Answer:
a. population
Explanation:
Gene flow cannot occur in hybrids because 99.9% of the time hybrids offspring are not fertile.
At genus level is not possible either because a hybrid is a cross breed at the genus level (horse-donkey).
A population is a group of species of 1 kind. A population of donkeys.
Explain how the Lac operon is regulated, including all negative and positive components of regulation.
Explanation:
Three enzymes for lactose metabolism are grouped in the lac operon: lacZ (β-galactosidase), lacY (permease) and lacA (trans-acetylase). The transcription of this operon occurs only when lactose is available to digest, presumably to avoid wasting energy. Apart from these protein-coding genes we have P(promoter), O(operator and CBS(Cap-binding site), sequences that work as binding sites for transcriptional regulation.
Negative components: An important regulator is lacI when four of these molecules assemble they form a repressor, this will bind to the promoter, this won't allow the operon to be transcribed as long as the operator is occupied by a repressor. LacI can also prevent lactose to bind to the operator by binding with it and changing its shape.
Positive components: cAMP binding protein is another molecule that when is bound to CBS improves the chance of RNA polymerase to bind to the promoter to initiate transcription.
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One theory of human evolution suggests that Homo erectus evolved directly into Homo sapiens (humans), after which point Homo erectus became extinct. If this is true, humans arose by ________.
a. allopatric speciation
b. cladogenesis
c. anagenesis
d. parapatric speciation
Answer:
The correct answer will be option-C.
Explanation:
The evolution of human species in which one species transformed into another species evolves by a mechanism known as anagenesis.
Anagenesis is the mechanism of evolution which transform one species into a different species within a lineage. This process is slow and takes time to form species, therefore, is also known as gradualism or phyletic transformation.
The Homo sapiens evolved from Homo erectus where Homo sapiens overwrites the ancestral species and caused the species to become extinct.
Thus, Option-C is the correct answer.
Your digestive system is equipped with a diversity of enzymes that
break the polymers in your food down to monomers that your cells can
assimilate. A generic name for a digestive enzyme is hydrolase. What is
the chemical basis for that name?
Answer:
Enzymes are named according to the reaction they catalyze. Polymers are made of subunits joined together by different types of bonds, forming a macromolecule.
Hydrolases are used by the organism to catalyze the hydrolysis of polymers so they can be easily manipulated as monomers. Hydrolysis means reacting with water, water can break the bonds of different polymers turning it into its constitutive monomers.
If someone shakes your hand, the proteins that change shape and are responsible for initiating the feeling of this hand touch pressure are called:
a. stretch receptor proteins
b. carboxy receptor proteins
c. cardiac receptor proteins
d. dynamin receptor proteins
Answer:
d. dynamic receptor proteins
Explanation:
In the skin there are several receptor cells that can pick up various stimuli such as temperature, pressure and pain. They can be classified into mechanoreceptor, thermoreceptor and pain receptor cells. When we are dealing with mechanical stimuli like pressure, mechanoreceptors respond to stimuli by sending the signal to the brain.
Light touches on the skin are received by nerve endings called Meissner corpuscles and Merkel discs. Already more intense pressures (such as a handshake) are received by receptors called Pacini corpuscles.
Explain how stem cells obtained from IVF leftovers and somatic cell nuclear transfers (SCNT) differ in terms of the sources of their genomes.
Answer:
In in-vitro fertilization, the oocytes from women were taken and inseminated with the donor sperm, post the process of insemination, the oocytes were fertilized successfully and were further allowed to develop into a blastocyst stage.
From the blastocyst, the inner cell masses were extracted, which further produces stem cell lines. From leftover in-vitro fertilization, the extraction of single blastomere becomes very easy.
In SCNT, the nucleus of a somatic cell is administered into the cytoplasm of an enucleated egg that then develops into a zygote and is then permitted to further give rise to a blastocyst stage.
Because of gall stones, Fred had his gallbladder surgically removed. Which of the following is the most likely side effect?
a. difficulty with protein digestion
b. difficulty with feces production
c. difficulty with vitamin B12 absorption
d. difficulty with fat digestion
e. difficulty with bile production
Answer:
d. difficulty with fat digestion
Explanation:
The gallbladder is an accessory gland that stores bile produced in the liver, so option "e" is not possible. Bile is a type of enzyme that helps to break down fats which makes option "d" as the best option. Protein digestion is mainly by different enzymes known as proteases. Feces are formed in the short intestine.
Write down symbols for the alleles. (These may be given in the problem.) When represented by single letters, the dominant allele is uppercase and the recessive is lowercase.
Answer: A (uppercase letter) is used to refer to dominant alleles.
a (lowecase letter) is used to refer to recessive alleles.
Explanation:
Alleles are different forms of a gene, and they can be dominant or recessive. Each gene codes for a specific trait.
