2. more birds results in increased biomass
Explanation:
Since birds mostly feed on insects. The increase in number of birds will cause a decrease in insect population.
Now if we consider plant eating insects, the increase in plant eating insect population will result in decrease in biomass of plants because they eat up the plant tissues.
Now relating the above situation we can conclude that if the population of bird increase, they will feed on more and more insects and thus the population of insect will decrease. and in turn, the survivorship or biomass of plants will also increase.
The predicted effect of bird presence on the survivorship or biomass of the plants is B. more birds results in increased biomass.
It should be noted that in a predator-prey relationship, the presence of a predator is vital in controlling the population of the prey.
Based on the complete information, the birds are predators while the grasshoppers are the prey. Therefore, the predators keep control of the population of the prey. Therefore, more birds result in increased biomass. They indirectly control the foraging of grass by the grasshoppers.
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Fill in the blanks using the option below:
The lipid bilayer consists of ______ rows of ______, with the ______ hydrophobic ______ forming the center of the bilayer. The ______ hydrophilic ______ are situated along the outside of the bilayer. This bilayer creates a separation between the contents of a cell and its surroundings.
Options:
tails, heads, non-polar, two polar, phospholipids, three fatty acids, steroids
Answer:
two, phospholipids, non polar, tails, polar, heads.
Explanation:
This is just basically illustrating one of the components of the lipid bilayer memebrane. The membrane is made up of two lipid layers with the polar head groups facing either the cytosol or the outer surroundings while the non polar tails facing each other. Such that this bilayer act as a barrier between the cytosol of the cell/organelle eg endoplasmic reticulum and the outer surroundings.
The lipid bilayer consists of two rows of phospholipids, with the non-polar hydrophobic tails forming the center of the bilayer. The polar hydrophilic heads are situated along the outside of the bilayer. This bilayer creates a separation between the contents of a cell and its surroundings.
The lipid bilayer consists of two rows of phospholipids, with the non-polar hydrophobic tails forming the center of the bilayer. The polar hydrophilic heads are situated along the outside of the bilayer. This arrangement creates a barrier between the cell's interior and the external environment. The hydrophobic tails shield the inner part of the bilayer from the surrounding aqueous environment, while the hydrophilic heads interact with water molecules on both sides of the membrane. This selective permeability allows the cell to control the movement of molecules in and out of the cell, maintaining homeostasis and enabling essential cellular processes to occur. The lipid bilayer also plays a crucial role in cell signaling, adhesion, and recognition, contributing to the overall function and integrity of the cell.
The following statements describe different types of lipids. Fill in the blanks with the following wordsa.triacylglycerolsb. glycerolc. glycerophospholipidd. sphingolipidse. glycolipidsf. fatty acidsg. steroids1.) compounds that contain a fused ring systerm are called __. These have three 6-membered rings and one 5-membered ring. some of these compounds are found in biological membranes.2.) ___ are the building blocks for many lipids, and they generally contain an even number of carbon atoms and an unbranched hydrocarbon chain3.) ___ are formed when a carbohydrate is glycosidically linked to a hydroxyl group of a lipid. examples include gangliosides and cerebrosides. these are also found in biological membranes.4.) ___are the storage form of lipids, accumulating in adipose tissue, and they can be used as metabloic fuel. these compounds have a polar part, made of three ester groups, and a nonpolar fatty acid tail.5.) ___ are made up of a long-chain amino alcoho joined, either by a glycosidic linkage or a phosphodiester linkage, to a fatty acid. these do not contain ___. they are abundant in the nervous system.6.) When glycerol esterified to two fatty acids and a phophoric acid molecule a ___ is formed. these are found in biological membranes.
Answer:
1.) Compounds that contain a fused ring systerm are called _steroids_. These have three 6-membered rings and one 5-membered ring. some of these compounds are found in biological membranes.
2.) _Fatty acids__ are the building blocks for many lipids, and they generally contain an even number of carbon atoms and an unbranched hydrocarbon chain
3.) _Glycolipid__ are formed when a carbohydrate is glycosidically linked to a hydroxyl group of a lipid. examples include gangliosides and cerebrosides. these are also found in biological membranes.
4.) _Triacylglycerols_are the storage form of lipids, accumulating in adipose tissue, and they can be used as metabloic fuel. these compounds have a polar part, made of three ester groups, and a nonpolar fatty acid tail.
5.) _Sphingolipids__ are made up of a long-chain amino alcohol joined, either by a glycosidic linkage or a phosphodiester linkage, to a fatty acid. these do not contain _glycerol__. they are abundant in the nervous system.
