Explanation:
Due to the Vroman effect, albumin will initially attach and eventually be replaced by the IgM, which has a higher affinity for the material
The higher concentration of the albumin results in a greater initial surface concentration via diffusion, but it will eventually be displaced by the proteins with greater surface affinity (first the transferrin, and finally the IgM)
If more addition of albumin occurs, which has less affinity for the material surface, will have minimal effect if IgM is already adsorbed to material surface
Third-base wobble allows some tRNAs to recognize more than one mRNA codon. Based on this chapter's discussion of wobble, what is the minimal number of tRNA molecules necessary to recognize the following amino acids?
a. leucine
b. arginine
c. isoleucine
d. lysine
Answer:
a. leucine
Explanation:
In the case of the amino acid leucine alone, there are six different codons (TTA, TTG, CTT, CTC, CTA, and CTG) that can recognize and translate tRNA. Since there is only one codon to be translated, only one tRNA is required. The anti-codon in the tRNA identifies complementary positions on the codon, and due to the characteristic of the third-base motif, there is no 1: 1 ratio between the number of codons present and the required anti-codon.A micrograph of dividing cells from a rat, a diploid animal, shows a cell where 21 pairs of sister chromatids are being pulled apart.
Which stage of cell division could such a photo have been taken?
a) Anaphase 1 of meiosis
b) Anaphase II of meiosis
c) Anaphase of mitosis O
d) Telophase 1 of meiosis
e) Telophase of mitosis
Answer:
b) Anaphase II of meiosisExplanation:
1. Meiosis is the process of cell division in which one cell is divided into four daughter cell, each contains equal number of chromosome, half the number of chromosomes as compared to parental cell.
2. In meiosis I, DNA duplication occurs but the sister chromatids are not separated, only homologous pair of chromosomes are separated, so this is called reductional division.
3. In meiosis II, chromatids are pulled apart and and are separated into different chromosomes, so it is called equational division. There is no DNA duplication in meiosis II.
The correct stage of cell division for the photo showing 21 pairs of sister chromatids being pulled apart in a rat cell is Anaphase of mitosis.
Explanation:The correct answer to this question would be c) Anaphase of mitosis. During the anaphase stage of mitosis, the sister chromatids of each chromosome are being pulled apart to opposite ends of the cell. We know it must be mitosis because the cell in the micrograph is a somatic (body) cell with a diploid number of chromosomes. In a rat, the diploid number is 42 (21 pairs), matching the number of pairs of sister chromatids being pulled apart.
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Which of the following describes empirical evidence?
A
dogs are good pets
B
dogs can speak English
C
dogs have two eyes and two ears
D
pet dogs should be allowed at school prom
Empirical evidence is information that is verifiable through observation or experimentation. Out of all the presented options, the fact that 'dogs have two eyes and two ears' qualifies as empirical evidence.
Explanation:Empirical evidence refers to information that can be verified by observation or experimentation. In the listed options, C 'dogs have two eyes and two ears' best describes empirical evidence. This statement is factual and can be supported through observation and experiment, unlike claims that 'dogs are good pets', which is subjective, or that 'dogs can speak English', which is inaccurate, or that 'pet dogs should be allowed at school prom', which is an opinion.
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The back of a water droplet acts as a mirror. When light hits the back of the water droplet it bounces back to our
eyes. This is called
refraction
evaporation
conduction
reflection
Answer:
reflection
Explanation:
Please help. This is bio question.
Answer:
D: 1:2:1
Explanation:
what is the complementary RNA called?
Answer:
It is miRNAs or microRNA
I hope this helps
Answer: Complementary RNA (cRNA) is a copy of a strand of RNA that will bind to the appropriate region of the original molecule.
Explanation: If the original RNA stand had a base sequence of AUU, for example, the sequence of the cRNA strand would be UAA. I hope this helps :).
Like DNA, RNA has
bases.
RNA also contains four different bases. Three of these are the same as in DNA: adenine, guanine, and cytosine. RNA contains uracil (U) instead of thymine (T).
Final answer:
RNA, like DNA, consists of four nitrogenous bases, where thymine (T) in DNA is replaced by uracil (U) in RNA; adenine (A), cytosine (C), and guanine (G) are found in both.
