Answer:
The answer is letter A.
Explanation:
The tropical rain forest is most likely environment where this leaf was growing
Final answer:
The leaf with a large surface area and stomata on the underside is adapted for a tropical rainforest environment, where it maximizes photosynthesis and minimizes water loss under a light-limited canopy.
Explanation:
The leaf described, being dark green, thin, having a large surface area of 10 square meters, and having stomata on the lower surface only, is most likely native to a tropical rainforest. In such an environment, large leaf surface areas are crucial for maximizing photosynthesis under the canopy where light is limited. The stomata, being primarily on the underside, minimize water loss while still allowing for gas exchange away from intense sunlight. Additionally, the thin nature of the leaf is characteristic of plants that do not need to retain large amounts of water, which is consistent with the moist conditions of a tropical rainforest.
Contrastingly, leaves from dry, sandy regions or from plants like cacti have adaptations for minimizing water loss and typically present a smaller surface area. The leaf's characteristics do not align with those one would find in the other environments presented, such as floating on a large, still pond (where we would find wide, flat leaves that can float) or growing in a dry, sandy region (where smaller, thicker leaves with a reduced surface area are advantageous).
____________ is an entity set that contains the commonly shared characteristics of it entities or subtypes. It includes characteristics that are not common to all entities within the set and becomes the parent to one or more subtypes in the hierarchy.
Answer:
A supertype entity
Explanation:
A supertype is a form of entity which has one or even more subgroups to relate to. A subtype is an entity type sub-group of entities that is relevant to the organisation and shares similar characteristics or relations which are different from certain sub-groups.
The entity supertype is a different form of entity type which has relationship with one or even more subtypes and includes subtypes ' of specific attributes.
What must be true for a male individual to be a carrier of a Y-linked recessive allele?
A. The gene has no other alleles.
B. The gene is only expressed in females.
c The gene is silenced on the Y chromosome.
D. The gene is also found on the X chromosome.
Answer:
c The gene is silenced on the Y chromosome.
The genotype of chromosomes classified the humans as males and females. The females have XX chromosomes, whereas the males have XY chromosomes.
The male individual can be the carrier for a Y-linked recessive allele only when the gene on the Y chromosome is silenced.
The Y-linked recessive allele is:
Y-linked diseases such as webbed toes, porcupine man, and hypertrichosis are the diseases of the Y-linked recessive allele. The Y-chromosome does not have the same genetic makeup as an X-chromosome. The women cannot be a carrier for a Y-linked disease as the women have purebred XX chromosomes.Thus, the male can only be a carrier when the gene on Y-chromosome is silenced.Therefore, the correct answer is Option C.
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A purebred plant that produces yellow seeds is crossed with a purebred plant that produces green seeds. The F1 plants have yellow seeds. What is the expected phenotypic ratio of seed color of the offspring of an F1 × F1 cross?
Answer:
3 yellow : 1 green
Explanation:
Let yellow seed trait be represented by Y allele and green seed trait by y allele.
Purebred yellow seed plant = YY
Purebred green seed plant = yy
YY x yy = Yy, Yy, Yy, Yy (all yellow)
This thus means that yellow seed trait is dominant over green seed trait.
F1 x F1 = Yy x Yy, resulting in YY, Yy, Yy, yy offspring.
Since Y is dominant over y,
YY, Yy, Yy = Yellow
yy = green
Hence, the expected phenotypic ratio of seed color of the offspring of F1 x F1 cross is 3 yellow : 1 green
In a Mendelian genetics cross where yellow seed color is dominant over green, the expected phenotypic ratio of an F1 x F1 cross is 3 (yellow):1 (green).
Explanation:The question is about predicting the phenotypic ratio of seed color in the F2 generation of a cross between two F1 plants, which came from a purebred yellow seed plant and a purebred green seed plant. In the classic Mendelian genetics framework, yellow and green seed colors are controlled by a gene with two different alleles. In this case, yellow is the dominant trait and green the recessive trait.
For the F1 x F1 cross, each F1 plant is heterozygous, meaning they carry both the yellow (Y) and green (y) allele. If we use a Punnett square to determine the expected outcome from their cross, we would have:
YY, Yy, and yy.