Diploid organisms, such as human beings, have two alleles for each gene. If an organism is heterozygous for a certain trait, it means it has one dominant and one recessive allele. And since the dominant allele masks the effects of the other one, the dominant trait is expressed in the phenotype. So, dominant alleles show their effect even if the organisms only has one copy of it.
If an organisms is homozygous, it has two dominant alleles or two recessive alleles. Then it can be homozygous dominant, where both are dominant alleles; or homozygous recessive where both are recessive. Recessive alleles only show their effect if the organisms has two copies of it.
For example, if we want to study the gene that codes for hair color, we could say that dominat allele "A" codes for brown hair, and recessive allele "a" codes for blonde hair. If a person has brown hair, the genotype could be either AA or Aa, becuase AA is homozygous dominant and it has two copies of brown color. And Aa is heterozyogus, but the person only needs one copy of the allele to be expressed. On the other hand, if the person has blonde hair, the genotype can only be aa because recessive alleles only show their effect if there are two copies of the allele.
In genetics, alleles are symbolized by letters, with uppercase for dominant alleles and lowercase for recessive alleles. A homozygous dominant genotype is written with two uppercase letters (VV), a homozygous recessive with lowercase (vv), and a heterozygous genotype combines both (Vv).
Explanation:Understanding Allele Symbols in Genetics
In genetics, the symbols for alleles are represented by letters, with the dominant allele typically denoted by an uppercase letter and the recessive allele by a lowercase letter. For instance, in the example of pea plant flower color, if violet is the dominant trait, then 'V' would symbolize the dominant allele, while 'v' would represent the recessive allele for a white flower color. Therefore, a homozygous dominant pea plant with violet flowers would have the genotype 'VV', a homozygous recessive plant with white flowers would be 'vv', and a heterozygous pea plant with violet flowers would have the genotype 'Vv'.
Dominance was discovered by Mendel, who observed that certain traits, like the height in pea plants, showed a dominant and recessive pattern. A plant with genotype 'TT' (tall) would be homozygous dominant, while 'tt' (dwarf) would indicate a homozygous recessive plant. A heterozygous plant with the genotype 'Tt' would also be tall, displaying the dominant trait. Furthermore, when two heterozygous pea plants are crossed (Tt x Tt), Mendel deduced a characteristic 3:1 ratio of dominant to recessive phenotypes in the offspring.
Besides a high voltage shock, what is another method to make E. coli competent to take up "naked" DNA?
a. high concentrations of calcium ions followed by high temperature
b. high concentrations of calcium ions and several hours on ice
c. large amounts of DNA added directly to a bacterial culture growing at 37C
d. high concentrations of minerals followed by high temperature
e. a high voltage shock is the only way to make E. coli competent
Answer:
The correct answer is a. high concentrations of calcium ions followed by high temperature.
Explanation:
Apart from high voltage shock, calcium chloride heat shock method can be used to make E. coli competent to take up naked DNA. As both DNA and lipids contain phosphorus and phosphorus contains negative charge, therefore, ice-chilled CaCl₂ makes cation bridge between phosphorylated lipid of cell membrane and phosphorus present in the DNA.
This allows the naked DNA to attach to the cell surface of bacteria. Then cells are given heat shock which creates gaps(holes) in the membrane and makes the cells competent to take up the naked DNA by these holes.
Then the transformed cells can be selected by screening methods like growing transformed cells on media containing a suitable antibiotics.
E. coli cells can be made competent to take up DNA by incubating with calcium ions and then applying heat shock, which allows DNA to enter the cells through pores formed in the membrane. So the correct option is a.
Explanation:Besides a high voltage shock, another method to make E. coli competent to take up DNA is by using high concentrations of calcium ions followed by a heat shock. This process involves incubating E. coli cells with an ice-cold solution of 50 mM calcium chloride at 4°C for 30 minutes, which allows the DNA molecules to attach to the cell exterior. Following this, the cells are briefly incubated at 42°C, which causes the formation of pores in the membrane, allowing DNA to enter the cell.
Sex steroids are secreted by the __________ cells of the ovary and the ___________ cells of the testes.
Answer:
The sex steroids are the essential hormones for the proper function and development of the body, they monitor sexual differentiation, sexual behavior patterns, and the secondary sex characters. There are five prime categories of steroid hormones. These are estradiol (estrogen), testosterone (androgen), aldosterone, and cortisol.
Of these testosterone, estradiol, and progesterone comes under the category of sex-steroids. The progesterone and estradiol are produced by the ovarian granulosa cells of the ovary. On the other hand, the testicular Leydig cells are the location of testosterone synthesis.
In mechanism, photophosphorylation is most similar to
a. substrate-level phosphorylation in glycolysis.
b. oxidative phosphorylation in cellular respiration.
c. carbon fixation.
d. reduction of NADP+.
