Answer:
Gene c is closer to a.
Explanation:
The process by which the bacteria take up the genetic material which is foreign from the environment is called as bacterial transformation and it is a type of horizontal gene transfer. When during transformation there is a simultaneous transfer of two or more genes then such type of transformation called co-transformation.According to the question, there is co-transformation occurring for both the genes a and b as well as for a and c, however, the frequency of cotransformation of a and c is higher (55%) than a and b (2%).A higher frequency of co-transformation indicates that the two genes are located more closely on the chromosome and hence, are usually transferred together, thus a and c are much closer as compared to a and b.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.
What might be a use for fluorescence in situ hybridization (FISH)?
a. For identification of a specific gene in a DNA extraction by hybridization to a DNA probe
b. For identification of a specific gene by hybridization to a DNA probe within live cells that have had their DNA denatured by heat.
c. For identification of an mRNA within an RNA extraction by hybridization to a DNA probe.
d. For identification of both mRNA and DNA in cellular extracts using an RNA probe
e. None of the above
Answer:
B.
Explanation:
Flourescent insitu hybridization-This technique permits the direct analysis of sequences of DNA or RNA in tissues
-Through this technique various chromosomal abnormalities can be determined as it focuses on that part of the DNA for which a complementary probe has been used.
-The method involves the denaturation of DNA in a tissue section followed by the application of a labeled probe which is complementary to the sequence of interest.
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.
Define virus and virion, and describe the parts of a virion.
Answer:
A virus is a tiny infectious biological agent that can only replicate or duplicate inside the host cell. These infectious agents can infect all different types of living organisms ranging from animals and plants to microorganisms and archaea and bacteria.
Virions are ineffective particle or form of the virus outside of the host cell, with RNA or DNA and a protein capsid.
The main role of these infectious agent virions is to transfer the DNA or RNA genome from itself to the cell of host and expressed the gene which means produce proteins from the genome transferred to the host cell.
Define mutagen and describe how mutagens are used in genetic research.
A mutagen is a physical or chemical agent that permanently changes genetic material, usually DNA, in an organism and increases the frequency of mutations. Mutagens are used in genetic research to induce mutations and study their effects on organisms.
Explanation:A mutagen is a physical or chemical agent that permanently changes genetic material, usually DNA, in an organism and increases the frequency of mutations above the natural background level. Mutagens are used in genetic research to induce mutations and study the effects of these mutations on organisms. They are used to investigate gene function, understand the causes of genetic diseases, study the mechanisms of evolution, and develop new treatments for diseases.
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.
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.
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.
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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Smooth muscles of the digestive system move food by the process of peristalsis.
a. True
b. False
Answer:A
Explanation:
Answer: A. True
Explanation: "Peristalsis. Peristalsis is a series of muscular contractions that propel food through the small intestine. These kinds of controls help you get the most nutrients out of your food. ... This coordinated contraction of smooth muscle keeps food moving on its one-way path through your digestive system."
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'.
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.Explain how mitotic recombination leads to the mosaic condition termed twin spots.
Answer:
If an individual Drosophila has two or more populations of cells comprising different genotypes from one single egg then it is termed as twin spots or mosaic.
Explanation:
There might be different reasons for mosaic to occur like
Nondisjunctioning of the chromosomes
Lag in anaphase
Endoreplication
Mutations in a single cell
Mitotic recombination:
One of the major ways through which mosaic or twin spots arise is the mitotic recombination. It is also termed as somatic cross over. Twin spot or mosaic generally occurs if there is linking of heterozygous genes in repulsion. The recombination generally happens among the centromeres from the adjacent genes.
A common example of the mitotic recombination is the Bloom's syndrome. Bloom's syndrome is caused due to the mutation that occurs in the blm gene. As a result, there are defects in the BLM protein produced.
Plant species A has a diploid chromosome number of 12. Plant species B has a diploid number of 16. A new species, C, arises as an allopolyploid from A and B. The diploid number for species C would probably be
a. 14. b. 16. c. 28. d. 56.
Answer:
c. 28
Explanation:
Plant A has a dipoid chromosome number of 12, so its gametes will have an haploid number of 6 chromosomes.
Plant B has a dipoid chromosome number of 16, so its gametes will have an haploid number of 8 chromosomes.
Plant A Plant B2n = 12 2n = 16 n = 6 n = 8The new species C arises as an allopolyploid from A and B. An allopolyploid usually originates from the breeding of two different species.
In this case, a gamete from plant A combines with a gamete from plant B to form a hybrid with 14 chromosomes (6 from A and 8 from B). These chromosomes are unpaired, so the hybrid is sterile.
In order to become a fertile diploid individual of species C, the most common mechanism is polyploidization, where the genome duplicates. That way, the resulting plant C has a diploid number of 14 x 2 = 28 chromosomes, of which 12 are A and 16 are B.
Species C, being an allopolyploid of species A and B, would likely have a diploid number that is the sum of the diploid numbers of species A and B. In this case, that is 28.