Explanation:
Like DNA, RNA has nitrogenous bases that are essential for storing and transferring genetic information. DNA contains four bases: adenine (A), thymine (T), cytosine (C), and guanine (G). RNA also consists of four bases, but with a crucial difference: instead of thymine, it has uracil (U). The bases adenine, cytosine, and guanine are common to both DNA and RNA. In the structures of DNA and RNA, adenine pairs with thymine/uracil, and guanine always pairs with cytosine. This base pairing is essential for the replication of DNA and the synthesis of proteins, as RNA uses the information from DNA to assemble amino acids into proteins.
Roots also play an important role in water transport. Which plays the most important role in the movement of water through a plant--the absorption of water by the roots or the evaporation of water from the leaves
Answer: plant--the absorption of water by the roots
Explanation: plants root aids the absorption of water and nutrients from the soil. Water is an important constittuents in plants when's its not adequate it could lead to wilting of the plant. Attach to the root is root hair this is what help in absorption of water and nutrients from the soil before it is been transported round the cell by xylem tissue. Water is moved through osmotic pressure from the region of higher concentration to low. The process of photosynthesis in plant requires water without the function of the root for absorption plants cannot actively photosynthesis. Hence root play an important role in water transport in plant .
Transpiration is loss of water from leave surface. The water been loss is
how is the geologic time scale divided?
Answer: Geologic time is divided into units. Major changes in the earth's surface or climate and the extinction of species help to divide the time scale into smaller units.
Explanation: Rocks grouped within each unit contain a similar fossil record.
g You are performing DNA sequencing reactions as part of your job as a employee at GeneTech, Inc. In your current project, you are attempting to sequence a 700 bp DNA fragment sent by a customer. After the reaction has begun, you realize that due to a decimal error, you accidentally added the fluorescently labeled di-deoxyribonucleoside triphosphates (ddATP, ddGTP, ddCTP, ddTTP) at ten times higher than the desired concentration. What do you expect will be the consequence of this mistake?
Answer:
The consequence of this error will be the waste of material, since several nucleotides were placed in the reaction and will not be used.
Explanation:
There are several DNA sequencing protocols and although we don't know which one the GeneTech, Inc. employee is using, we can say that the fluorescence-labeled di-deoxyribonucleoside triphosphates will be used to make countless copies of the DNA particle that he wants to sequence.
As we know, DNA replication separates the DNA strands and uses each strand to make copies of new DNA molecules. These copies are made with the DNA polymerase enzyme by adding deoxyribonucleosides to the new tapes. As the enzyme is highly regulated and has a DNA strand as a template for the new strands, it will not add nucleotides beyond what is necessary.
Therefore, by placing excess di-deoxyribonucleoside triphosphates in the reaction, the consequence will be just the waste of product.
In guinea pigs, black fur is dominant and white fur is recessive. A homozygous dominant black guinea pig is bred with a homozygous recessive white guinea pig. The dominant allele is represented by F, and the recessive allele is represented by f. Create a Punnett Square and answer the following question: What is the probability of having white fur guinea pigs with this combination
Answer:
The following cross represents the parent genotype for the offspring whose phenotype is to be determined.
Homozygous dominant black guinea pig FF
X
Homozygous recessive white guinea pig ff
The probability of having white fur guinea pig offspring is zero in the first generation because all of the offspring will be Ff, which is the genotype for black fur.
Punnett Square:
F F
f Ff Ff
f Ff Ff
Hope that answers the question, have a great day!
Certain substances pass from the blood in the glomerular capillaries into the ultrafiltrate under normal circumstances. Other substances will only appear in the ultrafiltrate under abnormal circumstances. Categorize the following substances' appearance in the ultrafiltrate into normal and abnormal.
1. Albumin (a protein)
2. Amino acid
3. Glucose
4. NaCl
5. Plasma
6. Red blood cell
7. White blood cell
Final answer:
Under normal circumstances, amino acids, glucose, NaCl, and plasma appear in the ultrafiltrate. Albumin, red blood cells, and white blood cells are considered abnormal.