However, because the yellow (Y) is dominant, both YY and Yy will have a yellow phenotype. Thus, the phenotypic ratio expected of seed color in the offspring from F1 x F1 would then be 3 (yellow): 1(green).
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The reaction-center chlorophyll of photosystem i is known as p700 because
The actions of the rhomboid major on the scapula do not include __________.
Answer:
The answer is the rhomboids palpation
Explanation:
Subjects hand placed in the small of the back the rhomboids can be palpated through trapezium when the hand is moved backwards . Felt between the medial border of the scapula and vertebral column.
A 30-year-old male who manages his type 1 diabetes with glyburide presents at the emergency room reporting headache, confusion, and tachycardia. He has come from a party at which he drank two beers to celebrate running his first half-marathon. Which of the following is likely to be the cause of his complaints?A)Blood glucose levels are primarily a result of the timing, quantity, and character of food intake.B)Ingested glucose that is not needed for cellular metabolism circulates in the blood until it is taken up to meet cellular needs.C)Blood glucose levels are kept in a steady state by selective excretion and reuptake by the kidneys.D)Glucose that exceeds metabolic needs is converted and stored by the liver.
Glucose that exceeds metabolic needs is converted and stored by the liver
Explanation:
The patient is suffering from type 1 diabetes and takes glyburide which is an oral hypoglycemic.
In the glucose pathway, excess glucose will be converted to glycogen and will be stored in the liver. Under normal conditions, when the body needs excess glucose, like in the above case after a marathon run, glycogen converts back to glucose by the processes of glycogenolysis and gluconeogenesis.
Glyburide acts against increasing blood glucose levels by inhibiting glycogen conversion to glucose in the liver.
Hence, the blood lacks the necessary amount of glucose leading to hypoglycemic conditions. Further, an imbalance of glucose efflux and influx also lead to hypoglycemia.
Hyoglycemia can cause neurogenic and sympathoadrenal symptoms like headache, confusion, tachycardia. or postural orthostatic tachycardia syndrome.
Barbara is tired all the time and feels overwhelmed by her workload. She is disengaging from her work and fellow employees. Her production is slowing down. She complains of headaches a lot. She is probably suffering from:
Barbara is likely suffering from job burnout, a condition characterized by emotional exhaustion, cynicism, and a reduced sense of personal accomplishment, often associated with symptoms like headaches, depression, anxiety, and decreased productivity.
Explanation:Barbara's symptoms of feeling tired all the time, feeling overwhelmed by her workload, disengaging from work, complaining of headaches, and a slowdown in production suggest that she could be suffering from job burnout. These symptoms fall in line with the general sense of emotional exhaustion, cynicism towards one's job, and a diminished sense of personal accomplishment that characterize job burnout. Factors such as work overload, little positive feedback, unsafe work environments, and frustration dealing with bureaucracy contribute to this condition. Moreover, symptoms like headaches, depression, anxiety, and reduced mental performance are commonly associated with job burnout.
It's also important to note that Barbara could be experiencing co-occurring health issues such as depression, as severe job burnout can lead to depressive disorders. Being exposed to chronically stressful work conditions is a significant risk factor for job burnout. In certain fields, particularly human service jobs, job burnout occurs frequently and is a serious concern.
An organism is discovered that thrives in both the presence and absence of oxygen in the air. Curiously, the consumption of sugar increases as oxygen is removed from the organism's environment, even though the organism does not gain much weight. This organism _____.
Answer: Facultative anaerobe
Explanation: Facultative anaerobes are organisms that can live with or without oxygen supply. this organism processes carbohydrates and proteins as oxygen is taken away from its enviroment in order to derive oxygen.
The organism is a facultative anaerobe, which prefers oxygen for aerobic respiration but shifts to anaerobic respiration or fermentation when oxygen is scarce, increasing sugar consumption without significant weight gain.
Explanation:The organism described is a facultative anaerobe, which is an organism that grows better in the presence of oxygen but can also proliferate in its absence. The increased consumption of sugar as oxygen is removed from the organism's environment is typical for facultative anaerobes as they switch from aerobic respiration, which is more efficient and generates more ATP, to fermentation or anaerobic respiration when oxygen is unavailable. Despite consuming more sugar via less efficient metabolic pathways in the absence of oxygen, the organism does not gain much weight due to the continuous loss of matter as carbon dioxide and water, products of cellular respiration.
