Even when your asleep, it makes sure that the body is in the right temp, and that all the organs help stay safe.
Answer:
The brain!
Explanation:
The body rests during sleep, not the brain. The brain remains active, gets recharged, and still controls many body functions!
a particular gene can have two forms, called alleles. which statement best describes the difference between the dominant allele and the recessive allele?
(a)only the recessive allele is expressed when both alleles ate inherited.
(b)the recessive allele can be expressed only if it moves to a different chromosome.
(c)only the dominant allele is expressed when both alleles are inherited.
(d)the dominant allele can be expressed only if two homologous chromosomes have it.
Answer:
The answer is D I'm pretty sure
A scientist discovers a cell that has chloroplasts, cytoplasm, DNA, and a cell membrane. Which statement best describes how
the cell could be classified?
Answer: The cell is eukaryotic because it has chloroplasts. A scientist is studying a cell and can clearly see that it has ribosomes and mitochondria.
Explanation:
Answer:
The cell is eukaryotic because it has chloroplasts. A scientist is studying a cell and can clearly see that it has ribosomes and mitochondria.
Explanation:
Just took the test 2020
A botanist has acquired a group of pea plants. All of the pea plants have yellow pea pods (the recessive form of this trait) except for one, which has green pea pods (the dominant form of this trait). The botanist decides to use a test cross to determine the genotype of the green pea pod plant for this trait. The botanist performed the test cross and found the following: 50% of the offspring had yellow pea pods and 50% of the offspring had green pea pods. Based on this information, what was the genotype of the initial green pea pod plant
Answer:
Hetezygous
Explanation:
The genotype of the initial green pod plant would be heterozygous.
Let pea pod color be represented by C (c) alleles.
First, let us assume that the genotype of the green pea pod plant is homozygous dominant, CC. This was crossed with homozygous recessive, cc.
CC x cc
Cc Cc Cc Cc
All the progeny will have Cc genotype with phenotypically green color. This is in contrast to the result obtained from the test cross in the illustration, hence the genotype of the initial green pea pod plant cannot be homozygous dominant.
Now, let us assume that the genotype is Heterozygous, Cc.
Cc x cc
Cc Cc cc cc
50% of the progeny has Cc genotype with phenotypically green color while the remaining 50% has cc genotype with phenotypically yellow color. This is consistent with the result from the illustration.
Hence, the genotype of the initial green pea pod plant is heterozygous.
2. The part of brain that is called ________ plays an important role in the consolidation of memories.
Answer:
Explanation:
The principal player in memory consolidation research, in terms of brain regions, is the hippocampus.
hope it helps
how does the nonpolar nature of lipids contribute to their insulating quality
SCENARIO ABiological anthropologists have discovered a previously unknown fossil species namedAustralopithecus sediba. The species lived about 2 million years ago in Africa and has an interesting mix of traits. For example, its arms are adapted for climbing in trees and its legs are adapted for walking on two legs on the ground. It does not directly resemble any of the living ape species, which suggests that the living ape species (including humans) have each become adapted for their own environmental contexts over time.What is the primary type of biological anthropology addressed in this research?A. Human biologyB. Forensic anthropologyC. PaleoanthropologyD. PrimatologyDoes the research also touch on topics that might be relevant to researchers in disciplines outside anthropology, such as biology, geology, psychology, medicine, sociology, or criminology? Which of the following disciplines would be most relevant?A. PsychologyB. BiologyC. CriminologyD. SociologyHow does this research relate to human evolution? In other words, what can we learn about human evolution from research along these lines?
Answer:
In scenario A A. Human biology
Which of the following disciplines would be most relevant?
B. Biology
Explanation:
In scenario A Australopithecus sediba arms are adapted for climbing in tree and its legs are adapted for walking on two legs. Australopithecus is related to homo habilis and also homo erectos. They are different human primate.
B. Biology because there are fossil remains dated to about 2 million years ago.
is heat involved in producing all three classes of rock.
Answer: yes. metamorphic, sedimentary, and igneous
Explanation:
Sedimentary and igneous rock is transformed into metamorphic rock inside the Earth by heat, pressure, and melting, so heat is involved in producing all three classes of rock.
What are the classes of rock?From eroding parent material, sedimentary rock is created. The source material may be sedimentary, metamorphic, or igneous in origin.
Strong heating causes hot liquid rock (magma) to break through the Earth's surface and solidify as igneous rock. This rock is worn and eroded throughout time, restarting the cycle.
The existing rock must stay solid and not melt in order to produce metamorphic rock. The rock will melt and turn into the lava if there is too much heat or pressure. An igneous rock, not a metamorphic rock, will develop as a result of this. Think about the shape changes in granite.
Therefore, sedimentary and igneous rock is transformed into metamorphic rock inside the Earth by heat, pressure, and melting.
Learn more about rock, here:
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Industrial melanism refers to the dark pigmentation that evolved in some insects giving them protective coloration on vegetation darkened by soot in heavily industrialized areas prior to air pollution regulation. Assume that in one heavily polluted area near Birmingham, England in 1956, 79% of the moths of the species Biston betularia were black due to the presence of a dominant gene for melanism. Estimate the frequency of the dominant allele in this population, and the proportion of black moths that are heterozygous.