Explanation:
Under normal circumstances, the appearance of substances in the ultrafiltrate is as follows:
Normal: Amino acids, glucose, NaCl, and plasma
Abnormal: Albumin (a protein), red blood cells, white blood cells
In normal conditions, substances like albumin and blood cells should not be found in the ultrafiltrate. If they are present, it indicates abnormal circumstances such as kidney damage or disease.
7. How are a white dwarf and a planetary nebula related?
Final answer:
A white dwarf is the core remnant of a star that has shed its outer layers to form a planetary nebula, and eventually cools into a white dwarf. There are more white dwarfs than planetary nebulae because nebulae are short-lived. In stellar evolution, white dwarfs can be involved in events like novas and Type Ia supernovae when in binary systems.
Explanation:
A white dwarf and a planetary nebula are related in the final stages of a star's life cycle. When a star of a certain size exhausts its nuclear fuel, it expands and sheds its outer layers, creating a planetary nebula. The core that remains eventually becomes a white dwarf, which is the dense remnant of the star's core. Over time, the planetary nebula disperses into space, leaving the white dwarf. This explains why there are more white dwarfs than planetary nebulas in the Galaxy since the planetary nebulas are ephemeral, while white dwarfs can exist for billions of years before cooling to become black dwarfs.
In the Orion Nebula, an active star-forming region, it would be unlikely to find white dwarfs because this nebula is relatively young, and stars within it have not yet evolved to that stage of their life cycle. Additionally, in binary star systems, a white dwarf can receive material from its companion star, potentially leading to novas or even Type Ia supernovae if certain conditions are met, such as exceeding the Chandrasekhar limit or the merger of two white dwarfs.
White dwarfs differ from neutron stars in that a white dwarf is supported against gravity by electron degeneracy pressure, whereas a neutron star is supported by neutron degeneracy pressure. Both are dense, collapsed stars, but neutron stars are the denser and smaller remnants of supernova explosions of more massive stars, while white dwarfs are the remnants of less massive stars.
Tryptophan synthase is one of the enzymes synthesized from the trp mRNA. In wildâtype E. coli, the trp mRNA has a short halfâlife, but the tryptophan synthase halfâlife is much longer. To investigate how changes to the stability of the enzyme or its mRNA affect enzyme activity, two strains of E. coli were engineered. Strain A stabilizes the trp mRNA and strain B rapidly degrades tryptophan synthase. The wildâtype, A, and B strains were grown in a medium with glucose as the sole carbon source. After several generations, tryptophan was added to all three cultures and tryptophan synthase activity was measured periodically.
Select the statements that describe the expected change in tryptophan synthase activity after the addition of tryptophan
Note: Evaluate each condition as a simple model, where changes in the stability of trp mRNA or tryptophan synthase do not elicit secondary effects in the cells
A. In the wild-type strain, trp mRNA is unstable and will be degraded rapidly. As the cells continue to divide, trýptophan synthase will be diluted out and enzyme activity per cell will decrease
B. In strain A, the trp mRNA is stable and tryptophan synthase will continue to be synthesized However, as the cultures grow the tryptophan synthase will be diluted and the amount of enzyme activity will slowly decrease
C. In the wild-type strain, the trp mRNA is degraded rapidly. However, since the tryptophan synthase is stable, the enzyme activity per cell will remain unchanged as the cells divide
D. In strain A, the trp mRNA is stable but the presence of tryptophan in the medium will cause the degradation of tryptophan synthase and abruptly decrease enzyme activity
E. In strain B, since both the trp mRNA and tryptophan synthase are rapidly degraded, the enzymee activity will drastically decrease.
Answer:
Explanation:
we have three strains.
the first is the wild type which has tryptophan mRNA which gets degraded rapidly and tryptophan synthase which is stable. therefore, in the presence of tryptophan, no new mRNA is formed and whatever tryptophan synthase was present would be diluted out as cells continue to divide.
Strain A has stable mRNA and stable tryptophan synthase. so since the mRNA is present, new enzyme will continue to form from it. However, since no new mRNA is formed, the enzyme activity will get diluted as cells divide, but at a much slower rate than the wild type.
in Strain B , synthase is rapidly degraded and no new mRNA is formed. so the tryptophan synthase activity would decrease drastically.