Therefore, the correct answer to the organism's behavior when oxygen is removed is B. The organisms are facultative anaerobes.
In the late 1960s, Robert Paine conducted landmark studies on diversity in the rocky intertidal zone, comparing the species diversity in control plots with diversity in experimental plots from which he removed the top predator, sea stars. After 5 years, 15 species of intertidal invertebrates lived in the control plots, while the experimental plots were dominated by only two species, one mussel and one barnacle. The process MOST likely responsible for the loss of species diversity in the experimental plots was:A. mutualism.B. predation.C. competitive exclusion.D. parasitism.E. resource partitioning.
Answer:
C: competitive exclusion
Explanation:
The competitive exclusion principle states that organisms living in the same community while competing for the same resources cannot coexist at a constant population rate.
Once some of the species within the community get a slight competitive edge over other species, they become dominant and this might lead to the extinction of the weaker species in the long run.
In the experimental plot, the removal of sea stars provided mussel and barnacle with a competitive advantage over other species within the community (sea stars are predators of mussels and barnacles). This led to the dominance of mussel and barnacle and the eventual extinction of other species within the experimental plot as compared to the control plot.
The correct answer is C.
Compared with rods, cones are__________.a more sensitive to any light and less sensitive to fine detail.b less sensitive to dim light and less sensitive to fine detail.c more sensitive to dim light and less sensitive to fine detail.d more sensitive to dim light and more sensitive to fine detail.e less sensitive to dim light and more sensitive to fine detail.
Answer:
e less sensitive to dim light and more sensitive to fine detail.
Explanation:
Rods and cones are light sensitive receptors located at the back of the eye. The human eye contains between 5 million and 7 million cones and 110000000-130000000 bars.
Function
Cones and canes are essential for vision. Together they are able to detect movement, light and color, and transmit that information back to the brain.
Rods
The rods are highly sensitive cells located in the outer area of the retina (the lining of the back of the eye). They are used in low light areas and are more acute to light, shape and movement changes. Rubs do not detect color.
Cones
The cones are located in the central fovea (central area of the retina). They are less sensitive than rods and require bright lighting. Cones are fundamental to our ability to see color.
The cones have a high resolution and can detect the color. Its maximum sensitivity is located at a wavelength around 555 nm (yellow light). This is the so-called photopic vision, the one used during the day to see things in detail and color, making use of direct vision, the one with the highest definition. During astronomical observation, cones can also be used in the case of bright objects, such as planets or stars, whose light is intense enough to detect color. Sometimes we will also notice it in deep sky objects, such as planetary nebulae with high surface brightness.
Darkness arrives, when the light diminishes, the canes begin to work. This is the so-called scotopic vision. The rods are located outside the optical axis, with their area of greatest sensitivity located at an angle of approximately 20 ° around the fovea, and a maximum sensitivity at 507 nm (green light) The rods are responsible for night vision , also being the most sensitive motion detectors. The well-known peripheral vision technique uses the rods to detect weak objects at the time of observation.
Here we can notice that the maximum sensitivity between the cones and the rods is different, 555 nm for the former and 507 nm for the latter. This has the effect that during the transition from vision with cones (photopic) to vision with rods (scoopic) the maximum response moves at shorter wavelengths. This is called the Purkinje effect, and it means that for weak sources the eye is more sensitive to blue, while for bright sources it is more red.
Cones are less sensitive to dim light and more sensitive to fine detail compared to rods. They are responsible for color vision and high visual acuity. Rods, on the other hand, work well in low light conditions and are involved in peripheral vision and motion detection.
Explanation:Compared with rods, cones are less sensitive to dim light and more sensitive to fine detail. Cones are specialized photoreceptors located in the fovea, where images are focused, and are responsible for color vision and high visual acuity. In contrast, rods are distributed throughout the retina, function well in low light conditions, and are involved in peripheral vision and motion detection.
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Which of the following statements is correct? a)All animals share a common ancestor. b)Sponges are diploblastic animals. c)Eumetazoans have three embryonic tissue layers. d)Most animal phyla belong to the clade Radiata. e)The origin of all extant animal phyla can be traced to the Cambrian explosion.
All animals share a common ancestor.
Explanation:
According to various phylogenetic gene sequence analysis, there are various evidences that proved all animals originated from a common ancestor.