Answer:
The frequency of the dominant allele, p = 0.542The proportion of black moths that are heterozygous 2pq = 0.496Explanation:
According to Hardy-Weinberg, the allelic frequencies in a locus are represented as p and q, referring to the allelic dominant or recessive forms. The genotypic frequencies after one generation are p² (Homozygous dominant), 2pq (Heterozygous), q² (Homozygous recessive). Populations in H-W equilibrium will get the same allelic frequencies generation after generation. The sum of these allelic frequencies equals 1, this is p + q = 1.
In the same way, the sum of genotypic frequencies equals 1, this is
p² + 2pq + q² = 1
Being
p the dominant allelic frequency,q the recessive allelic frequency,p² the homozygous dominant genotypic frequencyq² the homozygous recessive genotypic frequency2pq the heterozygous genotypic frequencyIn the exposed example, 79% of the moths of the species Biston betularia were black due to the presence of a dominant gene for melanism.
If the genotypic frequency of back moths is 0.79, then, by performing the following equation we can get the not-black moths genotypic frequency:
p² + 2pq + q² = 1
where p² is the homozygous dominant genotypic frequency, q² the homozygous recessive genotypic frequency, and 2pq is the heterozygous genotypic frequency.
As 0.79 is the phenotypic frequency of black moths, then this frequency equals p²+2pq.
Clearing the equation:
p² + 2pq + q² = 1
0.79 + q² = 1
q² = 1 - 0.79
q² = 0.21
The genotypic frequency of non-black moths is 0.21. So, from here we can calculate the allelic frequency:
q² = 0.21
q= v 0.21
q = 0.458
If 0.46 is the allelic frequency of non-black moths, then by clearing the equation p + q = 1, we can get the p allelic frequency:
p + q = 1
p + 0.458 = 1
p = 1 - 0.458
p = 0.542
The genotypic frequency p² = (0.542)² = 0.294The heterozygote genotypic frequency2 x p x q = 2 x 0.542 x 0.458 = 0.496
Finally, we can check this answer by clearing the following equation:
p² + 2pq + q² = 1
0.294 + 0.496 + 0.21 = 1
Fat mobilization is the breakdown of fats in adipose tissue into metabolic products that enter the bloodstream place the steps of fat mobilization in order
Actual Answer:
1) body releases epinephrine
2) epinephrine binds to fat cells in adipose tissue
3) tricylglycerols are hydrolyzed to glycerol & fatty acids
4) metabolic products enter the bloodstream
Answer:
The complete steps are:
1. Body needs energy for activity
2. body releases epinephrine
3. epinephrine binds to fat cells in adipose tissue
4. triacylglycerols (triglycerides) are hydrolyzed to glycerol and fatty acids
5. Fatty acids transported to tissue
Explanation:
Mobilization of fat is a gluconeogenetic process (a process that sequesters energy from a compound other than carbohydrates), in which fat molecules in the adipose tissues are metabolized for energy production.
The process starts when the body is in an energy-deprived (hypoglycemic) state, leading to the release of glucagon by the pancreas and the release of epinephrine (adrenaline) from the adrenal medulla of the brain.
These hormones activate gluconeogenesis and glycogenolysis (breakdown of glycogen) in the liver, it also activates lipolysis in the muscle cells and liver, as well as inhibiting fatty acid synthesis. This is mediated through the binding of these hormones to specific receptors on the surface of fat cells, leading to the activation of lipases (enzymes catalyzing lipolysis). Lipases catalyze the hydrolysis of triglycerides (common fat in diets) to glycerol and three molecules of free fatty acids. These free fatty acids are released into the blood where they attach to serum albumin; a protein for transporting the hydrophobic fatty acids to the muscle cells where they are taken up and utilized to produce energy and C0₂.
What are the fundamental parts of a typical cell
Answer:
the parts include Cell Membrane, Mitochondrion, Lysosome, Rough Endoplasmic Reticulum.
Explanation:
the cell membrane Is made out of phospholipids and proteins
the Mitochondrion Site of cellular respiration "power house"
the Lysosome S.uicide Sacks that contain digestive enzymes
rough Endoplasmic Reticulum contains Ribosomes, transports proteins
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You are monitoring the metabolism of two different cultures of the same species of yeast, which you have labeled Culture A and Culture B. One of your observations is that Culture A is using up about 10 times the amount of glucose per unit of time as Culture B. Which one of the following conclusions could you reasonably draw regarding these cultures?
A. Culture A would be accumulating lactic acid.
B. Culture A is being grown aerobically; Culture B is being grown anaerobically.
C. Only Culture A is recycling its NADH back to NAD+.
D. Culture A and Culture B are synthesizing the same amount of ATP per unit of glucose in glycolysis.
E. None of the above conclusions could be drawn from the information given.
Answer: D
Explanation: The experiment was set up using the very same species of yeast labeled differently. What this means is that they are of the same type or kind and thus would exhibit similar features from feeding to the metabolism of the food they take and subsequent production of by-products of the fermentation process. In this regard, therefore, the reasonable conclusion that can be drawn regarding these cultures is that both cultures A and B are synthesizing the same amount of ATP per unit of glucose in glycolysis (the cellular degradation of the simple sugar glucose to yield pyruvic acid, and ATP as an energy source).
What physiological alterations occur at both the vasculature and heart functioning level? (really stuck on this one, please help.)
Answer:
Most physiological alterations during aging and it is usually characterized by the deposition of a pigment called liposfuscin.
The valves of the heart also becomes stiff and thickened as a result of the pigment mentioned above. This makes the flow of blood slower than normal and contributes to cardiovascular diseases such as Stroke, high blood pressure etc.
