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Answer:
I posted some pictures, I hope it helps!
Related Questions
g The greater tubercle of the humerus is analog to the _______________. Group of answer choices lesser tubercle of the humerus lesser trochanter of the humerus greater trochanter of the femur lesser trochanter of the femur
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Answer:
greater trochanter of the femur
Explanation:
Analogous structures refer to those that are similar morphologically and even have a similar function in the organism, but their genetic origin differs.
The origin and embryological development of these organs are not the same.
Both structures are muscles insertion points or surfaces. These muscles are related to the extremity movement.
Which of the following statements is false? Choose one: A. Some Bcl2 family members promote apoptosis, whereas others inhibit apoptosis. B. Bax and Bak are death-promoting members of the Bcl2 family that induce the release of cytochrome c from mitochondria into the cytosol. C. Some death-inhibiting members of the Bcl2 family inhibit apoptosis by blocking cytochrome c release from mitochondria. D. The death-promoting members of the Bcl2 family include Bcl2 itself.
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Answer: The statement that is FALSE is D (The death-promoting members of the Bcl2 family include Bcl2 itself.)
Explanation:
Apoptosis is a process found in multicellular organisms whereby cells follows a order of events which eventually leads to their death without releasing harmful substances into the surrounding area of the cells. Through this process, old cells, unnecessary cells, and unhealthy cells are eliminated from the body.
The process of apoptosis can be regulated by cell regulator proteins such as Bcl-2 family.
Bcl-2 stands for B-cell lymphoma 2 which are encoded in human cells by BCL2 gene. The Bcl-2 family are capable of regulating apoptosis by either inducing it or inhibiting it.
Bcl-2 and Bcl-XL, PREVENT apoptosis by preventing the release of mitochondrial apoptogenic factors such as cytochrome c and AIF (apoptosis-inducing factor) into the cytoplasm
Part of the Bcl-2 family is the Bax and Bak and they are the core regulators of the intrinsic pathway of apoptosis. They PROMOTE apoptosis by inducing the release of cytochrome c from mitochondria into the cytosol. Therefore, the death-promoting members of the Bcl2 family does not include Bcl2 itself.
function of graffian follicle
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Answer:
It provides for the maturation and release of a fertilizable oocyte
Explanation:
let me know if you have any questions
6. Compare the different measurements of the thermometer which are used.
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There are many types of contact and contact-free thermometers to choose from. The ages of your household members, as well as personal preference, can help you determine which types to buy.
Like any other device in your home, it’s important to understand how your thermometer works. Not every type works the same way or is designed to provide the same temperature readings.
Erythrocytes have no nuclei. Without nuclei, erythrocytes can carry a higher amount of haemoglobin explain how this is an advantage in exchanging and carrying oxygen
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Answer:
The correct answer is - hemoglobin binds with respiratory gases such as oxygen so more hemoglobin will carry more oxygen .
Explanation:
Red blood cells have adapted the trait of having no nucleus for several reasons but the major reason is to increase the amount of hemoglobin so this can carry more oxygen to various parts of the body from lungs. If there will be more hemoglobin then it will allows RBC to transfer more oxygen.
Hemoglobin is the respiratory pigment that has four heme group, containing iron ion, arround globin group, each iron ion in the heme can bind to one oxygen molecule; So one molecule of hemoglobin can transport four oxygen molecules as it binds with oxygen.
Identify the energy carrier molecule ATP and its importance
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is the citric acid cycle the same thing as fermentation
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Answer:basically
Explanation:
How can algal blooms be
harmful to an aquatic
environment?
A. by blocking all sunlight and killing the bottom plants
causing no oxygen to go into the water
B. by putting oxygen up into the air for the
surrounding plants
C. by absorbing the phosphorus in the atmosphere and
having a symbiotic relationship with cyanobacteria
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Answer:
(A) By blocking all sunlight and killing the bottom plants
causing no oxygen to go into the water
You prepare a gel mobility assay with the following samples:
Lane 1 Radiolabeled lacO DNA + Lac Repressor protein preincubated with allolactose
Lane 2 Radiolabeled lacO DNA + Lac Repressor protein preincubated with lactose
Lane 3 Radiolabeled lacO DNA + Lac Repressor protein missing its DNA binding domain
Lane 4 Radiolabeled lacO DNA + Lac super-repressor (Is) preincubated with allolactose
Lane 5 Radiolabeled lacO DNA + lac repressor protein
On which lanes do you expect to see two bands?
a. 3
b. Lane 1 and 2
c. Lane 2, 4, and 5
d. 2 and 4
e. 1 and 5
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Answer:
c. Lane 2, 4 and 5.