Based on the conditions presented, we can expect a decrease in tryptophan synthase activity in all strains upon the addition of tryptophan, due to factors like mRNA instability, enzymatic dilution, and rapid enzymatic degradation.
Explanation:To evaluate the potential changes in tryptophan synthase activity upon the addition of tryptophan, the following can be considered:
A. In the wild-type strain, the trp mRNA is degraded fast, leading to the dilution of tryptophan synthase as cells divide and a subsequent decrease in enzyme activity per cell. B. In Strain A, trp mRNA is stable, hence, continuous synthesis of tryptophan synthase is expected. However, the increase in cell count due to the growth of the cultures can lead to dilution, thus enzyme activity can decrease over time.E. For Strain B, both trp mRNA and tryptophan synthase are rapidly degraded. This leads to an almost immediate and drastic reduction in enzyme activity.Learn more about Tryptophan Synthase Activity here:https://brainly.com/question/28136635
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Which do you think is most important about sleep? Write 2-3 sentences to explain
Answer:Sleep helps keep your heart healthy
A regular sleep pattern can help to lower the levels of stress and inflammation to your cardiovascular system, which in turn can reduce your chances of a heart condition.
Explanation:
Hope this helps
Stay Safe
plz give me brainiest
Answer:
sleep helps your body get energy for the next day.
It is true beacuse you learn it in school and the next day when you wake up you dont feel exhausted or weak. That only happens when you run outside too much. you shouldnt do that during this time, even during quarinten
Explanation:
The saying "Don’t judge a book by its cover." could be applied to the topic of evolution. For example, humans share 75% of their DNA with chickens. Biologists point to this as evidence that humans and chickens once shared a common ancestor. The advent of DNA technology has given scientists the tools with which to examine how closely related certain species are. DNA analysis allows scientists to construct phylogenetic trees whose branches link together the relatedness of different organisms.
Answer:
The DNA analysis is very important in evolutionary biology because the level of homology among sequences can be used for tracing the evolutionary relationships among populations, species, genera, families, orders, etc.
Explanation:
The DNA is key for phylogenetic inference because the order of the nucleotide bases in a DNA sequence enables to classify the biodiversity in the tree of life
Final answer:
The saying "Don't judge a book by its cover" underscores the importance of DNA analysis in understanding evolutionary relationships. It illustrates that species with similar appearances may be genetically distant, while those looking different can share significant genetic overlap, such as humans and chimpanzees sharing around 98% of their DNA, indicating a common ancestry.
Explanation:
The saying "Don't judge a book by its cover" applies to evolution by highlighting that the external appearance of species can be deceptive when it comes to their genetic relatedness. With the aid of DNA technology, scientists can directly compare the genes of different species. For instance, humans and chimpanzees share approximately 98% of their DNA, which suggests a recent common ancestor, supporting the theory of evolution.
Phylogenetic trees are constructed using DNA analysis to illustrate the connections between species. The more similar the DNA sequences are, the more recently two species likely shared a common ancestor. Mutations serve as a molecular clock to trace back the lineages and construct these trees. These genetic similarities and differences are powerful tools for understanding the vast evolutionary journey of life on Earth.
What is one half or one rod of the
condensed chromatin called?
A DNA
B chromatin
C chromatid
D chromosome
Final answer:
One half or one rod of condensed chromatin is known as a c) chromatid. The two identical halves of a chromosome are sister chromatids, and they are connected at a structure called the centromere. The first level of DNA organization is aided by histones.
Explanation:
One half of condensed chromatin is called a chromatid. During the cell cycle, specifically after the DNA has replicated during the S phase, eukaryotic DNA condenses and coils into an X-shaped structure known as a chromosome. This chromosome is composed of two identical parts referred to as sister chromatids. These sister chromatids are held together at a region termed the centromere. Before the cell proceeds to division through mitosis, these chromatids are attached at their centromeres, and they will eventually separate to be allocated to each daughter cell.