6.) When glycerol esterified to two fatty acids and a phophoric acid molecule a _glycerophospholipid_ is formed. These are found in biological membranes.
Final answer:
The types of lipids are identified and placed correctly in the blanks: steroids, fatty acids, glycolipids, triacylglycerols, sphingolipids, and glycerophospholipids, each with their respective characteristics and functions in biological systems.
Explanation:
The statements describe different types of lipids and require the correct lipid types or components to be placed in the blanks:
Compounds that contain a fused ring system are called steroids. These have three 6-membered rings and one 5-membered ring. Some of these compounds are found in biological membranes.
Fatty acids are the building blocks for many lipids, and they generally contain an even number of carbon atoms and an unbranched hydrocarbon chain.
Glycolipids are formed when a carbohydrate is glycosidically linked to a hydroxyl group of a lipid. Examples include gangliosides and cerebrosides. These are also found in biological membranes.
Triacylglycerolsare the storage form of lipids, accumulating in adipose tissue, and they can be used as metabolic fuel. These compounds have a polar part, made of three ester groups, and a nonpolar fatty acid tail.
Sphingolipids are made up of a long-chain amino alcohol joined, either by a glycosidic linkage or a phosphodiester linkage, to a fatty acid. These do not contain glycerol. They are abundant in the nervous system.
When glycerol is esterified to two fatty acids and a phosphoric acid molecule, a glycerophospholipid is formed. These are found in biological membranes.
In reality, the PCR reactions from the crime investigation would be carried out with primers that fluoresce. The products would then be run through an agarose gel, producing peaks like those seen in the paternity test, below.
1. What is the meaning of one peak instead of two at a given locus?
2. Referring to the paternity test below, what determines the position along the horizontal line of the various peaks?
3. What does the difference between two separate peaks of the same color signify?
4. In order to determine whether two people are siblings, their STRs have to be compared to the alleged parents. Why doesn't direct comparison of the siblings' STRs work for this purpose?
Answer:
Short Tandem Repeats (STR) intensification is utilized for the discovery of wrongdoing, Paternity testing or if there should arise an occurrence of chimer-ism and other. The pair rehashes change from individual to person. STR design in a youngster is the blend of their folks STR.
1: Homozygous allele for that specific quality: Peak on the gel is speaking to the kind of allele of a particular STR. Each quality comprises of two alleles, if the arrangements of both the alleles are a similar they are known as the homozygous allele for that quality and their demeanor on gel goes ahead a similar spot. On the off chance that there is yielding in STR of the alleles, at that point two separate groups for that qualities accompany some distinction in position.
2. The flat hub of gel speak to the DNA section size: With the expansion of STR piece size, the groups move towards the privilege of the gel in any case, with decline towards left.
3. Heterozygous alleles with two distinctive STR part size
4. Since youngster STR is the allelic blend STR example of their folks: that is STR of one allele originate from mother anyway second originates from mother for a similar quality. The STR example of kin be the equivalent (all the kin are a blend of their folks)
How many rounds of replication does it take for an incorporation error to be established as a base pair change?
Answer:
Just one round without DNA repair
Explanation:
When an error occurs before or during replication, in a normal system, there is always proofreading and correction. If the error is not corrected during replication, it can then be corrected by the DNA damage repair system. But if not also corrected, the error becomes fixed and subsequent replication will produce this error
What type of enzyme regulation happens when a molecule similar in shape to the original substrate binds to the active site preventing the enzyme from reacting with the substrate?
Answer: Competitive inhibition
Explanation:
Competitive inhibition is a form of enzyme control in which an inhibitor molecule, very similar in structure to the normal substrate of an enzyme, becomes reversibly bound to the active site.
A good example is malonate (an inhibitor) depriving succinate from binding to the active site of the enzyme, succinic acid dehydrogenase
Answer: Competition inhibition.
Explanation:
Competition inhibition happen when inhibitor compete with the substrate. This occur when an inhibitor which look like substrate bind to the enzyme at the active site thereby withholding the substrate from binding. Inhibitor are analogos because their structure look like the substrate. Example include antineoplastic drug methotrexate.
9. Which of the following statements are accurate?
A. The liquid portion of the blood is known as the hematocrit
B. Platelets function in blood clotting
C. Lymphocytes are the only type of white blood cells and are immune cells.
D. Red blood cells are known as erythrocytes
E. Hemoglobin would be found in white blood cells and function in carrying oxygen to the body tissues
Answer: Option B and D are accurate statements
Option B) Platelets function in blood clotting
Option D) Red blood cells are known as erythrocytes
Explanation:
Option A is not accurate because the liquid portion of the blood is known as the plasma.