Explanation:
Two bands will be seen when a fraction of DNA added to gel is bound by Lac repressor protein. Radiolabeled LacO DNA will have two bands. Lac repressor is encoded with lacl gene. It has binding ability due to allolactose formation.
What role does mutation play in the evolution of a species and the development of new characteristics?
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Answer:
Mutation plays a vital part in evolution; it is the greatest source of genetic variation. In evolution, it creates a new DNA sequence for a particular gene, creating a new allele. Additionally, recombination can produce a DNA sequence (a new allele) for a specific gene through intragenic recombination.
Explanation:
Hope this helps!
what is cyclic and noncyclic photo-phosphorylation
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Explanation:
The photophosphorylation process which results in the movement of the electrons in a cyclic manner for synthesizing ATP molecules is called cyclic photophosphorylation. ... During cyclic photophosphorylation, the electrons are transferred back to P700 instead of moving into the NADP from the electron acceptor
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How does energy acquisition in the deep sea differ from energy acquisition near the ocean’s surface?
a.
Organisms in the deep sea acquire energy directly from the sun.
b.
Organisms near the ocean’s surface rely on chemosynthesis rather than photosynthesis.
c.
Organisms in the deep sea do not have direct access to sunlight.
d.
Organisms in the deep sea have no means to obtain energy unless they travel to the ocean’s surface.
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Answer:
Organisms in the deep sea acquire energy directly from the sun.
in some plants pink flowers are dominant over yellow flowers if Mario crosses two hybrids what happens
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What is photosynthesis????
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Explanation:
The process of green plants make their own food by present sunlight is called photo synthesis
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Photosynthesis is the reaction that converts light energy to chemical energy in sugar and carbohydrates. Humans can't eat sunlight, so we eat plants. Plants convert the energy in the sunlight into chemical energy using chlorophyll. Chlorophyll is a molecule produced by plants, algae, and cyanobacteria, which aids in the conversion of light energy into chemical bonds. It is the green pigment found in the chloroplasts of higher plants. They are a part of our ecosystem, and they are lower on the food chain.
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Have a great day! :)
Explain in detail what caused the population of peppered moths to change from mostly light-colored variations to mostly dark-colored variations in the 1800s. Explain how the peppered moth mutation is an example of natural selection. Explain what would happen to the populations of both variations of peppered moth once the pollution disappeared. Determine which variation of the peppered moth would be more successful in your neighborhood, and explain in detail your reasoning supporting your conclusion. Predict what may happen to the population of peppered moths if they were relocated to a lush, green rainforest ecosystem. How would this impact natural selection and their struggle to survive?
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Answer:
Natural selection caused the population of peppered moths to change from mostly light-colored variations to mostly dark-colored variations in the 1800s. Pepper colored moths camouflaged among trees but after industrial revolution the couldn't because of soot, which camouflaged black moths; hence, they were preyed upon more.
From a chemical view, how is an amino acid is being recognized by its specific aminoacyl tRNA synthetase?
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Answer:
An aminoacyl-tRNA synthetase (aaRS or ARS), also called tRNA-ligase, is an enzyme that attaches the appropriate amino acid onto its corresponding tRNA. It does so by catalyzing the transesterification of a specific cognate amino acid or its precursor to one of all its compatible cognate tRNAs to form an aminoacyl-tRNA.
If you were rolling dice in a casino and observed results comparable to those observed in Part 2, would you assume the dice were faulty or rigged? Explain your answer using data from your chi-square analysis.
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Answer:
Yes, I would guess that the dice are faulty or rigged.
Explanation:
Yes, I would guess that the dice are faulty or rigged.
This is because the chi-square analysis indicates that there is a less than 5% probability that the difference between the expected and observed results were due to pure chance.
Match the following description with the appropriate type of respiration:
a. occurs in the mitochondria
b. resting muscles depend on this type of respiration
c. lactic acid builds up in muscle fibers
d. rapidly produces ATP for short time periods
e. produces large quantities of ATP but takes longer to synthesize
1. anaerobic respiration
2. aerobic respiration
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Answer:
The correct answer is -
1. c, and d.
2. a, b, and e.
Explanation:
Anaerobic respiration is the short and quick way of producing energy, however, it produces less amount of energy than aerobic respiration and produces lactic acid as a byproduct. It takes place in the cytoplasm of the cell and working or acting muscles depend on such respiration.