Answer:
b. oxidative phosphorylation in cellular respiration.
In mechanism, photophosphorylation is most similar to oxidative phosphorylation in cellular respiration. Both processes involve establishing a proton gradient across a membrane to generate ATP.
Explanation:In the biological process of energy production, photophosphorylation parallels most closely to oxidative phosphorylation in cellular respiration as option b suggests.
During oxidative phosphorylation, electron transport chain drives the creation of a proton gradient across a membrane, and the subsequent flow of protons down this gradient is used to power ATP production. Similarly, during photophosphorylation, the energy of sunlight is used to generate a proton gradient, which again drives the synthesis of ATP. So, in essence, both processes use a membrane-bound proton gradient to drive ATP synthesis.
This is different from substrate-level phosphorylation, carbon fixation and the reduction of NADP+. Substrate-level phosphorylation refers to direct ATP production during a specific reaction in the metabolic pathway, while carbon fixation is the process of converting CO2 into organic compounds; and the reduction of NADP+ is about adding electrons to NADP+ to convert it to its reduced form, NADPH.
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Enzymes that break down DNA catalyze the hydrolysis of the covalent bonds that join nucleotides together. What would happen to DNA molecules treated with these enzymes?
a. The two strands of the double helix would separate.
b. The phosphodiester linkages of the polynucleotide backbone would be broken.
c. The pyrimidines would be separated from the deoxyribose sugars.
d. All bases would be separated from the deoxyribose sugars.
Answer:
The correct answer is option b. "The phosphodiester linkages of the polynucleotide backbone would be broken".
Explanation:
The phosphodiester linkages of the polynucleotide backbone is what binds each nucleotide to each other in the DNA molecules. These linkages are covalent bonds that take place between 3' carbon atom of one sugar molecule and the 5' carbon atom of another. The enzymes that break down DNA catalyze the hydrolysis of the phosphodiester linkage, which results in DNA cleavage within the backbone at specific or unspecific nucleotides.
Final answer:
Enzymes that break down DNA cleave the phosphodiester bonds in the DNA polynucleotide backbone, leading to the degradation of the DNA molecule into its constituent nucleotides. The correct choice is b - the phosphodiester linkages would be broken.
Explanation:
When enzymes that break down DNA are applied to the DNA molecule, they catalyze the hydrolysis of the covalent bonds that join nucleotides together. Specifically, these enzymes target the phosphodiester bonds in the polynucleotide backbone. Hydrolysis of these bonds would lead to a cleavage of the polynucleotide chain. This would result in the degradation of the DNA molecule into its individual nucleotides, where each phosphate group would be associated with a deoxyribose sugar and a nitrogenous base.
The correct answer to the student's question is: b. The phosphodiester linkages of the polynucleotide backbone would be broken. This process is different from denaturation, where hydrogen bonds between complementary bases break; breaking phosphodiester bonds involves a chemical reaction that cleaves the DNA backbone itself.
Enzymes such as nucleases perform this function, specifically endonucleases which cleave phosphodiester linkages within a DNA strand. Conversely, enzymes like DNA ligases can repair these breaks, forming a new phosphodiester linkage, though in this context, we are discussing the breaking of the DNA molecule by hydrolysis, not the joining of fragmented DNA.
Discuss the evolutionary significance of increasing complexity from unicellular to multicellular organisation?
Answer:
Explanation:
A unicellular organism has only a single cell thus is unable to perform the diversity of functions. The complexity in multi-cellular organisms is more in terms of structure as well as in function thus the evolution of the multi-cellular organism is significant as as the time passes the changes in the morphology, physiology and genetic make up of the organisms will become appreciable.
Uterine muscle responsiveness to oxytocin
a. is enhanced by an increased in estrogen/progesterone ratio
b. is prevented by high prostaglandin levels during pregnancy
c. is decreased by down-regulation of oxytocin receptors during the third trimester
d. is enhanced by decreased estrogen/progesterone ratio
e. is independent of estrogen/progesterone ratio
Answer:
a. is enhanced by an increased in estrogen/progesterone ratio
Explanation:
The average duration of human pregnancy is about 9 months or we can say 38 weeks/266 days. This period is called as the gestation period. At the end of pregnancy vigorous contraction of the uterus result in expulsion or delivery of the baby. It is called parturition. Parturition is a neuroendocrine mechanism.
During pregnancy uterine contractions are inhibited due to the high progesterone levels. Progesterone maintains the endometrium and prevents contraction of myometrium. At the end of last trimester, the progesterone levels plateau and then drops whereas estrogen levels continue to rise.
As a result the E/P ratio increases which makes the myometrium more sensitive to contraction stimuli.
The decreases levels of progesterone may lead to Braxton Hicks contraction which is nothing but false labor. Meanwhile, oxytocin hormone is secreted by the posterior pituitary gland which induces the contraction of myometrium.
