Answer: a. True
Explanation: These postulates were formulated by Robert Koch as a result of his experiments with healthy and infected mice. The postulates are often used to determine if a phatogen causes a disease. These are:
The pathogen must be present in sick individuals but not in healthy ones. The pathogen must be isolated from individuals and cultivated in a pure culture.The pathogen cultivated must cause sickness when it's injected in suceptible inviduals.The pathogen must be isolated from the injected individuals and must be exactly as the first one.
How many lobes are found in Bufo Marinus liver?
a. One
b. Two
c. Three
d. Four
Answer:
c. Three
Explanation:
There are 3 lobes found in Bufo Marinus liver.
The right lobe, the left anterior lobe, and the left posterior lobe.
The liver is not primarily an organ of digestion, it does secrete a digestive juice called bile.
What type of behavior is a bird song, learned or innate?
Answer:
The tendency of the bird or the inclination to sing is innate in birds, however, the song, which is sung is learned. There exists an innate tendency to sing a song whether the bird is in seclusion or not. The birds possess sensitive/critical periods when they can produce memories of the songs they learn. The anterior forebrain pathway or the anterior vocal pathway is the template for learning a song by the bird.
A woman comes to your genetic counseling center because she knows that Huntington disease occurs in members of her family. Her paternal grandfather was afflicted, but so far her father shows no symptoms. Her two great-great grandmothers on her father's side were healthy well into their 90s, and one of her great-great grandfathers died of unknown causes at 45. Testing for Huntington disease is extremely expensive, but she is concerned that she may fall victim to this disease and wants to plan her life accordingly. After examining her pedigree you advise her to A. not get tested because there is no possibility that she is homozygous. B. not get tested because only males in her family get the disease. C. get tested because her father could be a carrier. D. not get tested because her father is only a carrier and it is very unlikely her mother is a carrier. E. not get tested because her 40-year-old father shows no symptoms.
Answer:
C. Get tested because her father could be a carrier
Explanation:
As Huntington Disease is an inherited dominant disease, it means that is enough the presence of one allele to develop illness. Her paternal grandfather was affected but not her father great-great grandmothers. Is her father, carrier of an HD allele? The problem here is that we can’t be sure her father is not a carrier, because although he’s 40 years old and doesn’t show symptoms of illness, HD uses to appear between 30 to 50 year old ages, but it can appear at any age. Eventually a possibility for the disease to appear, is still possible.
Because they alter the reading frame of all base-pair triplets, base-pair additions and deletions are collectively referred to as:
a. inversions
b. transversions
c. framshift mutations
d. point mutations
e. codon mutations
Answer:
The correct answer is option c. "frameshift mutations".
Explanation:
The reading frame of a gene is based on base-pair triplets, starting from the start codon until the ribosome encounters with the end codon. Base-pair additions and deletions are collectively referred to as frameshift mutations because they alter the reading frame of the gene. Base-pair additions and deletions break down the original sequence of the gene triplets, which alters the open reading frame and usually results in the production of non active proteins.
The antibiotic ciprofloxacin often is prescribed for serious cases of food poisoning caused by the bacterium Campylobacter jejuni, which is common in the intestines of farm animals and is not harmful to them, but may cause acute food poisoning in humans. Which of the following is correct given prolonged use of this antibiotic?
(A) Prior to antibiotic treatment, most Campylobacter are ciprofloxacin-resistant.
(B) Ciprofloxacin treatment kills or halts the growth of the sensitive strains, yet the resistant strains survive.
(C) Repeating the treatment of the same patient or a population multiple times results in a strain of Campylobacter that is more susceptible to the antibiotic.
(D) Treatment with ciprofloxacin causes the patient to evolve resistance such that the next time the patient gets an infection, it will no longer be effective.
Answer:
(B) Ciprofloxacin treatment kills or halts the growth of the sensitive strains, yet the resistant strains survive.
Explanation:
Campylobacter jejuni is the causal agent of the food-borne infection with the highest incidence in Europe. Both poultry and wild birds are a major reservoir of the bacteria, the bacteria lives in the intestines of warm blooded animals but is only quite harmful to humans. It has been identified that the diversity fo genes is responsible for its resistance to antibiotics.
