Answer:
Lipase
Explanation:
Lipase enzymes break down fat into fatty acids and glycerol. Digestion of fat in the small intestine is helped by bile, made in the liver.
The same sentences are written with a substitution, deletion, or insertion, much as a gene may have a substitution, deletion, or insertion mutation. Identify which sentence represents a substitution, which represents a deletion, and which represents an insertion.
Answer:
The question is incomplete, the complete question has been added as an attachment.
THE DGR AN represents deletion mutation
THE RED ANA TRA N represents insertion mutation
THE FOX WTS RED represents substitution mutation
Explanation:
Mutation is a change in the nucleotide sequence of a gene. It can be of different types depending on how it occurs. A deletion mutation is that which involves the removal of one or more nucleotide base from the sequence.
An insertion mutation is the addition of one or more nucleotide bases to the sequence.
A substitution mutation is the replacement of one or more nucleotide base in a sequence.
The original sentences were:
1. THE RED ANT RAN
2. THE FOX WAS RED
3. THE DOG RAN
The three sentences that mutated/changed are:
1. THE DGR AN- in this sentence, letter O which represents one base is missing due to removal, making the original sentence (THE DOG RAN) alterated. Hence, this represents an example of deletion mutation.
2. THE RED ANA TRA N - in this case, Letter A has been added in between N and T. This makes the original sentence (THE RED ANT RAN) altered. Since addition of letter (base) occurred, this represents an example of insertion mutation.
3. THE FOX WTS RED - In this case, letter A in "WAS" was replaced by T, making the original sentence (THE FOX WAS RED) altered. Since a replacement of letter (base) is the case here, it represents an example of substitution mutation.
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.
Which type of mutation is likely to have the least affect on an organism
Answer:
Point Mutations
A point mutation—the change of a single nitrogen base in a DNA sequence—is usually the least harmful type of DNA mutation
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How is global climate change is impacting our ecosystems ( in your own words not from google )
Answer:
Climate change impacts where species live and how they interact, also including the timing of biological events. Climate change can immense the capacity of ecosystems to extreme events and disturbance, such as wildfires, floods, and droughts.
Explanation:
Identify the structure represented in the picture . Then, identify the parts labeled A—D.
Given structure is that of a virus, which includes the viral capsid, nucleic acid, etc.
A= spike
B= Envelop
C= Nucleic acid
D= Capsid
What are Viruses?Small infectious agents known as viruses can multiply only in the living cells of other animals. They are composed of genetic material, which can be DNA or RNA, encased in a protein shell called a capsid. Some viruses also have a lipid-based outer coat.
Because they cannot perform metabolic functions on their own and require a host cell to reproduce, viruses are not considered living organisms. They spread infection to various living things including bacteria, plants, animals and people.
A virus replicates its genetic material and forms new viral particles when it infects a host cell using the host cell's machinery. This can lead to the death of the host cell and result in the spread of the virus to other cells of the host.
Therefore, given structure is that of a virus, which includes the viral capsid, nucleic acid, etc.
A= spike
B= Envelop
C= Nucleic acid
D= Capsid
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The results of gel electrophoresis are shown below. What can you determine
about the DNA from looking at the results of this test?
electrode
A. Strand 1 is the smallest.
B. Strand 2 is larger than strand 3
C. Strand 3 is larger than strand 4.
D. Strand 4 is the smallest
Answer:
A
Explanation:
Gel electrophoresis is an amazing technique that allows for DNA separation based on size (as in, length). Using electricity (and keeping in mind that DNA has a negative charge because of the phosphate groups), the shortest pieces will migrate the farthest.
So we can tell that the pieces are in ascending order of size. 4>3>2>1, in length.
A. Strand 1 is the smallest.
Answer:Given results of gel electrophoresis shows that strand 1 is smallest. The pieces are arranged in ascending order of size 4>3>2>1 . Thus,option A is correct.
Explanation:
What is Gel electrophoresis?
Gel electrophoresis is a laboratory technique used to separate mixture of DNA ,RNA or protein according to their molecular size.
The gel used in this technique is composed of polyacrylamide or agarose.
The main application of gel electrophoresis has been in the seperation of molecules and in restriction mapping.
The main advantage of gel electrophoresis is its capability to
simultaneously separate various samples in one round.
Therefore, strand 1 is the smallest according to the result of gel electrophoresis shown in figure. Thus, option A is correct.
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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
In each of the scenarios, an individual heterozygous for the indicated gene or genes is crossed with an identical heterozygous individual. Predict the number of different phenotypes that would be observed among the progeny of such a cross.
Scenario 1 : R is completely dominant to r.
Scenario 2: R is codominant with r.
Scenario 3: R and S are completely dominant to rand s, respectively; R and S are linked by 15 m.u. and do not interact with each other.