Initially, it was stated that all organisms descended from a single cell which then gave rise to multicellular organisms. Organisms that descend from a common ancestor are closely related and grouped.
The lineage of the common ancestor can be traced in the neoproterozoic era.
The last common ancestor or the basal animal was sea sponge according to some researchers. The last universal common ancestor is called as the concestor.
If a neurotransmitter attaches to a ligand-gated channel and creates a potential that causes the inside of the excitable cell membrane to become less negative, and this potential travels only a short distance before dying out, it is considered to be a(n)______.
Answer:
If a neurotransmitter attaches to a ligand-gated channel and creates a potential that causes the inside of the excitable cell membrane to become less negative, and this potential travels only a short distance before dying out, it is considered to be a depolarized graded potential.
Explanation:
We talk about a graded potential when a small change in membrane potential occurs is proportional to the size of the stimulus, usually they do not travel long distances along the neuron membrane, but a short distance and decrease as they propagate to end disappearing. If a graded potential is strong enough it can lead to action potential.
The depolirization occurs when a stimulus reaches the neuron producing an electrochemical change or disturbance where ions (Na) enter the cell, and in this way the membrane potential becomes less negative (more positive). It propagates along the membrane, but when it reaches the axon, the release of neurotransmitters occurs, which when in contact with the adjacent neuronal membrane exerts on it the action of a stimulus (that is, ionic changes are made).
In skeletal muscle, a muscle fiber that generates a larger contractile force has a larger number of:
a. mitochondria present.
b. crossbridges formed.
c. actin molecules present.
d. nebulin and titin molecules present.
e. thick and thin filaments present.
Answer:
b.crossbridges formed
Measuring its costs and benefits in terms of energy spent and/or gained, predation:
A. results in a gain for one individual and neither a gain nor a loss for the other.
B. results in a gain for both individuals.
C. results in a gain for one individual and a loss for the other.
D. is a lose-lose interaction.
E. None of the answer options is correct.
Answer:
The correct answer is C. results in a gain for one individual and a loss for the other.
Explanation:
Predation results in one individual gaining energy, the predator, and as the principle of energy conservation says, therefore, the prey represents a loss of energy.
In predation, there is an energy transfer from the prey to the predator. The act of predation can thus be described as a gain for one individual (the predator), and a loss for the other (the prey). So, the correct answer is option C.
Explanation:The act of predation is a biological interaction where a predator, an organism that is hunting, kills and eats its prey, the organism that is attacked. Thus, in terms of energy spent and gained, predation results in a gain for one individual and a loss for the other. The predator uses energy to catch its prey, but in return, gains more energy from the prey itself, which is usually more than what was expended in the hunting process, thus resulting in a net gain. However, the prey loses both its life and any energy it had. So, the correct answer to this question is C.
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Which molecules can easily diffuse across a plasma membrane?
Gases like [tex]O_2[/tex] and [tex]CO_2[/tex] can easily diffuse across a plasma membrane.
Small, nonpolar compounds diffuse easily through plasma membranes. Examples of molecules are:
Oxygen : A tiny, nonpolar molecule that may pass through the lipid bilayer and diffuse from high concentrations (e.g., breathed air) to low concentrations (in cells).[tex]CO_2[/tex]: Another tiny, nonpolar molecule generated by cellular respiration. It diffuses from cells into the circulation and is exhaled from the lungs.Lipid-Soluble Molecules: Vitamins A, D, E, and K and steroid hormones like cortisol and testosterone can diffuse through the plasma membrane.These compounds can cross the plasma membrane's lipid bilayer without transport proteins. Simple diffusion happens down the concentration gradient from high to low concentrations.
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Small nonpolar molecules such as oxygen and carbon dioxide can easily diffuse through the lipid bilayer of the plasma membrane due to their size and lack of charge, while polar molecules, ions, and larger molecules require the help of transport proteins.
Explanation:The substances that can easily diffuse across a plasma membrane are typically small nonpolar molecules, such as oxygen and carbon dioxide. These molecules have no charge and can pass through the lipid bilayer of the cell membrane by simple diffusion. This is because the structure of the lipid bilayer allows these types of molecules to slip through the hydrophobic internal layer. However, polar substances, ions, and larger polar molecules like simple sugars and amino acids cannot easily cross the plasma membrane because their charge or size creates a barrier.