Name one enzymatic step of the gluconeogenic pathway that constitutes a phosphoryl group transfer. ___________
Answer:
Final step: Glucose-6-phosphate is converted to glucose
Explanation:
Gluconeogenesis is the synthesis of glucose occuring during hypoglycaemia. The final step is catalyzed by glucose phosphatase which removes the phosphate group from glucose-6-phosphate.
Prokaryotes that are round are called spirochetes.
a. True
b. False
Answer:
False
Explanation:
Spirochetes have a spiral morphology, they are prokaryotes specifically they are bacteria. Spirochetes have flexible walls
Bacteria that are round in shape are called cocci.
Which are the likely consequences of insufficient fat in the diet of a victim of an eating disorder?
A. Weight loss
B. Bone thinning
C. Disruption of the menstrual cycle
D. Sterility
E. All of the above
The correct answer is E. All of the above
Explanation:
Eating disorders include multiple mental disorders such as anorexia, bulimia, rumination disorder or binge eating disorder in which individuals have negative eating habits that have serious consequences on their health including both mental and physical aspects.
In terms of physical health most eating disorders imply the individual restricts its diet or does not consume the nutrients or substances that are necessary to be healthily including insufficient fat, this leads not only to weight loss but also to disruption in the menstrual cycle as this lead to hormonal imbalances and therefore this might also be linked to sterility. Besides this, individuals with eating disorders are more prompt to develop conditions such as kidney or osteoporosis that occurs as bones thin or lose density due to the lack of nutrients. Thus, all of the options are likely the consequences of insufficient fat in eating disorders.
Biologists think that endosymbiosis gave rise to mitochondria before plastids partly because
a. the products of photosynthesis could not be metabolized without mitochondrial enzymes.
b. all eukaryotes have mitochondria (or their remnants), whereas many eukaryotes do not have plastids.
c. mitochondrial DNA is less similar to prokaryotic DNA than is plastid DNA.
d. without mitochondrial CO2 production, photosynthesis could not occur.
Answer:
b. all eukaryotes have mitochondria (or their remnants), whereas many eukaryotes do not have plastids.
Explanation:
It is a fact that among eukaryotes, all the reresentatives of the Domain (Eukarya) have mitochondria (more or less modified) but not all of them posses plastids.
Indeed, only a selected group of eukaryotes have acquired plastids. This group includes the autothropic algae (e.g. green algae or Chlorophytes, red algae or Rodophytes, brown algae or Phaeophyceae, etc), briophytes, and traqueophytes (vascular plants, pteridophytes, conifers and flowering plants). The endosymbiotic event that led to the formation of a plant cell initially involved an eukaryotic heterotrophic cell (that already had mitochondria and nucleus) which phagocyted a cyanobacterium. Over time, this cyanobacterium transformed into the plant cell's plastid.
Which of the following is true of unsaturated fats?
(A) They are more common in animals than in plants.
(B) They have double bonds in their fatty acid chains.
(C) They generally solidify at room temperature.
(D) They contain more hydrogen than do saturated fats having the same number of carbon atoms.
Answer:
The correct answer is option B.
Explanation:
The fatty acids are differentiated into unsaturated and saturated fatty acids on the basis of the double bonds they contain in their composition. If a fatty acid contains no double bond, it is known as saturated fatty acid, on the other hand, the double bond constituting fatty acids are known as unsaturated fatty acids.
These kinds of fatty acids are abundantly found in animals. An unsaturated fatty acid contains one or more double bonds in their composition. At room temperature, the saturated fatty acids are found in solid form, while unsaturated fatty acids come in liquid form. The number of hydrogen atoms is found more in a saturated fatty acid, in comparison to an unsaturated fatty acid.
Unsaturated fats are fats that contain at least one double bond in their fatty acid chains, which causes them to be generally liquid at room temperature and to contain fewer hydrogen atoms than saturated fats. They are more commonly found in plants than in animals.