Albino rabbits (lacking pigment) are homozygous for the recessive c allele (C allows pigment formation). Rabbits homozygous for the recessive b allele make brown pigment, while those with at least one copy of B make black pigment. True-breeding brown rabbits were crossed to albinos, which were BB. F1 rabbits, which were all black, were crossed to the double recessive (bb cc). The progeny obtained were 34 black, 66 brown, and 100 albino. a. What phenotypic proportions would have been expected if the b and c loci were unlinked? b. How far apart are the two loci?
Answer:
a. If the b and c loci were unliked the expected phenotypic proportions would be:
- 1/4 Brown
- 1/4 Black
- 1/2 Albino
b. The estimated distance between the two loci is equal to 17 centimorgans (cM).
Explanation:
a. If the loci b and c are unlinked or they are not in the same chromosome, it means that b and c are segregated independently (following Mendel's laws). So, the first step is to state the genotypes of the rabbits.
Albino rabbits are homozygous for c (cc)
Brown rabbits are homozygous for b (bb).
Black rabbits have a copy of B (Bb or BB).
But, rabbits have the two genes (b and c), so they could be:
Albino: BBcc/Bbcc/bbcc.
Brown: bbCC/bbCc
Black: BBCC/BBCc/BbCC/BbCc
A cross was made with true-breeding brown rabbits( that means "pure" or homocygotes) bbCC and albino BBcc.
So, we have :
bbCC X BBcc
This is a dihybrid cross with only one possible genotype in F1: BbCc. -> Black rabbits.
After this, these rabbits (BbCc) were crossed with double recessive (albino) rabbits, like this:
BbCc X bbcc
To know the possible outcomes of F2, it is necessary to make a Punnett square 4x4. Following the independent assortment principle, we will have the following possible gametes:
Black Rabbits: Cb, CB, cB,cb.
Double recessive rabbits: cb, cb, cb, cb.
After doing the Punnett square, we will have 4 possible genotypes and 3 possible phenotypes in F2:
1. bbCc = Brown
2 BbCc = Black
3. Bbcc= Albino
4. bbcc= Albino
That means:
1. 1/4 Brown
2. 1/4 Black
3. 1/2 Albino
b. To estimate the distance between the two loci is necessary to use the data provided in the question.
In this case, we are assuming that b and c are in the same chromosome, so they don't follow Mendel's laws. The first step is to know which are the initial genotypes.
The initial cross was between a brown rabbit (bbCC ) and an albino rabbit (BBcc). If the two genes are linked, we can say that in the brown rabbit bC are always together and in the albino Bc will be together, because they're linked.
F1 will be black rabbits bCBc, (remember bC and Bc go always together).
So, after this, a new cross was made with a double recessive:
bCBc X bcbc
The possible outcomes are:
bCbc Brown
bCbc Brown
Bcbc Albino
Bcbc Albino
As you see, there are not black rabbits, so to find black rabbits, we need to find a BCbc genotype. This genotype can be produced by recombination or crossingover in the same chromosome in parentals. So, the distance between two loci is equal to the proportion of individuals with a recombinant genotype, in this case, BCbc or black rabbits.
So, we have:
34 black
66 brown
100 albino
Total: 200
D[tex]Distance=\frac{Recombinants}{Total } *100= \frac{34}{200}*100= 17 cM[/tex]
Estimated distances are measured in centimorgans (cM).
Microphylls are found in which plant group?
a. lycophytes
b. liverworts
c. ferns
d. hornworts
Answer:
a. lycophytes is the correct answer.
Explanation:
Microphylls are found in lycophytes.
lycophyte is a vascular plant, they have a unique type of leaves called microphylls.
The lycophytes belong to the division of Lycophyta, they are seedless plants and have vascular tissue.
The leaves(microphylls) present in the lycophyte have a single vein that is unbranched and narrow, which provides the water to the leaf and supplies nutrients to other parts of the plant.
Microphylls, which are characterized by a single vascular vein, are found primarily in lycophytes. Lycophytes are an ancient group of vascular plants and include species like club mosses and quillworts.
Explanation:Microphylls are a specific type of leaf that are typically characterized by only having a single vascular strand or vein. This contrasts with megaphylls, which usually have multiple veins. The answer to which plant group microphylls are found in is a. lycophytes.
Lycophytes are one of the oldest groups of vascular plants and are recognized by their microphyllous leaves. Notable examples of lycophytes include club mosses, quillworts, and spike mosses. The presence of microphylls is a defining characteristic of this plant group.
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How are transfusion reactions type II responses? What happens in a transfusion reaction?
Answer:
Explained
Explanation:
Type II response can be defined as an antibody-dependent process in which specific antibodies bind to antigens and results in tissue damage. Transfusion reaction is a type II response because here the mismatched RBC's are rapidly destroyed by specific preformed antibodies (anti-ABO or -Rh) and complement.