Transport proteins, including channel proteins and carrier proteins, provide a means for these substances to move across the membrane through a process known as facilitated diffusion. For example, glucose, which is a large and polar molecule, requires a specialized transport protein to enter the cell.
Which of the following diseases is NOT caused by an organism that enters the body through the gastrointestinal tract?
a.leprosy
b.polio
c.infant botulism
d.adult listeriosis
Answer:
The correct answer is A) Leprosy.
Explanation:
Leprosy is not caused by organisms entering the gastrointestinal tract. The polio virus is often transmitted by drinking water contaminated with the polio virus especially in regions with low hygiene standards. Infant botulism is caused by ingesting spores of the strain C. botulinum. In the small intestine, the spores start to colonize and multiply. The cause for adult listeriosis is in most cases the ingestion of food contaminated with L. monocytogenes.
Which statement about operons is false?
A) The structural genes in the operon are either all on or all off at a given time.
B) They consist of a cluster of genes with a single promoter.
C) Their DNA is transcribed into a single molecule of mRNA.
D) They are the units of transcriptional regulation in prokaryotes.
E) A repressible operon is turned off unless needed.
Answer: All the statement are correct.
Explanation:
In prokaryotes, an operon are unit of transcription regulation, operon are functioning unit of DNA of which consists of cluster of related genes with a single promoter and an operator.
The structural genes in the operon are either all ON or OFF at the same time, because they are controlled by a single promoter and an operator. Represible or inducible operon are all turned of off unless needed.
The DNA of an operon are generally transcribed into a single mRNA molecule which is polycistronic that codes for more than one protein
The statement 'A repressible operon is turned off unless needed' is false. Repressible operons are normally 'on' and are turned 'off' if the pathway they control is not needed.
Explanation:The statement 'A repressible operon is turned off unless needed' is false. In reality, a repressible operon is typically in the 'on' state and is turned 'off' only if there is no need for the pathway it controls. These operons are typically involved in anabolic pathways. The remaining options are all true: The structural genes in the operon are indeed all on or all off at a given time, they consist of a cluster of genes with a single promoter, their DNA is transcribed into a single molecule of mRNA, and they are the units of transcriptional regulation in prokaryotes.
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A woman who has blood type A positive has a daughter who is type O positive and a son who is type B negative. Rh positive is a trait that shows simple dominance over Rh negativeand is designated by the alleles R and r, respectively. A third gene for the MN blood group has codominant alleles M and N.
Which of the following is a possible phenotype for the father?
A) A negativeB) O negativeC) B positiveD) AB negativeE) impossible to determine
Answer:
C) B positive is the right answer
Explanation:
Codominance
ABO blood group is an example of co dominance. According to this
Two alleles are both dominant, So A and B are both co dominant, they will express and I allele which will be recessive.
Type A-------------- AA or Ai
Type B---------------BB or Bi
Type AB------------- AB
Type O--------------- ii
Rh antigen is dominant to Rh- (for positive and negative due to presence of Rh antigen
So for Rh+ = RR (homozygous dominant) and Rr (heterozygous dominant)
Rh-= rr (homozygous recessive)
Given conditions in the questions
Given type Alleles for type Allele for Rh antigen
Type A Woman A+ Ai Rr
Daughter Type O+ ii R?
Son type B- Bi rr
Father type Bi Rr or rr
• As mother is A positive, so definitely the son got his B allele from the father, so father should have allele.
• So as the daughter is ii, so it means the daughter is getting one i allele from mother and other i allele came from the father
• As the son have B allele which comes from father, so what is son getting from mother? Definitely he is getting i allele from mother because that the only way mother can donate to the son and not change his blood type.
• No we know that the son is rr for Rh antigen, which means that he is getting one r from mother because she must have a little r to donate to her son and one r from father
The father is either type B Positive BiRr or type B Negative Birr, but we have no other answer than type B positive in the multiple option. This is the possibility
The skeletal system is not static in structure. Bones are constantly being broken down and rebuilt.Complete the statement about bone remodeling. _______________ is/are responsible for breaking down bone during remodeling.
Answer:
Osteoblast, osteocytes and osteoclasts are responsible for breaking down bone during remodeling.