Explanation:In the context of Biology, unsaturated fats are fats or fatty acids that contain at least one double bond in their fatty acid chains. Therefore, the statement (B) 'They have double bonds in their fatty acid chains' is true of unsaturated fats. This double bond causes the fat molecules to bend, preventing them from packing closely together and making them generally liquid at room temperature, which contradicts statement (C). Unlike saturated fats, which are fully 'saturated' with hydrogen atoms in their molecular structure, unsaturated fats contain fewer hydrogen atoms due to the presence of double bonds, making statement (D) inaccurate. Moreover, unsaturated fats are more commonly found in plants than in animals, refuting statement (A).
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Key features of seed plants facilitating life on land include three of the following four traits. Select the exception.
(A) homospory
(B) pollen
(C) reduced gametophytes
(D) seeds
Answer:
(A) homospory
Explanation:
Homospory is a type of reproduction found in pteridophytes. This character does not appear in gymnosperm or angiosperm hence it could not favor life on land.
The exception from the four traits is : ( A ) Homospory
Key features of seed plantsThe seed plants survive on land when planted due to certain key features that seed plants possesses and some of these key features are
PollenReduced gametophytes and SeedsWhile Homospory is not a key feature of seed plants it is form of reproduction that is found in pteridophytes. therefore it cannot favor the growth of plants on land.
Hence we can conclude that The exception from the four traits is : ( A ) Homospory
Where is DNA stored in the cell?
a. in the cell membrane
b. in smooth endoplasmic reticulum
c. in rough endoplasmic reticulum
d. in the vacuole
e. in the nucleus
Answer:
in the cell nucleus
Explanation:
Most DNA is located in the cell nucleus where it is called a nuclear DNA.
n German cockroaches, bulging eyes (bu) are recessive to normal eyes (bu+) and curved wings (cv) are recessive to straight wings (cv+). Both traits are encoded by autosomal genes that are linked. A cockroach has genotype bu+ bu cv+ cv, and the genes are in repulsion. Which of the following sets of genes will be found in the most-common gametes produced by this cockroach?a. bu+ cv+b. bu cvc. bu+ bud. cv+ cve. bu cv+
Answer:
e. bu cv+
Explanation:
The genes bu+/bu and cv+/cv are autosomal and linked --> each homologous chromosome has an allele of both bu+/bu and cv+/cv genes.
The cockroach is heterozygous for both genes, and they are in repulsion.
That means that on one of the homologous chromosomes one of the genes has a dominant allele and the other gene has the recessive allele, and on the other homologous chromosome the alleles are arranged in the opposite way.
In this case, the genotype of the cockroach would be best written as:
[tex]\frac{bu+\ \ \ cv}{bu\ \ \ cv+}[/tex]The most common gametes will be the parentals, i.e. gametes in which crossing-over between homologous chromosomes did not happen and therefore have the same distribution of alleles as the parental individual.
For those reasons, the most common gametes will be:
bu+ cvbu cv+The most common gametes produced by this cockroach will be bu⁺ cv and bu , cv⁺. Among the provided options, the set containing these genes is e. bu cv⁺. The correct answer is e. bu cv⁺
To solve this problem, let's first understand what it means for the genes to be in "repulsion" and then determine the most common gametes produced by the cockroach with the genotype bu⁺ bucv⁺ cv.
Key Points:
Linked Genes in Repulsion: When genes are in repulsion, it means that each chromosome in the homologous pair carries one dominant and one recessive allele of the two genes.
Genotype of the Cockroach: bu⁺ bucv⁺ cv
This means the cockroach has one chromosome with bu⁺ cv and another with bucv⁺ .
Chromosome Arrangement:
The chromosomes can be represented as:
One chromosome has bu⁺cv
The other chromosome has bucv⁺
Most Common Gametes:
Since the genes are linked and in repulsion, the most common gametes will be those that reflect the original arrangement of alleles on the chromosomes without recombination. Therefore, the two most common types of gametes will be:
bu⁺ cv
bu , cv⁺
These combinations are produced without crossing over between the linked genes.
Differentiate among somatic cells, gametes, and zygotes with regard to the number and origin of their chromosomes.