Scenario 4: Genes R and S perform the same function; R and S are completely dominant to r and s, respectively; and R and S are unlinked.
Scenario 5: R and S are completely dominant to r and s, respectively; R and S are linked by 10 m.u. and show dominant epistasis.
Answer:
Explanation:
Scenario 2: R is codominant with r.
Scenario 3: R and S are completely dominant to rand s, respectively; R and S are linked by 15 m.u. and do not interact with each other.
Scenario 4: Genes R and S perform the same function; R and S are completely dominant to r and s, respectively; and R and S are unlinked.
Scenario 5: R and S are completely dominant to r and s, respectively; R and S are linked by 10 m.u. and show dominant epistasis.
Examples of stimulants include
a. Nicotine
b. Ecstasy
c. Dextromethorphan
d. A and B
Answer:
The answer to this question is d(A and B)
Explanation:
Hope this helps
Answer:
d but also c !
Explanation:
A and B are correct but also dextromethorphan has lower doses of stimulants
Why to use small beads for immobilizing enzymes
Answer:
Explanation:
Effects of bead size and bead amount on enzyme activity. Beads containing crude enzyme were prepared in the presence of sodium alginate (2% w/v) in CaCl 2 (5% w/v) solution. Each treatment was performed in triplicate.
You are studying a bacterium that utilizes a sugar called athelose. This sugar can be used as an energy source when necessary.Metabolism of athelose is controlled by the ath operon. The genes of the ath operon code for the enzymes necessary to use athelose as an energy source.You have found the following:The genes of the ath operon are expressed only when the concentration of athelose in the bacterium is high.When glucose is absent, the bacterium needs to metabolize athelose as an energy source as much as possible.The same catabolite activator protein (CAP) involved with the lac operon interacts with the ath operon.Based on this information, how is the ath operon most likely controlled?Drag the labels onto the diagram to identify the small molecules and the states of the regulatory proteins. Not all labels will be used.POSITIVE CONTROLa) Inactive Activator (not glowing, w/o yellow square)b) cAMPc) Active Activator (glowing, w/ yellow square)NEGATIVE CONTROLd) Active Repressor (glowing, w/o yellow square)e) athelosef) Inactive repressor (not glowing, w/ yellow square)
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.
The feces of birds and mammals can also be used to assess ecosystem health. Research this methodology and provide two examples of animals used for these studies and the types of materials that are identified in their feces
Answer:
Birds and leopard
Explanation:
It has been researched and discovered that the feces of birds and animals can be used to determine the state of health of an ecosystem.
For example, it has been discovered that hormones and other substances are excreted through urine and feces. And researchers have discovered from the feces of leopards, that a leopard that has cave or trees where it can hide from hunters are not likely to be stressed out like the one that does not have.
The first line of defense involves which structure(s)?
O T-cells
O skin
O blood
O B-cells
Mark this andretum
Next
Submit
Answer:
skin.
Explanation:
the first line of defense uses physical parts such as skin.
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.
Which of these statements are accurate? Hexokinase is a type of transferase that catalyzes the transfer of a phosphoryl group from ATP to a hexose. Hexokinase transfers the terminal phosphate of ATP to carbon 3 of glucose. The conformational shift that occurs when glucose, but not water, enters the active site prevents water from hydrolyzing ATP. Hexokinase consists of two domains, or lobes, that come together when glucose and the MgATP2–MgATP2– complex are bound. Most kinases require the presence of a monovalent metal ion cofactor to prevent ATP hydrolysis. Hexokinase is found in the mitochondrial membrane.
Answer:
A. The conformational shift that occurs when glucose, but not water, enters the active site prevents water from hydrolyzing ATP. True
B. Hexokinase consists of two domains, or lobes, that come together when glucose and the MgATP2–MgATP2– complex are bound. True
C. Hexokinase is a type of transferase that catalyzes the transfer of a phosphoryl group from ATP to a hexose. True
Explanation:
A. Hexokinase suffers and conformational changes by binding glucose in a reaction that prevents ATP hydrolysis.
B. Hexokinase is an enzyme with two domains that function by binding to the substrate (i.e., glucose). The region linking both protein domains is responsible for the catalytic activity.
C. Hexokinase is a type of transferase that catalyzes the transfer of a phosphoryl group from ATP to a hexose. During the first stage of glycolysis, the hexokinase transfers one phosphorous group from magnesium-ATP (Mg-ATP) to one hexose molecule, such as fructose, mannose or glucose.
False statements:
- Most kinases require the presence of a monovalent metal ion cofactor to prevent ATP hydrolysis. False: In canonical kinases, conserved amino acids bind to divalent metal ions before the transference of the phosphate group to their substrates.
- Hexokinase is found in the mitochondrial membrane. False: Hexokinase is found in the cytosol.