Option B is accurate because blood platelets, also known as thrombocytes aid in the clotting of blood.
Option C is not accurate because types of white blood cells include lymphocytes, neutrophils, basophils, eosinophils, and monocytes
Option D is accurate because Red blood cells are also known as erythrocytes and helps in the transport of oxygen in the body.
Option E is not accurate because hemoglobin is found ONLY in red blood cells where they function in carrying oxygen to the body tissues.
Thus, only Option B and D are accurate statements
Complete hydrolysis of a glycerophospholipid yields glycerol, two fatty acids (16:1(∆9) and 16:0), phosphoric acid, and serine in the molar ratio 1:1:1:1:1. Name this lipid and draw its structure.
Answer:
Phosphatidyl Serine
Explanation:
Phosphatidylserine is a phospholipid and is a major constituent of the cell membrane. It aids in cell cycle signaling, majorly in the aspect of apoptosis. It is a basic pathway for viruses to invade cells through apoptotic mimicry
Phosphatidylserine is implicated in Alzheimer's disease, age-related decrease in mental function, boost thinking ability in young people, attention deficit-hyperactivity disorder (ADHD), depression, reducing exercise-induced stress, and boosting an athletic performance.
Phosphatidyl serine (PS) is a phospholipid nutrient which is seen in fish, green leafy vegetables, soybeans and rice, and is important for the proper working of neuronal cell membranes and activates Protein kinase C (PKC) which play a major role in memory function.
Various types of damage can lead to acute inflammation, including cuts and abrasions, heat, and microbial damge. Some microbes have structures which can trigger the acute inflammatory repsonse when they invade tissues.
What structural characteristics common to Gram-negative bacteria may trigger the acture inflammatoty respone?
A. Teichoic acid
B. Mycolic acid
C. Lipid A
D. External polysaccharides
Answer:
C: Lipid A
Explanation:
Lipid A is a component of the endotoxin (also called the lipopolysaccharide) present in Gram-negative bacteria. The LPS has 3 component namely;
the O-antigen: they are antigenic determinants and are the outer carbohydrate chainsthe core polysaccharide: forms the centre core of the LPSThe Lipid A: forms the innermost part of the LPS and triggers acute inflammatory responses (e.g endotoxic shock) when released.Upon detection of an endotoxin which forms the component of the outer membrane of a Gram negative organism (in exception to the Gram-positive bacteria called Listeria monocytogenes), the innate immune defense system (e.g macrophages and T-helper cells) are alerted to initiate elimination strategies towards the invading organism. Lipid A binds to the CD14/MD2 receptor on macrophages and monocytes which activates the nuclear factor kappa- light chain enhancer (NF-κβ). The activation of NF-κβ protein triggers the production of inflammatory cytokines which includes Interleukin-1 (IL-1), IL-12, IL-18 as well as the Tumor necrosis factor- alpha (TNF-α).
Inflammatory responses are as a result of the release of these cytokines which sometimes leads to shock and death of the host
As in most areas of biology, the study of mitosis and the cell cycle involves a lot of new terminology. Knowing what the different terms mean is essential to understanding and describing the processes occurring in the cell.
Fill the blanks with terms below to correctly complete the following sentences.
CYTOKINESIS, SISTER CHROMATID(S), CENTROSOME(S), CHROMATIN, CENTROMERE(S), KINETOCHORE(S), INTERPHASE, MITOTIC SPINDLE(S).
1. DNA replication produces two identical DNA molecules, called ______, which separate during mitosis.
2. After chromosomes condense, the _______ is the region where the identical DNA molecules are most tightly attached to each other.
3. During mitosis, microtubules attach to chromosomes at the ______.
4. In dividing cells, most of the cell's growth occurs during _______.
5. The _______ is a cell structure consisting of microtubules, which forms during early mitosis and plays a role in cell division.
6. During interphase, most of the nucleus is filled with a complex of DNA and protein in a dispersed form called ______.
7. In most eukaryotes, division of the nucleus is followed by _______, when the rest of the cell divides.
8. The ________ are the organizing centers for microtubules involved in separating chromosomes during mitosis.
Answer:
1. DNA replication produces two identical DNA molecules, called SISTER CHROMATID(S), which separate during mitosis.
2. After chromosomes condense, the CENTROMERE is the region where the identical DNA molecules are most tightly attached to each other.
3. During mitosis, microtubules attach to chromosomes at the KINETOCHORE(S)
4. In dividing cells, most of the cell's growth occurs during INTERPHASE.
5. The MITOTIC SPINDLE is a cell structure consisting of microtubules, which forms during early mitosis and plays a role in cell division.