Explanation:The new species C, being an allopolyploid of species A and B, is formed by a combination of the whole sets of chromosomes from both species A and B. Thus, we would add together the diploid numbers of both species to determine the diploid number of species C. Plant species A has a diploid chromosome number of 12 and plant species B has a diploid number of 16. Adding these together gives us a total diploid number of 28. Therefore, the correct answer would be option c. 28.
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Q:- Short hair rabbit is goverened by a dominant gene (L)
andlong hair by its recessive alleles (l). Black hair results from
theaction of the dominant genotype (B) and brown from therecessive
genotype (b).
A. In cross between dihybrid short ,black and homozygousshort,
brown rabbits, what genotypic and phenotypic ratios areexpected
among their progeny?
B. Determine the expected genotypic and phenotypic ratios
inprogeny from cross LlBb*Llbb.
Answer:
A) he phenotypic ratio is
short & black : short & brown
8:8
1:1
Genotypic Ratio is
LLBb : LLbb: LlBb : Llbb
4:4:4:4
1:1:1:1
B)
Genotypic Ration
LLBb : LLbb: LlBb: Llbb: llBb: llbb
2: 2: 4: 4: 2:2
1:1:2:2:1:1
Phenotypic ratio
Short & Black : Short & brown: Long & black : Long & brown
6: 6: 2: 2
3:3:1:1
Explanation:
Given -
Allele for short hair (L) is dominant over long hair allele (l)
And Black hair allele (B) is dominant over brown hair allele (b)
A) Genotype for dihybrid short and black rabbit is LlBb
Genotype for homozygous short and brown rabbits is LLbb
Cross between LlBb and LLbb will result into following offspring
LB Lb lB lb
Lb LLBb LLbb LlBb Llbb
Lb LLBb LLbb LlBb Llbb
Lb LLBb LLbb LlBb Llbb
Lb LLBb LLbb LlBb Llbb
Hence, the phenotypic ratio is
short & black : short & brown
8:8
1:1
Genotypic Ratio is
LLBb : LLbb: LlBb : Llbb
4:4:4:4
1:1:1:1
B) LlBb*Llbb.
LB Lb lB lb
Lb LLBb LLbb LlBb Llbb
Lb LLBb LLbb LlBb Llbb
lb LlBb Llbb llBb llbb
lb LlBb Llbb llBb llbb
Genotypic Ration
LLBb : LLbb: LlBb: Llbb: llBb: llbb
2: 2: 4: 4: 2:2
1:1:2:2:1:1
Phenotypic ratio
Short & Black : Short & brown: Long & black : Long & brown
6: 6: 2: 2
3:3:1:1
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.
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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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.
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.
The branch of biology concerned with identifying, naming, and classifying all living things is taxonomy.
a. True
b. False
Answer:
True.
Explanation:
Taxonomy may be defined as one of the main branch of biology that helps in the classification of organisms. The organisms are classified on the basis of similarities and differences among them.
Taxonomy includes the identification of an individual by looking at its morphological characters. The nomenclature ( naming of organism) and its classification in to different groups.
Thus, the correct answer is option (a).
Why does each replication fork require both leading and lagging strand synthesis?
Select the 2 reasons below.
A) Because DNA synthesis is semi-conservative.
B) Because DNA polymerases must synthesize DNA 3' to 5'.
C) Because the DNA templates are antiparellel.
D) Because DNA polymerase only synthesizes DNA in one direction.
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.
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.
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.
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.
Describe the characteristics of the garden pea that made it a good organism for Mendel's analysis of the basic principles of inheritance. Evaluate how easy or difficult it would be to make a similar study of inheritance in humans by considering the same attributes you described for the pea.
Answer and explanation:
The garden pea had many advantages to be used as a model in inheritance studies. These plants grow very fast, so you can rise several generations in little time. This fundamental if you want to do this kind of studies. The study object needs to have a brief generational time, so they can be selected and their descendants can be raised again. Another advantage was that garden peas can be raised easily. They don't need any special care. Besides, they were common in the market. Mendel could find them and bought them any time that he wants. Also, they presented visible characteristics that could vary from one plant to another. Some pea gardens had violet flowers, while others had white flowers. Some were tall, while others were small. These made possible to select plants with specifics features and to study if they could transmit these features to the next generations.
It could be really difficult to do the same experiment in humans. several years would have to pass to observe new generations, so during the lapse time of the researcher, only a few human’s generations could be studied.
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.
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.
What binds to a stop codon on a mRNA during translation?
a. transcription factor
b. tRNA
c. termination factor
d. transcription initiator
Answer:
Termination factor.
Explanation:
Translation may be defined as the process of formation of the protein from the RNA molecule y the help of enzymes and translation factors. The translation occurs in 5' to 3' direction.
The three main steps of the translation are initiation, elongation and termination. The release factors or translation termination factors binds to the stop codons and results in the translation termination. Prokaryotes release factors include RF1, RF2, and RF3 that binds to the stop codons.
Thus, the correct answer is option (c).
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.