- Hexokinase transfers the terminal phosphate of ATP to carbon 3 of glucose. False: Hexokinase catalyzes the transfer of the terminal phosphate ATP to form glucose 6-phosphate.
Select all that apply. Which of the following are vertebrates? earthworms grasshopper lancelets octopus crab amphibians tunicates starfish
Answer: amphibians
Explanation:
earthworms, grasshopper, lancelets, octopus, crab, tunicates, starfish are all invertebrates.
lancelets(cephalochordata) and tunicates(urochordata) are chordates but do not have a backbone.
Hii!! The correct answer is amphibians. (:
8. The function of lipoproteins in blood serum is to:
a. aid in blood clotting
b. transport cholesterol
c. maintain pH
d. remove defective red blood cells
This solid layer is found between the inner mantle (mesosphere) and the lithosphere.
O A. crust
B. asthenosphere
C. inner core
O D. outer core
Answer:
The layer below the rigid lithosphere is a zone of asphalt-like consistancy called the Asthenosphere . The asthenosphere is the part of the mantle that flows and moves the plates of the Earth.
Explanation:
Congenital hyperinsulinism is a rare condition in which infants experience frequent episodes of hypoglycemia. Based on the functions of the two main hormones that control blood sugar homeostasis, which of the following would be effective at treating the hyperinsulinemia?
A. Glucagon
B. Glucocorticoids
C. Preventing insulin release or production
D. All of the choices are correct.
Answer:
Any treatment which involves reduction in the levels of insulin to curb excessive influx of glucose into the cells from the blood will be an ideal treatments.
Glucagon, which converts excess glucose to glycogen. Glucocorticoids induces insulin resistance thus raises glucose levels.
suppressing insulin release will be another option.
Therefore all options are correct.
Explanation:
phosphate from atp is removed to make adp and a free phosphate molecule
Answer:
ATP is a nucleotide consisting of an adenine base attached to a ribose sugar, which is attached to three phosphate groups. ... When one phosphate group is removed by breaking a phosphoanhydride bond in a process called hydrolysis, energy is released, and ATP is converted to adenosine diphosphate (ADP).
Technology is often used in farming and agriculture. For example, farmers often use airplanes to spray chemicals on crops rather than spraying smaller amounts by hand. Which of the following is a negative side of using this technology?
5
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Introduction to Biology: Mastery Test
5
A student notices that when bananas are kept near other fruits, the other fruits ripen faster. She wonders what causes
more quickly and thinks that bananas must have an influence on the ripening process. She decides to perform an expe
bananas and apples and places an apple by itself at location A and an apple next to a banana at location B. She then w.
records the how ripe each apple is.
What would be the next step in this experiment?
O A
ask questions and construct a hypothesis
B.
make observations and draw a conclusion
O c.
construct a hypothesis and record data
D.
analyze the results and make a conclusion
Answer:
Explanation:
The next step in this experiment is constructing a hypothesis and recording data. The correct option is (C).
Making observations and keeping track of data would be the experiment's next phase. The learner would observe and record the ripening of both apples at locations A and B in this stage. They would have to keep track of the maturity of the apples throughout time by routinely observing and recording the changes. This could entail examining the apples visually for color, texture, and overall maturity and recording any variations between the apples at locations A (by themselves) and B (next to a banana).
The student can analyze the results and develop conclusions regarding the impact of bananas on the maturity of other fruits by gathering information on the ripening of the apples in both locations. The information can be used to prove or disprove the initial theory and assist determine whether bananas actually influence the maturity of surrounding fruits.
Hence, the correct option is (C). Construct a hypothesis and record data.
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Which of the following removes waste from the body through perspiration? blood kidneys skin large intestine
Answer:
Kidneys, skin and large intestine. Organs of excretion include the skin, liver, large intestine, lungs, and kidneys. All of them excrete wastes, and together they make up the excretory system. The skin plays a role in excretion through the production of sweat by sweat glands.
Explanation:
Answer: SKIN
Explanation: Perspiration is the process of sweating so your skin does that
The genetics of a-thalassemia are similar to narrow sense heritability in that there is a gradation of phenotype, with each allele seeming to contribute equally to the trait. However, there are also differences from narrow sense heritability. In particular, as you can see from your answers above, it is difficult to predict offspring phenotype based only on parental phenotype. What two aspects of HBA1 and HBA2 combine to offset what would otherwise appear to be narrow sense heritability of a-thalassemia?
Answer:
In strict mode, the narrow-sense heritability is the proportion of the additive genetic variance that contributes to the total of the phenotypic variance. This value can be associated with the inheritance of the a-thalassemia .
Explanation:
A-thalassemia is a genetic disorder caused by mutations affecting four different genes that encode alpha-globin, thus affecting the hemoglobin production process and, consequently, oxygen transport. The mode of inheritance of the a-thalassemia may be associated with narrow sense heritability since the phenotype is manifested by gradation, i.e., each allele might contribute in similar mode to this genetic condition.