Transfusion reactions takes place when incompatible blood products are transfused into a patient's circulation. This triggers the patient's immune system and consequently donor RBC's are destroyed by antibodies in the recipient's circulation. This is usually seen when antigen-positive donor RBC's are transfused into a patient who has preformed antibodies to that antigen.
What is the total number of ATPs produced by cellualar metabolism (per glucose molecule)?
Answer:
38 ATP
Explanation:
One molecule of glucose on complete breakdown yields 38 ATP. The break up of APT production at different steps of cellular respiration is given below:
Glycolysis yields 4 ATP and 2 NADH is produced, but 2ATP is used. Net gain of 2ATP. During formation of Acetyl CoA, 2 NADH is produced. During Krebs cycle 2 ATP, 6 NADH, 2 FADH₂ are produced. During Electron transport chain, reduced coenzymes NADH and FADH₂ oxidised to release ATP.Each NADH releases 3 ATP and each FADH₂ releases 2 ATP. Altogether 10 NADH is produced which yield 30 ATP and 2 FADH₂ yields 4 ATP.
Therefore, on complete oxidation of one molecule of glucose yields 38 ATP.
Storage form of glucose in animal cells is:
a. starch
b. disaccharide
c. glycogen
d. glucose
Answer:
Glycogen. (Ans. C)
Explanation:
Glycogen is known as the storage form of glucose in animal cells. It a polysaccharide of glucose which is multi-functional and serves as an energy form storage in animals. Liver cells primarily stored glycogen, some glycogen also stored in muscle cells for immediate use if needed.
Glycogen molecule composed of many glucose molecules that are linked together with the help of the alpha acetal group. To create energy currency (adenosine triphosphate) glucose is the primary source used by every cell.
The storage form of glucose in animal cells is glycogen, a complex carbohydrate stored mainly in the liver and muscle cells. This is utilised as a source of energy when required.
Explanation:The storage form of glucose in animal cells is glycogen. This polysaccharide is a complex carbohydrate that animals and humans use to store glucose. It is found mainly in the liver and muscle cells. When your body requires energy, it transforms the stored glycogen into glucose which can be used as a source of energy. You can perceive it as animal's equivalent to plant's starch storage mechanism. So, the correct answer to your question is: c) Glycogen.
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In eukaryotic cells, transcription cannot begin until
a. the two DNA strands have completely separated and exposed the promoter.
b. several transcription factors have bound to the promoter.
c. the 5 ′ caps are removed from the mRNA.
d. the DNA introns are removed from the template.
Answer:
b. several transcription factors have bound to the promoter.
Explanation:
The initiation of RNA synthesis in eukaryotes includes assembly of RNA polymerase and several transcription factors at the promoter. The binding of TATA-binding proteins (TBP) to the TATA box serves to stabilize the TFIIB-TBP complex at the promoter.
The TFIIB is a transcription factor that is bound to both transcription factor binding proteins and DNA. This is followed by the binding of transcription factor TFIIF and the RNA Pol II enzyme to the TFIIB-TBP complex. Then the other transcription factors such as TFIIE, TFIIB and TFIIH also join the complex. The result is the formation of a closed complex.
Here, the function of TFIIA is to stabilize the TFIIB and TBP at promoter while TFIIB serves in the recruitment of RNA Pol II enzyme to the promoter.
TFIIE facilitates binding of TFIIH and has helicase activity to unwind the DNA duplex while TFIIF serves to prevent the binding of Pol II enzyme to the DNA sequences other than promoters.
Finally, the transcription factor TFIIH phosphorylates the enzyme RNA polymerase II and brings about a conformational change in the whole complex to facilitate the start of transcription.
In eukaryotic cells, the process of transcription, the creation of an RNA copy of a gene sequence, cannot begin until transcription factors have bound to the promoter region of the DNA.
Explanation:In eukaryotic cells, transcription, which is the process of making an RNA copy of a gene sequence, cannot start until several transcription factors have bound to the promoter. This process takes place in the cell nucleus. Preparatory steps include the DNA strands unwinding and the promoter region being exposed, but transcription only actually commences when transcription factors bind to the promoter, effectively marking the start site for RNA synthesis.
The options 'the 5 ′ caps are removed from the mRNA' and 'the DNA introns are removed from the template' pertain to post-transcriptional modifications and mRNA processing, not the start of transcription itself.
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Proteins, which have diverse functions in a cell, are all polymers of the same kinds of monomers amino acids. Write a short essay (100–150 words) that discusses how the structure of amino acids allows this one type of polymer to perform so many functions.
Answer:
There is various kind of proteins found in the human body, which exhibits an array of functions and applications. All the proteins are formed of similar monomers, that is, amino acids, though one can witness different kinds of proteins.