The correct answer is C. chromatid. Furthermore, in a eukaryotic cell, the first level of DNA organization is maintained by histones which help package the DNA into structural units called nucleosomes. The histones and chaperone proteins aid in condensing the DNA and preventing it from becoming tangled.
How many cells are produced in Mitosis
Answer:
Mitosis produces two diploid somatic cells that are genetically identical to each other and the original parent cell.
Explanation:
When a cell divides by way of mitosis, it produces two clones of itself, each with the same number of chromosomes.
Can someone help I’m really going crazy because of quarantine pleaseHELP!
same i miss my friends :((
Answer:same I miss everyone but I think we are going back to school soon
Explanation:
A, B, C and D are genes that are linked on the same chromosome. Given the following recombination frequencies, what is the gene order? A-B 15%, B-C 26%, C-D 30%, A-C 11%, A-D 19%, and B-D 4%. (Hint: what are recombination frequencies equivalent to?)
Answer and Explanation:
We need to know that 1% of recombination frequency = 1 map unit = 1cm. And that the maximum recombination frequency is always 50%.
The map unit is the distance between the pair of genes for which one of the 100 meiotic products results in a recombinant one.
The recombination frequencies between two genes determine their distance in the chromosome, measured in map units. So, if we know the recombination frequencies, we can calculate distances between the four genes in the problem and we can figure the genes order out. This is:
Recombination frequencies:
1% of recombination frequency = 1 map unit (MU)A-B 15% = 15 MUB-C 26% = 26 MUC-D 30% = 30MUA-C 11% = 11 MUA-D 19% = 19 MUB-D 4% = 4MUWe know that 15 plus 4 equals 19. So we can infer that we have the gen sequence A--B--D, because the distance between A-B= 15, and the distance between B-D=4, and finally the distance between A-D=19.
A----B----D
15↔4
19
We also know that 11 + 19 = 30. So we can assume that C is next to A at a distance of 11 MU.
C----A-----B---D
11 15 ↔ 4
11 ↔19
30
The gene order is C, A, B, D
The gene order on the chromosome, based on given recombination frequencies, is most likely B, D, A, C. The recombination frequency corresponds to the relative distances between the genes on the chromosome, and is measured in centimorgans (cM).
Explanation:In order to determine the gene order based on given recombination frequencies, we'll use the principle that the recombination frequency between two genes corresponds to their relative distances on the chromosome. Recombination frequencies are measured in centimorgans (cM), where 1 cM equals a 1% chance of recombination occurring between two genes.
First, sum the given frequencies: A-B (15%) + B-C (26%) + C-D (30%) = 71%. However, B-D is only 4%, implying that B and D are very close. The discrepancy can be explained by double crossovers, which are not counted. Thus, the order is most likely B, D, A, C, based on their respective distances from each other.
Next, to find the distance between genes, subtract the lesser frequency from the greater: B-A (15%) - B-D (4%) = 11 cM, and C-D (30%) - C-A (11%) = 19 cM. We can now confirm that the gene order (from least to greatest) appears to be B, D, A, C.
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Billy is an 11-year-old boy who was admitted to hospital with dehydration. His parents said that over the last few months, he had seemed tired and lacking in energy, and that despite eating well and drinking large quantities of fluids he had lost about 7 kg (15 lbs) in weight. He also seemed to be going to the toilet frequently, not only during the day but also during the night. On examination, there was evidence of dehydration (sunken eyes, loss of tissue turgor); his pulse rate was 115 BPM, blood pressure was 95/55 mmHg, respiratory rate was 20 breaths per minute, and his breath smelled of acetone.
1. Comment on the history and findings following the physical examination of Billy.
2. Comment on and interpret the test strip results. Are they consistent with your provisional diagnosis based on the history and physical examination findings?
Answer:
Billy showed signs of type 1 diabetes.This is because,
1.His dehydration was from polyuria- excessive urination.
2. because of failure of his pancreas to secrete insulin, for cellular utilization of glucose by beta cells- he retains glucose in his blood, and lost a lot of glucose in his urine, thus water follow by osmosis, leading to polyuria causing dehydration.