Aerobic respiration is the main and important respiration process that takes place in mitochondria and it takes time to produce ATPs that are much more than anaerobic respiration. Resting muscles get energy by this type of respiration.
Which of the following is an example of an enzymatic cycle?
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Answer:
Catabolism
Explanation:
The process of catabolism degrades the bacterial and fungal enzymes into simple inorganic molecules.
A substance, without being a reactant, which speeds up a chemical process is referred to as a catalyst. Enzymes are known as catalysts for biological reactions in living organisms. Although ribonucleic acid (RNA) molecules behave as enzymes, they are usual proteins. Enzymes.
Enzymes carry out the essential role of reducing the activated energy of a reaction — that is, the amount of energy needed to start the process. Enzymes work by attaching and retaining reactant molecules so that the chemical bonding and bonding activities are carried out more easily.
Which of the following describes a predator?
A a fish that is killed and eaten.
B a bear that kills and eats fish
C a worm that lives inside a bear
D a bear that has a worms in its gut.
Answers
Answer:
b
Explanation:
a predator is a bear that kills and eats fish
Answer:
B. bear that kills and eat fish
click on the link below to see the primate family tree diagram. Which of the following statements is true?
Answers
Answer:
biology is amazing it contains some reproductive topics
Orangutans evolved from an ancestor they share with gorillas is true regarding primate family tree diagram. This means that orangutans and gorillas have a common ancestor. Option C is the correct answer.
Evolution is the process through which species change over time. It occurs through the accumulation of genetic variations and natural selection. The primate family tree shows the evolutionary relationships between different primate species, including orangutans and gorillas. Option C is the correct answer.
Orangutans and gorillas share a common ancestor in their evolutionary history. This means that at some point in the past, there was a species that gave rise to both orangutans and gorillas. Over time, as the common ancestor of orangutans and gorillas diverged into separate lineages, each species developed its own unique traits and characteristics. Despite their distinct characteristics, orangutans and gorillas still share certain genetic similarities due to their common ancestry. These similarities can be observed in their DNA sequences.
Learn more about Primate here:
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The complete question is, "Which of the following statements is true regarding primate family tree diagram?
A. Orangutans evolved from lesser apes and Old World monkeys.
B. Gorillas evolved from orangutans.
C. Orangutans evolved from an ancestor they share with gorillas.
D. Gorillas are more closely related to lesser apes than to chimpanzees."
Read the information below then answer the questions which follow:
To grow, plants require water. They cannot get this water unless it is available in the soil. Plants
obtain water from the soil through their roots. It then passes up the stem and to the leaves and
flowers. The plant does not take all the water available in the soil. Much of the remaining water
evaporates into the surrounding air.
In an experiment, a stem which contained several flowers were placed in a beaker of water
containing red ink.
QUESTION 1
(1 mark)
What is the purpose of the experiment?
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Answer:
I believe the experiment was too see if the flowers that would grow from the stem would turn out to be red in color. (Which is the reason they put red ink.)
(hope this helped :P)
For those women who are not progressing at all or the baby is showing early signs of distress, the decision is
often made to induce labor. This uses a synthetic form of oxytocin, called Pitocin, which is given intravenously
(IV) to the woman to get the contractions going. It is often stopped after a few hours. Why might that be?
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Answer:
It is common for contractions to occur every often meaning a sign of labor might be happening soon
Explanation:
Simply meaning your brain is telling your system the baby is ready
which primitive organic molecule was essential to form lipid bilayer?
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Answer:
autocatalytic RNA is the primitive organic molecules was essential form liquid bilayer .
In another experiment, the leu gene was found to be located 5 minutes from the pro gene. In which location(s) in the gene map could the leu gene be located? (Note that the distances depicted between the genes on the map are not to scale.)
Answers
Answer: Hello your question has some missing data attached below is the missing data
answer:
positions : A and E
Explanation:
Given that the leu gene was found 5 minutes from pro⁺ gene hence it can be located either after the bio⁺ gene or before met⁺ genes
hence the locations in the gene map = A and E
attached below is the transfer of genes
In eukaryotes, miR-5 is a microRNA with sequence complementarity to the 3’UTR of the mRNA transcript encoded by the UCD gene. The UCD gene encodes for a protein that acts as an allosteric activator of the protein p101. The p101 protein is a DNA methyltransferase (DNMT) which is an enzyme that methylates DNA when the protein is in its active form.