This has been made possible because of the existence of different sequences of amino acids coded for in the DNA. Due to different sequences of amino acids, each protein will fold distinctly and therefore, will exhibit distinct functions inside the body.
Thus, the distinct sequences of amino acids permit for the great diversity of proteins and their functions in the human body even though each of the protein comprises similar kinds of monomers.
If you are about to fight or flight a dog that wants to bite you, the cellular signal inside your body that will trigger a signaling network for this response is called:
a. NADH
b. connexon
c. estrogen
d. adrenaline
Answer:
D. Adrenaline
Explanation:
Adrenaline hormone is secreted by adrenal medulla during stressful conditions. The hormone is involved in preparing the body for stressful conditions by intensifying the sympathetic response.
If a person is chased by a dog, the adrenaline hormone is released from the adrenal medulla. The hormone triggers the preganglionic neurons of the sympathetic division of the autonomous nerve system to generate the flight or flight response.
Adrenaline increases the heart rate, blood pressure, and blood flow to skeletal muscles and adipose tissues. The hormone also triggers the dilation of airways and increases the blood levels of glucose and fatty acids.
A bacterium is infected with an experimentally constructed bacteriophage composed of the T2 phage protein coat and T4 phage DNA. The new phages produced would have
a. T2 protein and T4 DNA.
b. T2 protein and T2 DNA.
c. T4 protein and T4 DNA.
d. T4 protein and T2 DNA.
Answer:
Option (a).
Explanation:
Virus are the acellular organism and can only acts as living organism when present inside the host organism. Virus can have DNA or RNA as their genetic material.
The bacteria is infected with T2 phage protein coat and T4 phage DNA. As only DNA can acts as the genetic material and can be inherited to the next generation. The new phages has the T4 DNA as their genetic material and T2 phage protein coat.
Thus, the correct answer is option (a).
Answer:
Option (a).
Explanation:
With respect to angiosperms, which of the following is incorrectly paired with its chromosome count?
a. egg—n
b. megaspore—2n.
c. microspore—n
d. zygote—2n
Answer:
Megaspore—2n.
Explanation:
Angiosperms are the fruit bearing plants and reproduce by the process of sexual reproduction. The chromosome number are specific at each stage of the cell cycle of the angiosperms.
Microspores , egg are haploid. Zygote is diploid in nature. Megaspores get germinate into the female gametophytes and these are haploid in nature. Megaspores are also haploid in angiosperms.
Thus, the correct answer is option (b).
The description 'megaspore—2n' is INCORRECTLY paired with its chromosome count.
In plants (including angiosperms) the egg cell is the haploid (n) female gamete.
After fertilization, the egg cell forms a diploid (2n) zygote that subsequently develops the embryo inside the ovule.
A microspore is a haploid (n) cell that gives a male gametophyte.
In conclusion, the description 'megaspore—2n' is INCORRECTLY paired with its chromosome count.
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Tom's brother suffers from phenylketonuria (PKU), a recessive disorder. Tom and the brothers' parents do not have PKU. What are the chances that Tom is a carrier of the recessive PKU allele?
a. 1/4
b. 1/3
c. 1/2
d. 2/3
e. 4/3
Final answer:
Tom has a 2/3 chance of being a carrier of the recessive PKU allele.
Explanation:
To determine the chance that Tom is a carrier of the recessive PKU allele, we need to understand the inheritance pattern of PKU. PKU is a recessive disorder, which means that an individual needs to inherit two copies of the recessive allele to have the disorder. Since Tom's parents do not have PKU, they must be carriers of the recessive allele. This means that each parent has one copy of the recessive allele and one copy of the dominant allele.
If both parents are carriers, there are four possible combinations of alleles that they can pass on to their offspring: two dominant alleles (NN), one dominant allele and one recessive allele (Nn), one recessive allele and one dominant allele (nN), and two recessive alleles (nn).
Since Tom does not have PKU, we know that his genotype is either NN or Nn. The chance that Tom is a carrier of the recessive PKU allele is 2/3, which corresponds to the possibility of him having the Nn genotype. Therefore, the correct answer is d. 2/3.
Which of the following is an example of post-transcriptional control of gene expression?
a. the addition of methyl groups to cytosine bases of DNA
b. the binding of transcription factors to a promoter
c. the removal of introns and alternative splicing of exons
d. gene amplification contributing to cancer
Answer:
Option (c).
Explanation:
Transcription may be defined as the process of formation of RNA molecule from the template DNA with the help of enzymes and transcription factors. The transcription occurs in 5' to 3' direction.
Post trancriptional modification occurs in the RNA molecule that plays an important role in transcription as well as translation. The introns are removed from the RNA transcript and exons joined together is known as splicing. The alternative splicing occurs in which different protein isoforms are formed by the single exons.