His urge to replace lost glucose and fluids leads to great thirst polydipsia, another symptoms of type1. However, his drop in weight is because, the solute concentration of blood glucose is high due to lack of insulin.Therefore large amount is lost in the urine, with large volume of fluids leading to drop in weight, and energy because cells lack glucose for biochemical reaction of energy production.
The dehydration leads to sunken eyes, loss of tissue turgor, and low fluid levels results in drop in blood pressure,-Hyportension.
Due to lack of glucose in the blood gluconeogensis sets in. Fats is converted to glucose by the liver to supplement the lost glucose. The major by product of this reaction is a ketone-acetone. This explains the reason for the smell of acetone in his breath.
2, The Test strip indicated large quantity of ketones. above ,3.0mmol/L .This is a confirmation of the Type 1 diabetes mellitus in Billy.As more fats molecules are broken down to glucose.
Explanation:
Billy's history and physical examination findings suggest he may have type 1 diabetes and be in a state of ketoacidosis. The test strip results are consistent with this diagnosis.
Explanation:The history and physical examination findings of Billy suggest that he may have type 1 diabetes. The symptoms of tiredness, weight loss, increased urination, and dehydration are common signs of diabetes. His elevated pulse rate and blood pressure along with the smell of acetone on his breath indicate that his body may be in a state of ketoacidosis, which can occur in uncontrolled diabetes.
The test strip results are consistent with the diagnosis of diabetes. The presence of glucose in the urine indicates that Billy has high blood sugar levels, which is a characteristic of diabetes. The presence of ketones in the urine confirms that he is indeed experiencing ketoacidosis.
Based on the history, physical examination findings, and test strip results, it is likely that Billy has type 1 diabetes and is currently in a state of ketoacidosis. He should be treated immediately to restore his fluid balance, stabilize his blood sugar levels, and prevent further complications.
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Identify each statement as TRUE or FALSE. a. Plants are defined as multicellular, eukaryotic, photosynthetic autotrophs. b. Plants are defined by their chloroplasts, which contain chlorophyll a and b. c. Charophytes and land plants share four derived traits that suggest they share a relatively recent common ancestor. d. Charophytes are embryophytes.
Answer:
A. True
B. True
C. True
D. False
Explanation:
A. Plants in general are multicellular organisms, they are often referred to as eukaryotic autotrophs because they are make their own foods. They carry out photosynthesis, meaning the obtain their energy from sunlight.
B. Plants can be characterized by the presence of chloroplast, an organelle that helps in absorption of sunlight, because of the presence of green pigment called chlorophyll. There two major forms of chlorophyll in plants which are chlorophyll a and b.
C. Charophytes are a type of green algae, but several studies shows they have similar traits to land plants compared to other algae, making them have a common ancestor as plants. Some of the traits similarities to land plants compared to other algae are: they have similar enzymes, they have similar type of cell division, also their sperm is also similar to that of land plants.
D. Charophytes are not embryophytes. Embryophytes are also referred to as land plants, they are multicellular in nature, they include ferns and mosses. Charophytes are a type of green algae similar in traits embryophytes. Although embryophytes are a sister group to charophytes, they are not the same.
Final answer:
The statements identify key characteristics defining plants as multicellular, eukaryotic, photosynthetic autotrophs with chlorophyll in chloroplasts, shared traits between charophytes and land plants, but incorrectly classify charophytes as embryophytes.
Explanation:
Identify each statement as TRUE or FALSE regarding plants:
a. Plants are defined as multicellular, eukaryotic, photosynthetic autotrophs. TRUEb. Plants are defined by their chloroplasts, which contain chlorophyll a and b. TRUEc. Charophytes and land plants share four derived traits that suggest they share a relatively recent common ancestor. TRUEd. Charophytes are embryophytes. FALSE
Plants, being multicellular eukaryotes, have distinctive features including chloroplasts containing chlorophyll for photosynthesis. Though charophytes share common traits with land plants pointing to a shared ancestry, they are not classified as embryophytes, highlighting the distinction and evolutionary pathway between aquatic and terrestrial plants.
What is the Function of cambium
Answer:
the function of the cambium: layer of actively dividing cells between xylem (wood) and phloem (bast) tissues that is responsible for the secondary growth of stems and roots (secondary growth occurs after the first season and results in increase in thickness).