Answer:
Somatic cells are the ones obtained by mitosis. They constitute the body tissues and have a variety of different functions. They are diploid, which means they have two copies of each chromosome (2n), one obtained from the father and one from the mother.
Gametes are obtained by meiosis. A diploid cell divides to form four haploid gametes, they have a single copy of each chromosome (n), obtained from the parent cell.
Zygotes are obtained by the fusion of two gametes, one from the father and one from the mother, they are diploid again having two copies of each chromosome.
Describe mechanisms by which new genes could arise.
Answer:
Exon shuffling.
Explanation:
Exon shuffling may be defined as the phenomena of in which genes are exchanged between the different exons. Types of exon shuffling are transposon shuffling and cross over at the time of gamete formation.
Exon shuffling is responsible for the formation of new genes. Different exons combine together to form the new genes. The duplication of exons also leads to the formation of new genes. The intron-exons structure is altered for the new gene formation.
Thus, the correct answer is exon shuffling.
If the body produces proteins naturally what happens when weconsume
more proteins?
Answer:
Body constantly break downs and repair or rebuild its own cells and tissues which is lead to the need for the protein. In under different conditions like intense physical activity or sickness or injury can also result in requirement of protein as in diet.
All these process to occur one need to consume enough protein. Excessive protein than requirement for cell or body requirement will normally metabolize to produce amino acids and energy.
Thus, the correct answer would be - it will provide essential amino acids for producing different hormones and building and repair cells and tissues and energy.
The most important function of meiosis is:
a. growth of the individual from a small body to a large body
b. the formation of sex cells
c. repair of wounds
d. both a. and c. are correct
e. none of the above is correct
Answer:
The formation of sex cells.
Explanation:
Meiosis is the process of cell division in which a single parent cell divides to form the four daughter cells. The meiosis division is also known as reduction division.
The meiosis is generally important in the sex cells. The sex cells contains the half number of chromosome as compared with the somatic cells. This type of reduction division occurs in the process of meiosis. The sex cells are haploid cells that are produced by the process of meiosis.
Thus, the correct answer is option (b).
Answer:
The most important function of meiosis is the formation of sex cells.
Explanation:
Meiosis is the process of cell division which produced haploid cells from diploid parent cells. It is called heterotypic cell division. This occurs in germ cells only.
In sexually reproducing organisms, male and female gametes fuse to form a diploid (2n) zygote. If the zygote had the same number of 2n chromosomes as somatic cells, then the zygote has a double number of chromosomes in each generation.
In reality, it does not happen. This is because of the meiotic division. It reduces the chromosome number into half. Thus fertilization does not double the chromosome number in the zygote. Meiotic division counteracts the effect of fertilization.
Meiosis occurs only in sexually reproducing organisms. In this division synapsis of homologous chromosomes takes place. The meiosis results in tetrads at the end and crossing over occurs in this division. The genetic constituents of daughter cells differ from the parent cell.
Growth of the individual from a small body to large body, and repair of wounds occur by mitotic cell division.
Describe the role of chiasmata in chromosome segregation during meiosis.
Explanation:
The phase of Prophase I of meiosis is very long and divided into 5 subphases: Leptotene, Zygote, Pachytene, Diplotene, and Diakinesis. During a phase of the Diplotene, the degree of condensation is high, which allows individualizing the sister- chromatids that remain attached by the cohesins. The synaptonemal complex disintegrates, and from the centromeres begins a repulsion between homologous chromosomes, which remain associated only with the places where they occur as permutations.
These sites are called chiasmas (Greek, crossed) because they show the crossover of homologous chromatids. Chiasmas represent the cytological finding of the occurrence of permutation. The presence of at least one bivalent chiasm is essential to ensure the correct segregation of the homologous chromosomes in anaphase I.
In glycolysis and the TCA cycle, glucose is _____ down to CO2; this process _____ lots of ATP and reducing power. In photosynthesis, CO2 is ______ back to sugar by the _______ of lots of ATP and reducing power.
A) reduced; requires; oxidized; production
B) oxidized; requires; reduced; production
C) oxidized; produces; reduced; input
D) reduced; produced; oxidized; input
Answer:
C) oxidized; produces; reduced; input
Explanation:
Glycolysis and TCA cycle are the first two stages of cellular respiration. Glycolysis breaks down glucose into pyruvate which then enters the TCA cycle in the form of acetyl CoA and is completely oxidized into CO2 and H2O.