Thus, the correct answer is option (c).
What is the basis for the difference in how the leading and lagging strands of DNA molecules are synthesized?
a. The origins of replication occur only at the 5′ end.
b. Helicases and single-strand binding proteins work at the 5′ end.
c. DNA polymerase can join new nucleotides only to the 3′ end of a pre-existing strand.
d. DNA ligase works only in the 3′ -> 5′ direction
Answer:c. DNA polymerase can join new nucleotides only to the 3′ end of a pre-existing strand.
Explanation:
The double stranded molecule of DNA having the pairing strands in the anti-parallel condition. One strand is remain in configuration as the 5’ to 3’ end, the complementary strand is positioned 3’ to 5’ end.
New DNA strand synthesis occurs from the 3’ to 5’ end as the enzyme DNA polymerase attach itself to the 3’ end of a DNA strand. It will add new nucleotides to the 3' end of the DNA.
Hence, the option (c) is the correct answer.
The primary reason for the difference in the synthesis of leading and lagging strands of DNA is that the DNA polymerase can only add new nucleotides to the 3′ ends of a pre-existing strand. DNA replication always occurs in the 5' to 3' direction, which influences the synthesis of both strands due to the anti-parallel structure of DNA.
Explanation:The basis for the difference in how the leading and lagging strands of DNA molecules are synthesized lies in the manner in which the DNA polymerase enzyme works. The key factor is that DNA polymerase can add new nucleotides only to the 3′ end of a pre-existing DNA strand.
This signifies that DNA replication always takes place in the 5′ -> 3′ direction. Considering the DNA double helix's anti-parallel structure, one strand (leading strand) is synthesized in a continuous fashion, as its orientation allows adding nucleotides at the 3′ end.
Conversely, the other strand (lagging strand) being oriented in 3′ -> 5′ direction, is synthesized discontinuously in small fragments called Okazaki fragments, as nucleotides can't be added to its 5' end. The enzyme DNA ligase then joins these pieces. Options a, b, and d may influence the process, but they're not the main reason for the leading and lagging strands' different synthesis.
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A researcher conducted crosses between two different strains of Drosophila. When true-breeding flies with singed bristles (s) and normal wings (L) were crossed to true-breeding flies with normal bristles (S) and vestigial wings (l), all F1 offspring had normal wings and normal bristles. The F1 offspring were crossed to flies with singed bristles and vestigial wings. Which F2 offspring is/are recombinant?
The best way of solving this is to draw a Punnett square.
You know the F0 had one parent with singed bristles (s), and normal wings (L), and the other parent is normal bristles (S) with vestigial wings (l).
If you do the cross ssLL x SSll you'll find 100% of the offspring is F1: SsLl, this means, all of them show the dominant traits: normal wings and normal bristles.
If you cross two parents from F1 to have F2, you'll find:
SsLl x SsLl = SSLL + SslL + sSlL+ ssll = 25% SSLL,all dominant traits. 50% SsLl is a recessive trait carrier but shows dominant traits. 25% ssll this one has all recessive alleles, which means, it will show vestigial wings and signed bristles.
The recombinant F2 offspring in this genetic cross of Drosophila are those that do not exhibit the parental trait combinations. Therefore, singed bristles/normal wings and normal bristles/vestigial wings are considered recombinant offspring. Option C is correct.
A researcher crossed true-breeding Drosophila flies with singed bristles (s) and normal wings (L) with true-breeding flies having normal bristles (S) and vestigial wings (l). The F1 offspring all had normal wings and normal bristles, indicating that these traits are dominant. When these F1 flies were crossed with flies having singed bristles and vestigial wings, the F2 generation showed various combinations of these traits.The F2 progeny would exhibit the following combinations: normal bristles/normal wings, singed bristles/vestigial wings, normal bristles/vestigial wings, and singed bristles/normal wings. The recombinant offspring are those that do not exhibit the parental combinations of traits. Thus, the recombinant F2 offspring would be singed bristles/normal wings and normal bristles/vestigial wings.Complete question as follows:
A researcher conducted crosses between two different strains of Drosophila. When true-breeding flies with singed bristles (s) and normal wings (L) were crossed to true-breeding flies with normal bristles (S) and vestigial wings (l), all F1 offspring had normal wings and normal bristles. The F1 offspring were crossed to flies with singed bristles and vestigial wings. Which F2 offspring is/are recombinant?
A.Singed bristles/vestigial wings only
B.Singed bristles/normal wings only
C.Singed bristles/normal wings and normal bristles/vestigial wings
D.Singed bristles/vestigial wings and normal bristles/normal wings
Prokaryotes have their chromosomes located in an area called the nucleoid.
a. True
b. False
Answer:
True
Explanation:
Unlike eukaryotic cells, which have a nucleus that contains the genome and is separated from the cytoplasm by a membrane, the prokaryotic nucleoid is not membrane-bound and is not considered an organelle. The nucleoid is simply the area within a prokaryiotic cell where its DNA is located.