Explanation:
Answer:
Your answer would be "B"
Explanation:
Select the correct match. Two plants in the same family are in the same _______. If two plants are in the same order, are they in the same family? If two plants are in the same family, are they in the same order? Do two plants more likely have a recent shared common ancestor if they are in the same order or in the same class? If plant species A and plant species B are in the same family, and plant species C is in a different family, which pair of species listed likely has the more recent common ancestor?
Answer:
The correct answers are "order; no; order; plant species A and plant species B".
Explanation:
Species are categorized in different taxonomic ranks according to physical and genetic features that establish the degree of similarity between species. For instance, two species that are in the same order have a more recent common ancestor than two species that have the same class. Also, two species that have the same genus share the other taxonomic ranks that are in an upper section. In this case, two plants in the same family are in the same order (the next taxonomic rank) however, two plants that are in the same order not necessary are in the same family.
Final answer:
Plants are classified into hierarchical taxonomic groups. Two plants in the same family are in the same order, but two plants in the same order are not necessarily in the same family. The more recent common ancestor can be determined based on whether the plants are in the same order or the same class.
Explanation:
In biology, organisms are classified into hierarchical taxonomic groups. Two plants in the same family are in the same order. However, two plants in the same order are not necessarily in the same family. The closeness of a common ancestor can be determined by looking at whether the plants are in the same order or in the same class. If plant species A and plant species B are in the same family, while plant species C is in a different family, the pair of species A and B likely has the more recent common ancestor.
Blood flow
The necessity of the normal pattern of blood flow through the heart is graphically illustrated in a pathology known as Transposition of the Great Vessels (TGV). About 4/10,000 infants are born with the congenital heart defect TGV. In this disorder, the origin of the aorta and the pulmonary trunk are switched, however the destinations of these vessels are normal. You may have to do some online research to find a diagram and a description in order to understand this pathology.
1. Select each answer choice that describes the NORMAL sequence of structures through which blood flows, starting with vena cavae. Note that not all structures are listed and that more than one answer may be correct.
a. Bicuspid → right ventricle → aortic valve → pulmonary circulation
b. Right atrium → semilunar valve right ventricle → AV valve → systemic circulation
c. Left atrium → bicuspid → left ventricle → AV valve → systemic circulation
d. Tricuspid → right ventricle → pulmonary valve → pulmonary circulation
Answer:
D. Tricuspid → right ventricle → pulmonary valve → pulmonary circulation
Explanation:
The circulatory system is the system of organs and blood vessels which is associated with the circulation of the blood in the body.
The circulatory system is divided into two portions: the systemic circulatory system (body) and the pulmonary circulatory system (lungs).
The vena cava brings the oxygen-poor blood from the body to the right atrium from where the blood is transferred to the same side of the ventricle controlled by the tricuspid valve (atrioventricular valve).
From the right ventricle, the blood is pumped to the pulmonary artery which transports the blood to the lungs where the oxygen will be exchanged.
Thus, Option-D is correct.
In the neuron, the sodium-potassium pump helps to
Answer:
The sodium-potassium pump carries out a form of active transport—that is, its pumping of ions against their gradients requires the addition of energy from an outside source. That source is adenosine triphosphate (ATP), the principal energy-carrying molecule of the cell.
Explanation:
The sodium-potassium pump is vital for establishing a neuron's resting potential by creating a negative charge inside the neuron and using ATP to pump ions against their concentration gradients.
Explanation:The sodium-potassium pump plays a crucial role in maintaining the resting potential of a neuron. This active transport mechanism uses ATP to move sodium ions out of the neuron and potassium ions into it, both against their concentration gradients. For every ATP consumed, three sodium ions are exported and two potassium ions are imported, which helps to establish a net negative charge inside the neuron. The difference in charge across the cell membrane is essential for the transmission of nerve impulses as it keeps the neuron ready to respond to stimulation.
The sodium-potassium pump is a mechanism of active transport that moves sodium ions out of cells and potassium ions into cells. It helps to maintain a difference in charge across the cell membrane of the neuron, creating an electrical gradient called resting potential. The pump moves both ions from areas of lower to higher concentration, using energy from ATP and carrier proteins in the cell membrane.