6. During interphase, most of the nucleus is filled with a complex of DNA and protein in a dispersed form called CHROMATIN.
7. In most eukaryotes, division of the nucleus is followed by CYTOKINESIS, when the rest of the cell divides.
8. The CENTROSOME(S) are the organizing centers for microtubules involved in separating chromosomes during mitosis.
Explanation:
One chromosome is composed of two chromatids. One chromatid is a chromatin strand that got thick after folding. The chromatin is a dispersed form of the DNI associated with histones. Before cellular division occurs, the chromatin strand condensates and generates a copy or clon so both of the daughter cells can get the same genetic information. These two strands keep joint together by a centromere. So, the chromosome is conformed by the chromatin strand and its copy, known as sister chromatids, and are joint by the centromere. It looks X-shaped. The spindle apparatus -microtubules- is the structure that distributes the chromosomes to each pole. Microtubules attach to the chromosomes by the kinetochores, which are laminar proteinic structures situated next to the centromere. Once in the poles, chromosomes became lax again and it occurs cytokinesis, which is the cytoplasm division. Each cell contains two centrosomes that separate during cell division and migrate to the poles facilitating the action of spindle apparatus.The question involves understanding various biological terminologies related to the cell cycle and mitosis. These terms include sister chromatids, centromere, kinetochore, interphase, mitotic spindle, chromatin, cytokinesis, and centrosomes.
The following is the correct completion of the sentences using the provided terms:
1. DNA replication produces two identical DNA molecules, called sister chromatids, which separate during mitosis.
2. After chromosomes condense, the centromere is the region where the identical DNA molecules are most tightly attached to each other.
3. During mitosis, microtubules attach to chromosomes at the kinetochore.
4. In dividing cells, most of the cell's growth occurs during interphase.
5. The mitotic spindle is a cell structure consisting of microtubules, which forms during early mitosis and plays a role in cell division.
6. During interphase, most of the nucleus is filled with a complex of DNA and protein in a dispersed form called chromatin.
7. In most eukaryotes, division of the nucleus is followed by cytokinesis, when the rest of the cell divides.
8. The centrosomes are the organizing centers for microtubules involved in separating chromosomes during mitosis.
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Consider the alleles for leaf color first. Drag the white labels to the white targets to identify the genotype of each F2 class. Remember that p (the pale mutant allele) and P (the wild-type allele) are incompletely dominant to each other.
Consider the alleles for leaf shape next. Drag the blue labels to the blue targets to identify the genotype of each F2 class. Remember that F (the forked mutant allele) is dominant to f + (the wild-type allele).
Answer:
a) PP
b) F_
c) Pp
d) F_
e) pp
f) F_
g)PP
h)f+f+
i)Pp
j)f+f+
k)pp
l)f+f+
Explanation:
True or False: Throughout your entire body, the function of acetylcholine is always to inhibit the activity of organs (slow things down), and the function of norepinephrine is to stimulate the activity of organs (speed things up).
Answer: False
Explanation:
Acetylcholine functions in the central nervous system and peripheral nervous system both as activator and inhibitor.
It can cause the skeletal muscle to contract and it also inhibits the activation of the cholinergic system.
Whereas, the norepinehrine functions to mobilize the brain and body for actions. It is released lowest at the the time of sleeping , it shoots up at the time we are awaken and is highest at the time of stress.
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A new mountain lake just formed by a retreating glacier. The lake starts out with little nutrients or life, but over time it begins to support a variety of species.
Which type of succession does this describe?
primary
secondary
aquatic
island
Answer: Primary succession
Explanation:
Primary succession occurs when living organisms colonize (appear on) a newly found area. So, since the retreating glacier (a large body of ice flowing downwards) resulted in a new environment, it tells that no living organisms previously existed in the area.
Therefore, the presence of nutrients in the lake would then support new life activities brought by the various species of living organisms.