Both glycolysis and the TCA cycle produce a few ATPs and a large number of reducing powers (NADH and FADH2). The oxidation of reducing powers during oxidative phosphorylation drives the synthesis of a large number of ATP.
Photosynthesis is the process wherein CO2 is reduced into glucose. The process includes light-dependent synthesis of ATP and NADPH which in turn are used during the Calvin cycle to produce glucose from CO2.
Pathogens are transmitted in only two ways: by direct contact and by vectors.
a. True
b. False
Answer: False
Explanation:
Pathogens can be transmitted in many ways. It can spread by direct contact, indirect contact, or by vectors.
The mode of transmission can be skin contact, airborne particles, touching a surface, bodily fluids, touched by an infected person.
The mode of transmission can be vector that carries disease and helps in disease transmission.
So, the pathogens can be transmitted by direct contact, indirect contact or by vectors and by many more ways.
Name one enzyme of glycolysis that catalyzes an essentially irreversible step: __________
Answer:
Phosphofructokinase-1
Explanation:
Phosphofructokinase-1 catalyzes the phosphorylation of fructose 6-phosphate into fructose 1,6-bisphosphate. The reaction is exergonic with a large negative free energy change to make it essentially irreversible.
Phosphofructokinase-1 is an allosteric enzyme with regulatory sites. Higher ATP concentration serves to inhibit the phosphofructokinase-1 by binding to the allosteric site of the enzyme and thereby reducing its affinity for the substrate (fructose 6 phosphate).
Hexokinase is a glycolytic enzyme that catalyzes an irreversible conversion of glucose to glucose-6-phosphate, which is critical for the continuation of the glycolytic pathway.
Explanation:One enzyme of glycolysis that catalyzes an essentially irreversible step is hexokinase. This enzyme is responsible for the phosphorylation of glucose, producing glucose-6-phosphate at the beginning of the glycolytic pathway. Such a reaction is biologically irreversible, meaning the enzyme cannot readily catalyze the reverse reaction. In fact, this step is so crucial that when hexokinase is inhibited, glucose can diffuse out of the cell, preventing it from becoming a substrate for cellular respiration.
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In Bufo marinus, the tongue is attached at the anterior part of the buccal cavity?
Select one:
a. True
b. False
Answer:
The correct answer is option a. "True".
Explanation:
Bufo marinus is the world's largest toad, it is native from South and Central America but it has been introduced to islands in Oceania and the Caribbean. Bufo marinus rapidly flipps its tongue to catch its prey for eating and defensive purposes. Bufo marinus has its tongue attached at the anterior part of the buccal cavity, while posterior margin of the tongue lies freely. This morphology allows Bufo marinus to make the quick movement needed to catch its prey.
While the Cenozoic is often incorrectly referred to as the "Age of Mammals", in what time period were synapsids the dominant terrestrial vertebrates?
a. permian
b. triassic
c. jurassic
d. cretacous
e. devonian
The correct answer is A. Permian.
Explanation
The synapsids were terrestrial vertebrate mammals or animals related to them rather than to reptiles or birds that lived during the Permian (a period that began about 299 million years ago and ended about 251 million years ago) and were the dominant terrestrial vertebrates during it. These animals survived until the Triassic; however, due to the Permian-Triassic extinction, they were not the predominant vertebrate as in the previous period. Later, during the Cretatic and Jurassic periods, their development was minimal and did not have great importance as reptiles were dominant in this period. So, the correct answer is A. Permian.
There are 25 individuals in population 1, all with genotype AA, and there are 40 individuals in population 2, all with genotype aa. Assume that these populations are located far from each other and that their environmental conditions are very similar. Based on the information given here, the observed genetic variation most likely resulted from
a. Genetic drift. c. nonrandom mating.
b. gene flow d. directional selection.