Use the diagram below to explain the difference between primary and secondary succession and give an example for each one
Answer:
Primary succession occurs following an opening of a pristine habitat. Secondary succession is a response to a disturbance.
Primary and secondary succession are two types of ecological succession, which describe the processes of ecosystem development and change over time. Here's an explanation of the differences between primary and secondary succession, along with examples for each:
Primary succession occurs in areas that are devoid of soil or lack any significant life forms. It starts from bare rock, lava flows, sand dunes, or newly formed land surfaces. The process begins with pioneer species, such as lichens and mosses, that can colonize the barren environment.
Secondary succession occurs in areas that have previously supported life but have been disturbed or disrupted by events such as fires, hurricanes, or human activities (e.g., deforestation, abandoned agricultural land).
Therefore, primary succession occurs in areas without soil or previous life, starting from bare rock or other non-living substrates. Secondary succession occurs in areas with existing soil and previous life, following disturbances that disrupt the established ecosystem.
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You have a 15kb plasmid. EcoRI cuts the plasmid into 7kb and 8kb pieces, and BamHI cuts the plasmid into 1kb and 14kb pieces. If you do a double digest with both EcoRI and BamHI on your plasmid, which of the following band patterns are possible on your gel?a. Bands at 1kb, 6kb and 8kb.b. Bands at 1kb, 7kb and 8kb.c. Bands at 1kb, 6kb and 14kbd. Bands at 7kb, 8kb and 14kb.
Answer:
a. Bands at 1kb, 6kb and 8kb
Explanation:
The EcoRI and BamHI are the restriction enzymes which cut the DNA sequence especially a plasmid at specific sites called the restriction sites.
The restriction enzymes produces bands of specific length therefore these restriction enzymes are used to estimate the approximate length of the DNA.
In the given question, the
1. EcoRI- produces two strands of 7 kb and 8 kb
2. BamHI- produces two strands of 1kb and 14kb
This shows that the length of DNA sequence is 15kb
But when the DNA strand are digested with both the enzymes simultaneously then it will produce three bands as:
i) 14 kb can be broken down in 2 bands of 6 kb and 8 kb
ii) 1 kb band is already produced by the Bam HI.
This shows that 1+6+8= 15 kb
Thus, Option-A is correct.
which of this is a cause if loss of genetic diversity?
1. climate change
2. invasive species
3. pollution
4. interbreeding of animals
Answer:
2 invasive species
Explanation:
If one species over grows the other there would be little to no genetic diversity. :)
Answer: Its D: Interbreeding of animals
Which of the following statements is not true regarding mutations? A. Mutations are always harmful B. Mutations may be helpful C. Mutations generate the raw material for natural selection D. Mutations create variety in the gene pool
Mutations are changes to an organism's DNA and can be harmful, advantageous, or neutral. They are not always harmful.
Explanation:Mutations are changes to an organism's DNA and are an important driver of diversity in populations. Some mutations are harmful, some are advantageous, and some are neutral. Advantageous mutations lead to changes that improve an individual's survival and/or chances of reproduction. For example, the mutation for resistance to insecticide in mosquitos improved their survival. Neutral mutations have no effect on survival or reproduction. The statement that mutations are always harmful (A) is not true, as mutations can be beneficial or neutral as well.
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The statement that mutations are always harmful is not true. Mutations can be beneficial, neutral, or harmful and are essential for evolution as they create genetic diversity upon which natural selection can act.
Explanation:The statement that is not true regarding mutations is: Mutations are always harmful. This is incorrect because mutations can have various impacts, including beneficial, neutral, or harmful effects on the organism. Mutations are the primary source of genetic variation, which is essential for evolution. They provide the raw material upon which natural selection acts. Most mutations are indeed neutral, having no significant effect on an organism's fitness. However, some mutations may confer an advantage in certain environments and will be selected for, thereby becoming more common in the population. Conversely, harmful mutations that negatively impact survival or reproduction are typically eliminated from the population by natural selection. In this way, mutations contribute to the diversity in a population's gene pool and play a critical role in the process of evolution.
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