Thus, primary succession occurs
The order of genes on a plant chromosome is A, B, and D, where A and B have a recombination frequency of 0.2, and B and D have a recombination frequency of 0.3. A pure-breeding plant that has the dominant A and B phenotypes with a recessive d phenotype is crossed to a pure-breeding plant that has the recessive a and b phenotypes with the dominant D phenotype. The resulting hybrid is crossed to a plant that has all three recessive phenotypes. Under the assumption that interference in this region is zero, what percentage of each progeny type would result from a single crossover between A and B
Answer:
It is given that RF (recombination frequency) between A and B= 0.2
RF between B and D= 0.3
Now, DCO (Double Crossover frequency) can be calculated as 0.2 X 0.3 = 0.06
Interference assumed= 0
We can now calculate for SCO as RF (B and D)- DCO frequency
= 0.3- 0.06 = 0.24
Since, the result of each recombinant event is two recombinant types, hence,
24/2 = 12% Abd/abd and 12%aBD/abd
Explanation:
Answer: 12% Abd/abd 12%aBD/abd
Explanation: When recombination frequency(RF) between A and B is given as 0.2
Recombination frequency between B and D = 0.3
Where double crossover frequency (DCO) is calculated as 0.2 × 0.3 = 0.06
The interference assumed as = 0
Single crossover (SCO) can be calculated as recombination frequency of (B and D) - Double crossover frequency which is 0.3 - 0.06 = 0.24
Each recombinant event has an overall result of two(2) types
Therefore; 24÷2= 12%
= 12% Abd/abd and 12%aBD/abd
Explain how the potential for specialization varies with cell type and how it varies over the life span of an organism.
Answer:
Explanation:
Cell specialization is the process of cell differentiation. In this process the generic or normal cells in the body change their specific forms and functions to perform certain specific tasks in the body. This varies with the cell type for example, in young development embryos the stem cells proliferate and differentiate to support the development of body parts and organs in the body whereas in the case of stem cells in adults are specialized to replace the worn out cells of the brain, bones, blood and heart. Thus the specificity of the stem cells varies during various stages of life span of an organism that is from birth till the death.
Cell specialization varies according to the type of cell and over the life span of the organism. This variation allows for increased efficiency of the cells to perform their respective roles in the organism. The process, known as cellular differentiation, involves varying gene expressions, with stem cells representing the unspecialized stage.
Explanation:The potential for specialization varies with cell type and over the life span of an organism based on the function they are required to perform. Organisms start their life from a single fertilized egg cell, which then divides to give rise to trillions of cells. Through a process known as cellular differentiation, these initially similar cells become specialized in both structure and function.
Cells in multicellular organisms like humans become specialized to allow for increased efficiency. Different sets of genes are expressed in different cells leading to their unique functions. For instance, a muscle cell is different from a liver cell, meeting their respective functional requirements for the organism. This variation in specialization continues over the lifespan of the organism, including the changes in gene expression to adapt to environmental changes.
Before a cell becomes specialized, it is known as a stem cell. These cells can divide into more stem cells or differentiate into various specialized cell types. While an already specialized cell generally remains as such, research is ongoing to coax these cells in the lab to become a different specialization.
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Consider two pairs of grandparents. The first pair has 4 grandchildren and the second pair has 32 grandchildren. Which of the two pairs is more likely to have between 40% and 60% boys as grandchildren, assuming that boys and girls are equally likely as children? Why?
Answer:
The second pair of grandparents have approximately 12 to 19 boys as grandchildren.
Explanation:
The probability of a grandchild being a girl or a boy is same, i.e., [tex]\frac{1}{2}[/tex].
The first pair of grandparents have 4 grandchildren.
The second pair of grandparents have 32 grandchildren.
If first pair of grandparents have between 40% to 60% boys as grandchildren then the number of boys are in the limit,
[tex](40\%\ of 4,\ 60\%\ of 4)=(1.6, 2.4)\approx(2, 3)[/tex]
Thus, the first pair of grandparents have approximately 2 to 3 boys as grandchildren.
If second pair of grandparents have between 40% to 60% boys as grandchildren then the number of boys are in the limit,
[tex](40\%\ of 32,\ 60\%\ of 32)=(12.8, 19.2)\approx(13, 19)[/tex]
Thus, the second pair of grandparents have approximately 12 to 19 boys as grandchildren.
So, it is clear that the second pair of grandparents are more likely to have between 40% and 60% boys as grandchildren.
Also according to the law of large numbers as the sample size increases the probability of an event gets closer to the theoretical probability.
The cytoplasm can be defined as everything that fills the space between the
and the
A.
cytosol; nucleus
B. chromosomes; ribosomes
C.
cytoskeleton; organelles
D. nucleus; DNA
E
plasma membrane; nucleus
Answer:
E, the plasma membrane and the nucleus or nuclear envelope
Explanation:
Answer: E
plasma membrane; nucleus
Explanation:
Cytoplasm is the content between the plasma membrane and the nuclear envelope.
It consist of 70-80% water, protein ,polysaccharides,potassium,nucleic acid,fart acid. It is the seat of metabolic reactions and cellular activity. It is enclosed by cell membrane.
The Hopi, Zuni, and other Southwest indigenous peopleshave a relatively high frequency of albinism resulting from homozygosity for a recessive allele, a. A normally pigmented man and woman, each of whohas an albino parent, have two children.a.What is the probability that both children are albino?b) What is the probability that at least one of the children is albino?