Which of the following is a type of action, assistance, or information that a genetic counselor can provide? Check all that apply.
Calculate the risk of a child inheriting a genetic disorder.
Look for patterns of inheritance in a client’s family history.
Prescribe a medication to help with a client’s anxiety.
Explain causes and inheritance patterns of a genetic disorder.
Recommend the best of several long-term options for a client.
Explain what genetic tests are available and what information they would provide.
Interview a client to obtain details of the family’s medical and genetic history.
Order a test to be done without the client’s approval if it would provide useful information.
Answer: Calculate the risk of a child inheriting a genetic disorder.
Look for patterns of inheritance in a client’s family history.
Explain causes and inheritance patterns of a genetic disorder.
Recommend the best of several long-term options for a client.
Explain what genetic tests are available and what information they would provide.
Interview a client to obtain details of the family’s medical and genetic history.
Explanation:
The genetic counseling will involve the following steps:
1. The counselors check the family medical history and medical records of parents to estimate the chances of inheritance of genetic disorder or disease in a developing fetus.
2. The counselors order genetic testing.
3. The counselors evaluate the results of the tests.
4. They counsel parents based on the results and recommend them the long term options to prevent the effect of disease.
Answer:
1 2 4 6 7
Explanation:
Through logic and the prosess of elimenation I found the answers.
In humans, oculocutaneous (OCA) albinism is a collection of autosomal recessive disorders characterized by an absence of the pigment melanin in skin, hair, and eyes. That is, normal pigmentation (A) is dominant over albinism (a). For this question, assume it is a single gene with two alleles. Assume it is a single gene with two alleles. If two people have normal pigmentation, what possible phenotypes may be observed in their offspring?
Answer:
The definition of the problem is listed in the explanation segment below.
Explanation:
The parent phenotype is an albino on every one of them. Albino gene seems to be located primarily while it has its whole genotypes in such a recessive state called "aa". When this trait becomes autosomal, it does have an equivalent amount of alleles across both parent members. The gametes including its albino genotypes.⇒ [tex]Mother \ aa\times Father \ aa[/tex]
⇒ [tex]a\times a[/tex]
Genotype including its offspring - albino. Well, all offspring will also have albino phenotypes.So that the above is the right answer.
Which happens during the second trimester?
A. Blood pumps through the heart.
B. The stomach produces digestive fluids.
C. The lips begin to develop.
D. Teeth begin to form.
Answer:
The lips begin to form. I know that during the second trimester ears and eyes begin to develop, so I assume that lips would also begin to develop too.
In pregnancy, during the second-trimester blood pumps through the heart. The correct option is A.
What is trimester?A pregnancy is divided according to the weeks namely trimesters, these are first trimester, second trimester, and third trimester.
A trimester usually occurs between 12 and 14 weeks and a full-term pregnancy occurs at around 40 weeks.
In pregnancy, during the second-trimester blood pumps through the heart.
Thus, the correct option is A.
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Arrange the isotopes of oxygen and carbon in order of decreasing number of neutrons.
Rank the isotopes from most o fewest neutrons. To rank items as equivalent, overlap them.
Answer:
C-12 > C-13 > C-14,O-16 > C-16,O-18
Explanation:
Generally, isotopes are described as atoms of the same elements with the same atomic number but different number of neutrons.
Hence,
For O-18, oxygen has an atomic number of 8 and neutron number of 10.
For C-14, carbon has an atomic number of 6 and neutron number of 8.
For C-13, carbon has an atomic number of 6 and neutron number of 7.
For C-16, carbon has an atomic number of 6 and neutron number of 10.
For carbon 12, carbon has an atomic number of 6 and neutron number of 6.
For O-16, oxygen has an atomic number of 8 and neutron number of 8.
Hence the rank from the fewest neutrons would be C-12>C-13>C-14,O-16>C-16,O-18
5. Find the LC.M of XX,X-1,X-X
Answer:
-1 I believe
Explanation:
When jared gets to work, he looks at the latest data on wind speed, temperature, and humidity. Then, he uses a computer model to help him predict where storms may occur. In what field of earth science does jared work?
A.astronomy
B.meteorology
C.environmental science
D.geology
Answer:
B. meteorology
Explanation:
Someone who predicts weather based on patterns, models, and current atmospheric conditions is a meteorologist.
╭☞ When jared gets to work, he looks at the latest data on wind speed, temperature, and humidity. Then, he uses a computer model to help him predict where storms may occur. In what field of earth science does jared work?
ANSWER:╭☞ B. meteorology
Explanation: Because meteorology is a branch of the atmospheric sciences that includes atmospheric chemistry and atmospheric physics, with a major focus on weather forecasting. like your question.– Meteorologists are scientists who study and work in the field of meteorology.
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