Explanation:
The process which is responsible during bone formation in equally in initial and later stages of remodelling, they are the bigger cells which are responsible for the synthesis and mineralisation of bone. Osteocytes are the rich in cells in the bone tissue, which is helpful in detection of the mechanical loading, also manages the in bone formation and bone resorption. Osteoclast are cells that disintegrate the bone to start normal bone remodelling and also contemplate bone loss in pathologic condition by growing its resorptive capacity.
As animals have evolved large body size, they have also evolved adaptations to improve the exchange of energy and materials with the environment. For example, in many larger organisms, evolution has favored lungs and a digestive tract with ________.
a. increased thickness
b. more branching or folds
c. larger cells
d. decreased blood supply
Answer:
b. more branching or folds
Explanation:
Animal size has increased during evolution on several animals, this has favored enhancements in gas exchange because this enlargement made diffusion too slow to cover internal distances in an efficient time, therefore branches and foldings were developed to cover distances quicker.
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An advantage of asexual reproduction is that
a) it enhances genetic variability in the species.
b) allows the species to endure long periods of unstable environmental conditions.
c) enables the species to rapidly colonize habitats that are favorable to that species.
d) produces offspring that respond effectively to new pathogens.
e) allows a species to easily rid itself of harmful mutations.
Answer:
c) enables the species to rapidly colonize habitats that are favorable to that species is the correct answer.
Explanation:
An advantage of asexual reproduction is that enables the species to rapidly colonize habitats that are favorable to that species.
Asexual reproduction where offspring is produced from the single parent and it carries the genes of that parent.
In asexual reproduction, an organism is formed without the exchange of genetic material
In the asexual reproduction there no fusion of gametes,
Examples of asexual reproduction are fragmentation,binnary fission, spore formation, budding.
Advantage of asexual reproduction
occurs very rapidly.one parent is required for reproduction.The population increases quickly when the conditions are suitable.A mutation that has risen to high frequency through a selective sweep shows a characteristic pattern in which only one allele is found for other loci that occur nearest the selected mutation. Which of the following contribute(s) to this pattern?
Answer:
Recombination is less likely to separate nearby allele from the favored one contribute(s) to this pattern.
Explanation:
Selective sweep is the reduction/elimination of the difference in the nucleotide (allele) in the mutation. This usually occurs in a rare allele. Mutation is the alteration in the existing gene of a species.Recombination is the exchange of DNA between the chromosomes of the parents. The mutation levels are prone to increase or decrease during recombination.Although the law of independent assortment is generally applicable, when two loci are on the same chromosome the phenotypes of the progeny sometimes do not fit the phenotypes predicted. This is due to__________.
Answer:
Linkage
Explanation:
The existence of two loci on the same chromosome (linkage) increases the chances of the genes on these loci to assort or be inherited together.
The closer the two loci on the chromosome, the higher the chances of the two genes being inherited together and further apart they are, the lower the chances because the two genes are more likely to be separated during recombination.
Hence, if two loci on the same chromosome produce progeny that do not fit the predicted phenotypes according to the law of independent assortment, it means the two loci are linked.
The answer is linkage.
Answer:Variation in single and multiple genes.
Explanation:
This is due to changes in the sequence of nucleotides in the GENE on specific loci located on the chromosomes of the DNA molecule.
The change may be due MUTATION. Therefore the expression of phenotype may be different from Madelia laws of independent assortments of genes, because the genes may be linked together and assort jointly,(linkage), or some genes may not have full dominance over the other(incomplete dominance and codominance). Some alleles of another gene may mask or conceal the expression of other allele
Example of these exceptions to the independent assortments are Incomplete dominance, codominance, sex –linkage, epistasis, complementary genes.
The adaptation that made possible the colonization of dry land environments by seed plants is most likely the result of the evolution of _____.
A. pollen
B. heterospory
C. cones
D. ovules
Answer:
The correct answer is A. pollen.
Explanation:
Pollen is a very efficient way to colonize and reach further. It's believed that this adaptation is the main reason why land plants could colonize so many environments. The fact that pollen is little and easily transported also was very important.
The key adaptation facilitating the colonization of dry land by seed plants was the evolution of pollen, enabling reproduction without the need for water.