Answer:
a. Zero percent
b. Fifty percent
Explanation:
As it has been mentioned in the question that both man and woman are normally pigmented and each one have an albino parent thus both will be heterozygous and their allelic makeup will be as follows -
Man - Aa
Woman - Aa
After a cross between Aa × Aa we get the following result-
One AA, Two Aa and One aa
It is clear that only the aa type of offsprings will be albino.
Out of 4 there is probability of only one to be albino so out of two there will be 50% probability of being albino.
The probability of both sons to be albino will be zero percent.
Apoptosis can occur in a cell when the cell is ________________. Group of answer choices no longer needed damaged infected by a virus all of the above (damaged, no longer needed, infected by a virus)
Answer:
all of the above
Explanation:
The programmed cell death which is a part of normal growth and development of the cell is called apoptosis. Apoptosis occurs in multicellular organisms.
By apoptosis, the body eliminated the old cell, unnecessary cells, damages cells, and the cells which are infected by viruses. Caspases are the enzymes that are activated during the time of apoptosis and mediates cell death.
It kills the cell by cleaving some protein present in the cytoplasm and nucleus of the cell. By apoptosis, the body maintains the healthy cell numbers. So the right answer is all of the above.
Match each cardiac valve with the two chambers it separates. Each pair of chambers may be used more than once, or not at all. tricuspid mitral bicuspid pulmonary semilunar aortic semilunar A. right atrium and right ventricle B. left atrium and left ventricle C. right ventricle and pulmonary artery D. left ventricle and aorta E. right ventricle and aorta F. left ventricle and pulmonary artery
Answer:
Explanation:
The tricuspid valve separates the right atrium from the right ventricle carrying deoxygenated blood from right atrium to right ventricle
The mitral/bicuspid valve separates the left atrium from the left ventricle carrying oxygenated blood from left atrium to left ventricle
Pulmonary semilunar valves separates the right ventricle from the pulmonary artery preventing bachelor of blood.
Aortic semilunar valves separates the left ventricle from the aorta preventing backflow of blood
The tricuspid valve separates right atrium and ventricle, the mitral/bicuspid valve separates left atrium and ventricle, the pulmonary semilunar valve separates right ventricle and pulmonary artery, and the aortic semilunar valve separates left ventricle and aorta.
Explanation:The tricuspid valve separates the right atrium and the right ventricle (Option A). The mitral valve or the bicuspid valve separates the left atrium and the left ventricle (Option B). The pulmonary semilunar valve separates the right ventricle and the pulmonary artery (Option C). Lastly, the aortic semilunar valve separates the left ventricle and the aorta (Option D). These valves control the flow of blood between different chambers to ensure it moves in the correct direction.
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Hair pigmentation in mammals follows the generic enzymatic pathway depicted below. Enzyme A synthesizes a black pigment from a precursor, and the contribution of enzyme B results in the final agouti pigment:
Precursor molecule (colorless) Black pigment Agouti pigment
Each enzyme is coded by a single allele with complete dominance. If a crossing design begins with true breeding parents for coat color with the genotypes AAbb and aaBB. What will be the expected F2 genotypic and phenotypic ratios of the eventual F2 progeny?
Answer:
Given,
Colourless molecule is converted to black pigment by enzyme A. Black pigment is converted to agouti pigment by enzyme B.
Parent 1: AAbb = Black
Parent 2: aaBB = Colourless
AAbb X aaBB = AaBb (F1) The entire progeny is agouti coloured since dominant alleles for both the genes are present.
When AaBb X AaBb according to dihybrid cross in F2 :
A_B_ = 9 ( Agouti )
A_bb = 3 ( Black )
aaB_ = 3 ( Colourless )
aabb = 1 ( Colourless )
Hence genotypic ratio is 9 : 3 : 3 : 1
and phenotypic ratio is 9 (agouti) : 3 (black) : 4 (colourless)
g In rabbits, an allele that produces black fur (B) is dominant over its allele for brown fur (b). The allele of another gene (E) produces long ears which is dominant to floppy ears (e). Suppose a female with brown fur and homozygous for long ears is mated with a floppy-eared male homozygous for black fur.
Expected genotype ratio = 100% BbEe
Expected phenotype ratio = 100% black fur and long ears
Genotype of a female with brown fur and homozygous for long ears = bbEE
Genotype of a floppy-eared male homozygous for black fur = BBee
Crossing the 2 in a 4 x 4 Punnet's square:
Genotype = 100 BbEePhenotype = 100% black fur with long earsThe result of Punnet's square is attached.