Answer:
a. Genetic drift
Explanation:
When there is a change in the frequency of genes in a small population due to chance event as a result of which a few alleles disappear from the population, is known as genetic drift. In the given situation, the population size is small i.e. 25 and 40 respectively and in both these populations, it can be seen that one of the population only has allele A whereas the other has allele a, and because of this it can be concluded that this situation must have arisen due to genetic drift.The populations got established due to random sampling and then the progenies inherited only the alleles of the organism that reproduced and thus, the frequency of the other allele decreased and this is merely a chance event.In a short essay (100–150 words), explain how genetic information—along with an understanding of the process of descent with modification—enables scientists to reconstruct phylogenies that extend hundreds of millions of years back in time.
Answer:
Evolutionary biology illustrates both the pattern and processes. The processes of evolution are natural selection and other mechanisms, which modifies the genetic structure of the populations. These processes result in evolutionary patterns, that is, the products generated by evolution with time.
Phylogeny refers to the evolutionary history of a species or a group of species. In order to redevelop phylogeny, the scientists use systematics, that is, an analytical method to categorize the diversity and finding the evolutionary associations between the extinct and the living species.
The evidence used to redevelop phylogenies can be attained from the fossil record and from the biochemical, morphological, and genetic similarities between the species. The scientists are functioning to develop a universal tree of all life, which will get refined with the gathering of new information.
Genetic information and an understanding of descent with modification allow scientists to reconstruct phylogenies extending back millions of years.
Explanation:In order to reconstruct phylogenies that extend hundreds of millions of years back in time, scientists use genetic information and an understanding of the process of descent with modification. By comparing the nucleotide sequence of a gene across different species, scientists can determine how closely related those species are. Genetic variations among species can be used as a measure of their evolutionary relationships. For example, by comparing the gene sequences of humans and chimpanzees, scientists have determined that these two species are closely related. This genetic analysis allows scientists to construct a timeline of the evolutionary history of life on Earth, known as the 'tree of life'.
Compare chromosome behaviors during mitosis and meiosis.
Answer:
Explanation:
he cell division of eukaryotes consists of 2 types of division - mitosis, and meiosis. Both the cell divisions have karyokinesis which follows the cytokinesis. It takes some hours and an indirect type of cell division. The mitotic chromosomes and meiotic chromosomes show different behaviors -
In mitosis, the cell divides one time while in meiosis the cell divides 2 times. DNA replication occurs during interphase in mitosis. In meiosis DNA replication happens in the first cell division and no DNA replication in the second cell division.
There is no synapsis in the mitotic chromosome. In meiosis, synapsis occurs in homologous chromosomes. It has seen in the prophase I of meiosis.
The 2 chromatids of the chromosome do not exchange their segments in mitosis. In meiosis the chromatids of 2 homologous chromosome exchange segments. This results in the crossing over between the 2 homologous chromosomes.
The mitotic chromosome, each chromosome joined by a centromere. The meiotic chromosome forms tetrads from bivalent. The bivalent consists of 2 centromeres where the tetrads are attached.
In mitosis, the chromosomes separate slowly during anaphase. but in meiosis short chromosomes separate early, and long chromosomes take some time to separate.
An RNA virus that infects plant cells is copied into a DNA molecule after it enters the plant cell. What would be the sequence of bases in the first strand of DNA made complementary to the section of viral RNA shown here? 5' CCCUUGGAACUACAAAGCCGAGAUUAA 3'
Answer: Thecorrect answer would be
3' GGGAACCTTGATGTTTCGGCTCTAATT 5'
Explanation:
DNA and RNA are nucleic acids, that is, polymers of nucleotide. DNA consists of adenine, guanine, cytosine, and thymine. In contrast, RNA contains uracil in place of thymine. Rest three nucleotide are same.
Please note that transcription (DNA to RNA) as well as reverse transcription (RNA to DNA) is done on the basis of base complementary nature.
The base pairs are:
adenine- thymine/uracilguanine- cytosineSo, if adenine is present in DNA then uracil will be incorporated in RNA and vice-versa.
For example, the initial 5 nucleotide of given RNA are 5' CCCUU...3'
So, in DNA, based on base complementary rule, G will be incorporated against C and A will be incorporated against U. Thus, the DNA sequence would be 3' GGGAA...5'.