A functional group on an amino acid that is polar and can become positively charged: ___________
Answer:
Amino group (NH3)
Explanation:
The amino acids have a central carbon atom to which four functional groups are bonded. These are namely a carboxyl group (COOH), an amino group (NH3), a hydrogen atom and an R group (it varies for different amino acids).
Amino group (NH3) is common to all the standard 20 amino acids. NH3 is a polar group since higher electronegativity of nitrogen atom allows it to pull the shared electrons of the covalent bonds towards itself. It makes the nitrogen atom of amino group partially negative and the hydrogen atoms carry a partial positive charge (a dipole is present).
Amino group (NH3) accepts one proton and becomes positively charged (NH4+).
The polar functional group that can become positively charged on an amino acid is typically found in basic amino acids like lysine and arginine.
Explanation:The functional group on an amino acid that is polar and can become positively charged is often found in basic amino acids. Basic amino acids such as lysine and arginine have side chains that contain a positive charge under physiological conditions. These amino acids are essential in the structure and biological activity of proteins. For example, in the classification of amino acids, those like lysine (Lys) and arginine (Arg) are recognized for the basic groups in their side chains which can participate in hydrogen bonding and other important biochemical interactions.
The functional group on an amino acid that is polar and can become positively charged is the amino group (-NH2). This group contains a nitrogen atom, which can accept a hydrogen ion (H+) to form a positively charged ammonium group (-NH3+). Amino acids that have an amino group in their side chain, such as lysine and arginine, are considered basic amino acids because they can donate a proton and become positively charged.
How many possible open reading frames (frames without stop codons) are there that extend through the following sequence? 5'... CTTACAGTTTATTGATACGGAGAAGG... 3' 3'... GAATGTCAAATAACTATGCCTCTTCC... 5'
According to the DNA sequence shown in the question above, we can see that there are 3 reading frames without stop codons.
You can find this answer as follows:
Isolate the 5'-3' Sequence: You will not need the 3'-5' Sequence, so you will need to isolate the 5'-3' Sequence. In this, you should observe the sequence of nitrogenous bases, identify the codons (set of three nitrogenous bases) and create three sequences. The first sequence will exclude the first nucleotide, the second sequence will exclude the second nucleotide, and the third sequence will exclude nothing. Based on this, you will have the three sequences below:5'... C TTA CAG TTT ATT GAT ACG GAG AAG G... 3' (no stop codon)
5'... CT TAC AGT TTA TTG ATA CGG AGA AGG... 3' (no stop codon)
5'... CTT ACA GTT TAT TGA TAC GGA GAA GG... 3' (with stop codons)
Create complementary sequences: You should do this based on the complementarity of the nitrogenous bases. In this case, it is necessary to remember that Adenine (A) is complemented by Thymine (T), Guanine (G) is complemented by cytosine (C), and vice versa. In this case, you will have the following sequences:3'... GAA TGT CAA ATA ACT ATG CCT CTT CC... 5'
3'... GA ATG TCA AAT AAC TAT GCC TCT TCC... 5'
3'... G AAT GTC AAA TAA CTA TGC CTC TTC C... 5'
Identify the inverse complement: Still taking into account the complementarity of the nitrogenous bases, you should find the inverse complementary sequences. In this case, you will find the sequences:CCT TCT CCG TAT CAA TAA ACT GTA AG (with stop codon)
CC TTC TCC GTA TCA ATA AAC TGT AAG (no stop codon)
C CTT CTC CGT ATC AAT AAA CTG TAA G (with stop codons)
From this, we can see which sequences have one of the stop codons, which are TAA, TAG, TGA. As you can see, only the first two frames of the 5'-3' sequences and the first two frames of the inverse sequences do not have these codons, so it is possible to observe 3 open reading frames without stop codons.
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The query's genetic sequence potentially has three different open reading frames, starting from the first, second or third position. However, without specific information about the position of the start codon AUG and the stop codons, we can't definitively verify how many open reading frames extend through this sequence.
Explanation:In genetics, the concept of an open reading frame (ORF) is of immense importance. An ORF is a sequence of DNA or RNA that could be potentially translated to give a protein. Understanding the open reading frames in a genetic sequence helps us predict how that sequence might be translated into amino acids.
For the given sequence, we can have three ORFs since you can start reading the sequence from three different positions (first, second, or third nucleotide) until you run into a stop codon. However, without knowing the exact position of the start and stop codons in this specific sequence, we cannot accurately say exactly how many open reading frames might extend through it.
The concept of AUG start codon used to initiate translation and stop codons that terminate protein synthesis are essential for determining open reading frames. The sequence AUG signifies the beginning of an ORF.
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