The complete question:
In rabbits, an allele that produces black fur (B) is dominant over its allele for brown fur (b). The allele of another gene (E) produces long ears, which is dominant to floppy ears (e). Suppose a female with brown fur and homozygous for long ears is mated with a floppy-eared male homozygous for black fur. What are the expected genotypes and phenotypes of their offspring?
You must read the New York Times article "Diabetes Study Ends Early With a Surprising Result" found in the pre-lab notes before answering the pre-lab quiz questions this week The hypothesis for the study presented in this article is a. Diet and exercise impacts likelihood of becoming diabetic. b. Diabetes impacts likelihood of developing heart disease. c. Diet and exercise impact blood sugar levels, blood pressure, and cholesterol levels in diabetics. d. Diet and exercise impact rates of heart attacks, strokes, and cardiovascular deaths in diabetics.
Explanation:
Diabetes is a disease which happens when our blood glucose or blood sugar amount in the body is too high.The main source of energy to carry out other activities of body is Blood Glucose. This Blood Glucose comes from the food we eat.The hormone named Insulin, which is made by the pancreas, helps the glucose which is taken from the food to get into the cells and is used for energy.The regular exercise and control diet reduces the rate of heart attacks, cardiovascular deaths in diabetics and strokes.
Jason Ormand was arrested on suspicion of sexual assault. The arresting officers did not give Jason his Miranda warnings. Jason confesses to the assault. The DNA evidence taken from the victim shows with 99% certainty that Jason was her attacker. a. Because of the violation of Jason’s rights in the officers’ failure to give the Miranda warnings, the case must be dismissed. b. The case can still proceed with the DNA evidence. c. The case can proceed but only with the victim’s testimony. d. none of the above
Answer:
B
Explanation:
Although the arresting officer did not give Jason a miranda warning (a notification of Jason's right to remain silence and decline to answer any questions or give information to the arresting officer), Jason already confessed to the crime. Above all the DNA evidence gave a compelling revelation of 99% certainty which is a full proof evidence for the case. Therefore the case can still proceed with the DNA evidence against Jason
Answer: The case can still proceed with the DNA evidence.
Explanation: In the U.S, the Miranda warning is a warning given by a law enforcement officer to criminal suspects in his custody advising them of certain constitutional rights, called their Miranda rights. This is their right to silence; that is, their right to refuse to answer questions or provide information to law enforcement or other officials.
Jason confessed to the assault without receiving his Miranda warning. Although he is guilty and the DNA evidence proved him guilty as charged. The case can still be proceeded with that evidence .
Please help !!What role might an antibiotic factor such as temperature play in the evolution of a species ?
Answer:
if we are talkin about temperature then it could change how a species live so if it's hot it could have scales but if it's cooled it could have fur whatever the animal or species is.
Answer:
Long term temperature change can result in selective pressure that are individual specific and which are better adapted to the new temperature, causing populations to evolve.
Temperature changes could also alter the availability of food type and thus create a selective pressure towards individuals that can take advantage of the different food sources.
W1 should not have fluoresced. Why? (I know it did not contain rGFP, but why?) W2-W4 should have fluoresced slightly. Give two possible reasons why. (Yes, I know it contained GFP…. but why?)
W1 did not fluoresce because it did not contain the gene for rGFP. W2-W4 fluoresced slightly, possibly due to low expression levels or environmental factors.
Explanation:The fluorescence of a substance is dependent on the presence of specific molecules or proteins that can emit light when excited by UV light. In this case, W1 did not fluoresce because it did not contain the gene for rGFP, which is responsible for fluorescence. The presence of GFP in W2-W4 allowed them to fluoresce slightly, potentially due to two reasons:
The GFP in W2-W4 may not have been expressed at high enough levels to produce a strong fluorescence signal.Other factors, such as the environment or interactions with other molecules, may have affected the fluorescence of GFP in W2-W4, resulting in a weaker fluorescence compared to transgenic mice with higher GFP expression levels.Learn more about Fluorescence here:https://brainly.com/question/37247128
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W1 did not fluoresce because it lacked rGFP. W2-W4 contained GFP, but fluorescence may have been faint due to low GFP concentration or suboptimal environmental conditions.
W1 should not have fluoresced because it did not contain rGFP, the protein responsible for fluorescence. Fluorescence occurs when certain molecules absorb light at one wavelength and emit light at a longer wavelength. Without the presence of rGFP, there would be no protein capable of fluorescing in W1.
W2-W4 should have fluoresced slightly because they contained GFP, which is known for its fluorescence properties. However, the fluorescence may have been faint for two possible reasons:
1. Concentration: The concentration of GFP in W2-W4 may have been lower than optimal for strong fluorescence. Higher concentrations of GFP typically result in brighter fluorescence, so if the concentration was too low, the fluorescence may have been less intense.