Similarly, the whole sequence can be worked out.
Final answer:
The complementary DNA sequence would be: 3' GGGAA CCTTGAUGUUUCGGCUCTAA 5'
Explanation:
To find the sequence of bases in the first strand of DNA complementary to the given RNA sequence, we need to use the complementary base pairing rules:
Adenine (A) pairs with Thymine (T) in DNA or Uracil (U) in RNA.
Thymine (T) pairs with Adenine (A) in DNA.
Cytosine (C) pairs with Guanine (G).
Given the RNA sequence:
5' CCCUUGGAACUACAAAGCCGAGAUUAA 3'
The complementary bases for DNA are:
Adenine (A) → Thymine (T)
Uracil (U) → Adenine (A)
Cytosine (C) → Guanine (G)
Guanine (G) → Cytosine (C)
So, the complementary DNA sequence would be: 3' GGGAA CCTTGAUGUUUCGGCUCTAA 5'.
Carbon skeletons to be broken down during cellular respiration can be obtained from
A) polysaccharides.
B) proteins.
C) lipids.
D) A and B.
E) A, B, and C
Cellular respiration breaks down carbon skeletons obtained from polysaccharides, proteins, and lipids for energy.
Carbon skeletons to be broken down during cellular respiration can be obtained from polysaccharides, proteins, and lipids. This means option E is correct.
As you read in this chapter, fungi have long formed symbiotic associations with plants and with algae. In a short essay (100– 150 words), describe how these two types of associations may lead to emergent properties in biological communities.
Answer:
Fungi show symbiotic association with algae and plants. With plants, they thrive as endophytes in a form of the symbiotic association. This symbiotic association results in the emergence of novel characteristics in the world of biology.
The lichens function as a tool in finding the quality of air, as they grow in the environment containing good air quality. The tolerance towards heat is another characteristic. Some of the endophytes are witnessed in the plants, which grows in very hot conditions.
At such conditions, no fungi or plant can thrive, however, in the symbiotic association, they possess the tendency to thrive. If one tries to separate them, it results in death of both.
Symbiotic associations between fungi, plants, and algae contribute to increased nutrient cycling, enhanced plant growth, and ecosystem stability, fostering emergent properties in biological communities.
Nutrient Exchange: fungi form associations with plant roots, facilitating the absorption of water and essential nutrients like phosphorus and nitrogen.
This enhanced nutrient uptake benefits both the plants and the fungi, leading to healthier and more productive vegetation.
As a result, the increased plant biomass supports a broader range of herbivores and, in turn, predators, contributing to the complexity of food webs within biological communities.
Improved Plant Tolerance: Symbiotic fungi can also improve a plant's tolerance to various environmental stressors, such as drought or heavy metal contamination.
This enhanced resilience can allow for the coexistence of diverse plant species in the same ecosystem, further increasing biodiversity.
Carbon Sequestration: Mycorrhizal fungi, in association with trees, contribute to carbon sequestration.
This sequestered carbon helps mitigate climate change by reducing atmospheric carbon dioxide levels.
Ecosystem Stability: The overall effect of these fungal-plant associations is an increase in ecosystem stability.
The presence of these relationships helps buffer ecosystems against disturbances and allows for the persistence of diverse species, making the community more resilient in the face of environmental changes.
The symbiotic associations between fungi, plants, and algae promote nutrient cycling, improve plant tolerance, sequester carbon, and enhance ecosystem stability, leading to emergent properties that foster greater biodiversity and complexity within biological communities.
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Inhibin secreted by granulosa cells selectively inhibits secretion of FSH by the pituitary
a. True
b. False
Answer:
The correct answer will be option-true.
Explanation:
Follicular stimulatory hormone or FSH and inhibin are the complex proteins which play a various biological role in the reproductive system.
The FSH stimulates the secretion of inhibin from the gonads in both males (Sertoli cells) and females (granulosa layer). The production of inhibin suppresses the secretion of FSH from the pituitary gland.
The suppression of FSH directly suppresses the secretion of further inhibin and thus, option-true is the correct answer.