A) In a lung cancer cell line, the p101 protein is overly active leading to misregulation of gene expression. In these lung cancer cells which of the following would be a correct prediction in regards to miR-5? [choose ALL that apply, note that incorrect selections will deduct points]
A) The gene encoding miR-5 is in a region of chromatin with hypoacetylation
B) The gene encoding miR-5 is in a region of chromatin with hyperacetylation
C) The gene encoding miR-5 in a region of euchromatin
D) The gene encoding miR-5 in a region of heterochromatin
E) The miR-5 transcript is translated at high levels
F) The miR-5 transcript is activating translation of the UCD gene
B) A different lung cancer cell line is found to have significantly decreased amounts of DNA methylation. Which of the following could explain why there is decreased DNA methylation in this cell line? [choose ALL that apply, note that incorrect selections will deduct points]
A) increased amounts of the p101 protein due to low levels of miR-5
B) increased amounts of the UCD protein due to high levels of miR-5
C) decreased amounts of the UCD protein due to low levels of miR-5
D) the absence of the UCD protein due to miR-5 inhibiting translation of the UCD mRNA transcript
E) the absence of the p101 protein due to miR-5 inhibiting transcription of the p101 mRNA transcript
Answers
Answer:
A) A) The gene encoding miR-5 is in a region of chromatin with hypoacetylation.
D) The gene encoding miR-5 in a region of heterochromatin.
B) D) the absence of the UCD protein due to miR-5 inhibiting translation of the UCD mRNA transcript
Explanation:
MicroRNAs (miRNAs) are small regulatory RNA sequences (approximately 20-24 nucleotides in length) that regulate gene expression by RNA interference (RNAi) mechanisms. These sequences (miRNAs) bind by base complementarity to messenger RNAs and thus inhibit protein translation and/or trigger mRNA degradation. In this case, miR-5 binds to the 3’UTR of the mRNA transcript of the UCD gene, thereby inhibiting/slowing protein UCD synthesis. The UCD protein is an allosteric regulator that binds and activates the expression of the p101 protein, thereby the miR-5 RNAi pathway also indirectly decreases the expression of the p101 gene. Moreover, hypoacetylation is an epigenetic mark generally associated with gene silencing (heterochromatin is a transcriptionally inactive state of chromatin).
In Drosophila melanogaster the recessive alleles for brown and scarlet eyes (of two independent genes) produce a novel phenotype so that bw/bw;st/st is white. If a pure-breeding brown is crossed to a pure-breeding scarlet, what proportion of the F2 will be white
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You learned in the previous lesson that T cells require two independent signals to become activated. When T cells are activated, they acquire effector functions and begin to proliferate. This two-signal mechanism prevents the expansion of self-reactive effector T cells. Similarly, B cell receptor signaling alone is not sufficient to activate B cells. B cells require assistance from helper T cells to become activated by monovalent protein antigens and gain certain immune functions. What changes do you suspect will occur to B cells when they are activated by protein antigen and receive T cell help? (select all that apply)
a) They will generate memory cells that can respond to a subsequent exposure to antigen.
b) Their immunoglobulin genes will be mutated in a way that alters antigen affinity.
c) The isotype of their immunoglobulin (Ig) genes will change.
d) They will proliferate and expand.
e) They will secrete antibodies.
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Answer:
a. They will generate memory cells that can respond to subsequent exposure to antigen.
d. They will proliferate and expand
e. They will secrete anti bodies
Explanation:
B cells are type of white blood cells who belong to lymphocyte subtype. They are activated by binding of antigen to receptors on cells surface, against which they initiate and antibody response. Some of these cell proliferate and divide while other become long lived memory cells that respond subsequent exposure to antigen at later stage.
The locus B on the X chromosome of a malaria-carrying mosquito shows a 49% recombination rate with respect to the locus M. Since a recombination rate of 50% is essentially indistinguishable from independent assortment, you might be tempted to look for a locus that falls between B and M. Before you decide to do all that work, you run a chi square test to determine the P value of your experiment.
Required:
What range of P values would tell you that you should accept the conclusion that locus B and locus M are, indeed, 49mu apart and that another locus is not necessary?
Answers
Answer:
The correct answer is - P = 0.45
Explanation:
The Chi-square test is a test that determines the assorted genes independently. Here p-values used to make conclusions in significance testing.
If the p value is less than the significance level we expect or choose, then this null hypothesis is rejected in favor of the alternative hypothesis. It p value is greater than or equal to the significance level, then we fail to reject the null hypothesis.
So in this case, as the recombination rate is 49% so the p value which satisfies the conclusion will be around 0.45 and values near to it.
If you measured the energy content of organisms in each trophic level in an ecosystem, which would have the greatest total energy content
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BLACKSHARK
