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:
Consider an advantageous allele segregating in a population as a major polymorphism. Which of the following would not generally slow the fixation of the allele?
a. The environment changes in such a manner as to reduce the selective advantage.
b. The population begins to receive immigrants from a population that maintains the same initial frequency of the alleles.
c. In addition to being advantageous the allele also exhibits overdominance.
d. The population begins to exhibit positive assortative mating for each of the genotypes.
e. The population size increases.
Answer:
C. In addition to being advantageous the allele also exhibits overdominance.
Explanation:
If the allele is advantageous and dominant over the other alleles, the individuals who own it will adapt better to the environment and most of their offspring will exhibit the attributes granted by the allele, which would increase its frequency in the population over time. In addition, individuals who own it will be more successful and more likely to reproduce than those who do not.
In order to caluclate the specific activity of the isolated
fractions, the protein concetration has to be determined. Why is
the protein concetration necessary for accurate comparison of the
enzyme activity of the fractions?
Answer:
Explanation:
This is evident that all enzymes are proteins but not all proteins are enzymes. The specific activity can be define as the number of enzyme units per milliliters that is divided by the available concentration of the proteins typically in mg/ml. Thus the value of the specific activity can be measured in units/mg.
In others words this can be said that how much enzymes units can be found in the 1 mg of the total protein. So in the total concentration of the proteins the estimation of the enzyme units is possible. Thus protein concentration is necessary for calculating the number of the enzyme units.
These bacteria are not pathogenic and normally reside in the vagina, creating an acidic environment.
A) Neisseria
B) Listeria
C) Lactobacillus
D) none of the above
Answer:
C) Lactobacillus
Explanation:
Lactobacilli form part of the natural flora of the vagina producing lactic acid and hydrogen peroxide which keep the pH acidic and prevent the growth of yeast.
Explain the meaning of the term "tetrad" as applied to the asci produced by certain fungi.
Answer:
Explanation:
In the members of the class ascomycetes such as yeast, two haploid yeast cells of two different stains or mating types fuse to form a diploid zygote. Zygote quickly undergoes meiosis or reductional division to form four haploid spores called a tetrad. The tetrad is contained in a sac-like structure called an ascus or fruiting body. Formation of ascus is the characteristic feature of ascomycetes. In the ascus tetrad are not arranged in any specific order. The tetrad are called random or unordered spores. The genetic analysis of the genotypes in the tetrad can tell us about the events during meiosis.
A gene (gene X) undergoes a mutation which converts if from wild-type to mutant. Then a second mutation in the gene Y occurs which causes the wild-type phenotype of the first gene (gene X) to be restored. Respectively, what are the appropriate designations for the two mutational events?
a. forward mutation, back mutation
b. forward mutation, suppressor mutation
c. reverse mutation, back mutation
d. reverse mutation, suppressor mutation
e. reverse mutation, forward mutation
Answer:
b. forward mutation, suppressor mutation
Explanation:
When the nucleotide sequence of an organism is altered, it is called as mutation. It can be caused by DNA damage or replication errors. In forward mutation, the wild type allele is converted to a mutant version such that the gene product is non functional or its not produced at all. Suppressor mutation is the second mutation which reverses the phenotypic effects of the previous mutation. This process is called as synthetic rescue.
Since here gene X was converted into a mutant form by the mutation, it had undergone forward mutation. When gene Y was mutated, the function of gene X was restored which ultimately also restored the phenotype hence it is an example of suppressor mutation.
The surgical removal of male reproductive organs is known as orchidectomy.
a. True
b. False
Answer:
The correct answer will be option-true.
Explanation:
Orchidectomy or orchi is the surgical procedure performed to remove the male reproductive organ of humans especially testicles.
This procedure is performed to treat testicular cancer, to manage prostate cancer and testicular torsion. Once a testicle is removed, the person becomes infertile, may lose sexual interest, high production of estrogen and breast enlargement.
Thus, option-true is the correct answer.
What are the nephrons? How are they utilized in filtration of the blood?
What are the nephron?
Nephrons are the functional unit of the kidney. There are about two million nephrons in each of our kidneys. Each nephron has a network of glomelural capillaries called glomerulus where blood filtration occurs, and the renal tabule which is where the filtered fluid is converted to urine.
How they work?
The nephrons act as a filter, cleaning our blood. Unwanted metabolites like urea and creatinine are taken from the blood, as well as high amounts of sodium. The filtered fluid flows from inside Bowman's capsule (epithelial cells surrounding the glomerulus) and from there into the proximal tubule (see attached figure at the end). From the tubule, fluid flows into several other ducts until it reaches the ducts where collectors will empty into the renal pelvis.