2. Environmental factors: Environmental conditions such as pH, temperature, and presence of certain ions can affect the fluorescence of GFP. If the environmental conditions in W2-W4 were not optimal for GFP fluorescence, it could have resulted in weaker fluorescence compared to ideal conditions.
Natural Selection is any natural process that changes the genetic composition of a population by ensuring that some individuals leave __________ offspring than others.
Answer: better
Natural Selection is any natural process that changes the genetic composition of a population by ensuring that some individuals leave better offspring than others.
Explanation:
Natural selection simply involves the elimination of less favourable traits such as slender body parts, lower resistance to diseases etc while highly favorable traits such as plump body parts, higher resistance to diseases that makes organisms better suited to the environment are retained, and then passed from parents to offsprings.
Thus, natural selection brings about the emergence of better offspring than others.
The following descriptions apply to either mitochondrial import or nuclear import. Choose all the statements that apply to mitochondrial import. a. The signal sequence is recognized and bound by a receptor protein in the organelle's outer membrane. b. ATP hydrolysis is required c. GTP hydrolysis is required d. Chaperone proteins are involved e. The imported protein enters through a pore that is large enough to allow ions and water molecules to traverse freely.
Answer:
Option-(A):The signal sequence is recognized and bound by a receptor protein in the organelle's outer membrane.
Explanation:
The mitochondria are also termed as the "power house's" of the cell structures, as there is more to create energy for the body inside the cell organelle. However, there are certain amino acid chains or simply the proteins which are required for sustaining the over all structure of the cell's organelle. As, there are some protein on the outside of the organelle known as porin. While, the organelle itself is involved in the electron transport chain that occurs inside the cell body for generation of the optimum level of energy for the organism to carry out various functions for its survival.Jack Friedman, a 41-year-old male, was in a fight at a soccer game and was hit in the head with a bottle, which caused some deep lacerations in his scalp. Dr. Girald performed a layered closure of the wounds: one 2.0 cm, one 4.5 cm, and two lacerations that were 1.0 cm each in length.
Answer:
Good for him :)
Index Wound Repair Scalp Intermediate 7.6 cm to 12.5 cm2.0 + 4.5 + 1.0 + 1.0 = 8.5 cm Rudy Porter, a 71-year-old man, had in a history of gastrointestinal problems, including colon polyps.
After taking home a special kit, he submits stool samples for an occult blood test by immunoassay.
What does laceration mean?
A laceration or laceration refers to a skin wound. Unlike an abrasion, none of the skin is missing. A cut is usually thought of as a wound caused by a sharp object, such as a piece of broken glass.Lacerations are usually caused by blunt trauma.Learn more about laceration here.
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21. Quiet inspiration will ____ thoracic and lung volume and _____ intrapulmonary pressure. A. increase, increase B. increase, decrease C. decrease, increase D. decrease, decrease
Answer:
B. increase, decrease.
21. Quiet inspiration will increase thoracic and lung volume and decrease intrapulmonary pressure.
Explanation:
Inspiration is the process by which air enters from the outside to the inside of the lungs. The communication of the lungs with the outside is done through the upper airways. Inspiration is the active phase of breathing, for it to occur it is necessary that different muscles contract in order to increase the size of the chest, which causes the lung to expand and atmospheric air tends to enter to equalize the pressure . During inspiration the vertical diameter of the thorax increases due to the decrease in the diaphragm, but the transverse and anteroposterior diameter also increases due to the action of the remaining muscles that raise the ribs.
In ______ , the spindle disappears as the nuclear envelopes form. The plasma membrane furrows to give two complete cells, each of which has the haploid, or N, number of chromosomes. Each chromosome has one chromatid.
Answer: telophase;
Explanation:
In the telophase stage meiosis 2 during cell division, the spindle disappears and the nuclear envelope begins to reform and the cell begins to split into two.
During cytokinesis, the final phase of cell division, the spindle disappears and nuclear envelopes form around each set of decondensed chromosomes. The plasma membrane then furrows to create two separate haploid cells, each with a single set of chromosomes.
Explanation:The subject of your question is referring to the final phase of cell division, which is known as cytokinesis. During cytokinesis, the spindle disappears and nuclear envelopes begin to form around each set of sister chromatids, which have decondensed into chromosomes. The plasma membrane furrows or forms a cleavage furrow to eventually split the cell into two separate cells. Each new cell is haploid, meaning that it has one complete set of chromosomes (N). Therefore, each chromosome indeed consists of a single chromatids.
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