The image to the left shows Earth's major plates. A
geologist is studying a plate boundary indicated by the
arrow
Fill in the blank to complete the statement about the
plate boundary
This boundary is between the African Plate and the
Plate

You can determine the optimum pH of an enzyme by graphing its pH activity and studying the results.

Enzymes each perform differently in distinct environments. Some require lower pH to function, while others achieve maximum performance at higher pH levels.  

This relationship between the productivity of the enzyme and the pH can be studied and graphed. Once done, we can expect this graph to look like a parabola, increasing steadily until maxed, then gradually decreasing back to zero at the other extreme. The vertice of this parabola will give us the optimum pH for the enzyme.

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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.

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

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).

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.

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.

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.

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:

Answer:

skin.

Explanation:

the first line of defense uses physical parts such as skin.

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

Answer:

Explanation:

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

In 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

        2 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                  

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:

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

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.

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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Answer:

-1 I believe

Explanation:

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.

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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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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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.

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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Given structure is that of a virus, which includes the viral capsid, nucleic acid, etc.

A= spike

B= Envelop

C= Nucleic acid

D= Capsid

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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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.

Answer:

The definition of the problem is listed in the explanation segment below.

Explanation:

        ⇒  [tex]Mother \ aa\times Father \ aa[/tex]

        ⇒  [tex]a\times a[/tex]

So that the above is the right answer.

Answer:

glandular epithelium are specialized for secretion. lining epithelium simply lines organs like the stomach. ... Endocrine glands release substances within the body and exocrine glands release substances outside the body.

Simple columnar epithelia are tissues made of a single layer of long epithelial cells that are often seen in regions where absorption and secretion are important features. The cells of this epithelium are arranged in a neat row with the nuclei at the same level, near the basal end ...

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:

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. (:

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.

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.

Answer:

squamous epithelium consists of a single layer of flat cells. It is found in lungs,heart and blood vessels.It allows the movement of materials across it.

stratified epithelium has many layer of flat cells.It is present in lining of oesophagus and mouth and also over the skin