Explanation:The adaptation that made possible the colonization of dry land environments by seed plants is most likely the result of the evolution of pollen. Pollen and seeds provided evolutionary advantages that enabled plants to reproduce and develop independently of water, which was crucial for their success on dry land. Pollen, as the male gametophyte, is encased in a protective coat to prevent desiccation and can be transported far from the parent plant, ensuring gene spread. This adaptation, along with the development of seeds, allowed seed plants to become widespread and diverse by being less reliant on moist environments for reproduction.
Concerning proton movement in mitochondria and chloroplasts, the __________ have a __________ H+ concentration. Choose one: A. lumen in chloroplasts; low B. matrix in mitochondria; high C. stroma in chloroplasts; high D. intramembrane space in mitochondria; high
Answer: The answer is D - Intramembrane space in mitochondria; high
Explanation:
Mitochondria is an organelle in the cell, it is known as the power House of the cell because of its role in the storage and production of energy. During electron transport chain the proton pumps generate a proton gradient which is stored in form of energy.
During proton pumping and electron transfer the electrons are passed down the chain from the matrix to the inter membrane of the mitochondria. The electrons move from higher concentration of level to lower energy level. Some of these are used to pump Hydrogen ions (H+), moving them out of the matrix into the inter membrane space.
The Intramembrane space of the mitochondria therefore has a a high concentration of Hydrogen ions (H+).
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A patient who reports nasal congestion is prescribed naphazoline, 2 drops in each nostril, every 6 hours, for 5 days. The patient reports increased nasal congestion after 10 days. On assessment the nurse learns that the patient had increased the drug frequency to every 3 hours for 10 days. Which statement describes what the nurse would interpret from this information?
Answer:
The patient has developed the condition called as rebound congestion
Explanation:
in rebound Condition, the situation of previous congestion becomes more severe and worse due to treatment of nasal congestion using medication .
rebound condition is caused due to overdose of the the medication used for treatment. Patient without knowing the consequences and trying to recover the condition quickly, increase the dose and frequency of nasal medication in the start of the nasal congestion which can result in severe condition and blockage of nasal breathing and can only be cured by surgery.
The _________________, which binds to a core promoter, consists of general transcription factors and RNA polymerase.
Answer:
The correct answer is basal transcription apparatus.
Explanation:
The basal transcription apparatus, which binds to a core promoter, consists of general transcription factors and RNA polymerase. This apparatus is made of proteins and synthesizes mRNA.
Collision of macromolecules in surface water produced particles that sink to the bottom, know as_______.a. Marine snowb. Macroalgaec. Nektond. Marine debris
Answer: It is called Marine snow.
Explanation:
Marine snow is the particulate matter that originate from the ocean. It is formed by collisions and combinations larger molecules(macromolecules) on surface water which later settles at the bottom. Marine snow may also be formed from decaying organic materials such a salp calony and house of a larvacean. Marine snow contains larger molecules of microorganism and protistans.
Black eyes are dominant to orange eyes, and green skin is dominant to white skin. Sam, a MendAlien with black eyes and green skin, has a parent with orange eyes and white skin. Carole is a MendAlien with orange eyes and white skin. If Sam and Carole were to mate, the predicted phenotypic ratio of their offspring would be _____.
a) 1 black eyes, green skin :
b) 1 black eyes, white skin :
c) 1 orange eyes, green skin :
d) 1 orange eyes, white skin
Answer:
ratio: 25% for each phenotype
Explanation:
See attached document for proper explanation of this case
All members of the plant kingdom have ________.A)oogamous sexual reproductionB)life cycle that includes both gametophyte and sporophyte generationsC)egg cells surrounded by a jacket of sterile cellsD)multicellular embryosE)All of the answer choices are correct.
Answer: Option D
All members of plants kingdom have multicellular embryos.
Explanation:
Kingdom plantae is the kingdom that includes all kind of plants, the seeds plant with and without flowers,green and brown algae, liverworts, mosses, ferns and so on. They all have multicellular embryos and they can undergo photosynthesids because of the presence of chloroplast.
This kingdom is divided into 5 divisions which are the peridophytes, bryophytes,thallophytes, gymnosperms and angiosperms. The bryophytes have life cycle of both sporophyte and gametophyte. Oogamus sexual reproduction is found in many protist of the order of algae and some bryophtes like fern and mossess. Egg cells surrounded by a jacket of sterile cells is found in Antheridia and bryophytes.