Nephrons are functional units in the kidney that play a vital role in filtering blood and forming urine. They undergo three principal biological processes: filtration, reabsorption, and secretion. Ultimately, these nephrons contribute to the creation of urine from the filtrate, maintain body homeostasis, and help excrete potential toxins.
Explanation:Nephrons are the functional units of the kidney, critical for the filtration of blood and the formation of urine. Composed of the renal corpuscle and the renal tubule, they take a straightforward filtrate of the blood and transform it into urine, a process that involves three principal functions: filtration, reabsorption, and secretion. The glomerulus is a significant part of the nephron; it functions as a specialized capillary bed that enables the filtration process.
The filtration process uplifts almost all solutes, except proteins, from the bloodstream into the glomerulus, a process known as glomerular filtration. Next, the filtrate is collected in the renal tubules where most solutes are reabsorbed in a process known as tubular reabsorption. The filtrate then interacts with solutes and water in the loop of Henle, where further reabsorption occurs. Subsequently, additional solutes and wastes are secreted into the kidney tubules during tubular secretion. The final product, urine, is then collected from the filtrate in the collecting ducts.
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What is the probability that each of the following pairs of parents will produce the indicated offspring? (Assume independent assortment of all gene pairs.)
(a) AABBCC × aabbcc->AaBbCc
(b) AABbCc × AaBbCc->AAbbCC
(c) AaBbCc × AaBbCc->AaBbCc
(d) aaBbCC × AABbcc->AaBbCc
Answer:
(a) AABBCC × aabbcc->AaBbCc → 1
(b) AABbCc × AaBbCc->AAbbCC → 1/32
(c) AaBbCc × AaBbCc->AaBbCc → 1/8
(d) aaBbCC × AABbcc->AaBbCc → 1/2
Explanation:
In such cases we calculate the probability of each allelic pair separately.
(a) AABBCC × aabbcc -> AaBbCc
Aa Aa Aa Aa Bb Bb Bb Bb Cc Cc Cc Cc
↓ ↓ ↓
4/4 = 1 4/4 = 1 4/4 = 1
In case of gene A, out of the 4 probable allelic combinations, 4 will be Aa so the probability is 4/4 = 1.
In case of gene B, out of the 4 probable allelic combinations, 4 will be Bb so the probability is 4/4 = 1.
In case of gene C, out of the 4 probable allelic combinations, 4 will be Cc so the probability is 4/4 = 1.
So, the total probability of getting AaBbCc = 1 x 1 x 1 = 1.
(b) AABbCc × AaBbCc -> AA bb CC
AA AA Aa Aa BB Bb Bb bb CC Cc Cc cc
↓ ↓ ↓
2/4 1/4 1/4
In case of gene A, out of the 4 probable allelic combinations, 2 will be AA so the probability is 2/4.
In case of gene B, out of the 4 probable allelic combinations, 1 will be bb so the probability is 1/4.
In case of gene C, out of the 4 probable allelic combinations, 1 will be CC so the probability is 1/4.
So, the total probability of getting AA bb CC = 2/4 x 1/4 x 1/4 = 1/32.
(c) AaBbCc × AaBbCc -> AaBbCc
AA Aa Aa aa BB Bb Bb bb CC Cc Cc cc
↓ ↓ ↓
2/4 2/4 2/4
In case of gene A, out of the 4 probable allelic combinations, 2 will be Aa so the probability is 2/4.
In case of gene B, out of the 4 probable allelic combinations, 2 will be Bb so the probability is 2/4.
In case of gene C, out of the 4 probable allelic combinations, 2 will be Cc so the probability is 2/4.
So, the total probability of getting AaBbCc = 2/4 x 2/4 x 2/4 = 1/8.
(d) aaBbCC × AABbcc->AaBbCc
Aa Aa Aa Aa BB Bb Bb bb Cc Cc Cc Cc
↓ ↓ ↓
4/4 = 1 2/4 4/4 = 1
In case of gene A, out of the 4 probable allelic combinations, 4 will be Aa so the probability is 4/4 = 1.
In case of gene B, out of the 4 probable allelic combinations, 2 will be Bb so the probability is 2/4.
In case of gene C, out of the 4 probable allelic combinations, 4 will be Cc so the probability is 4/4 = 1.
So, the total probability of getting AaBbCc = 1 x 2/4 x 1 = 1/2.
a. The probability for a cross of AABBCC × aabbcc to produce AaBbCc is: 1.
b. The probability for a cross of AABbCc × AaBbCc to produce AAbbCC is: [tex]\frac{1}{32}[/tex]
c. The probability for a cross of AaBbCc × AaBbCc to produce AaBbCc is: [tex]\frac{1}{8}[/tex]
d. The probability for a cross of aaBbCC × AABbcc to produce AaBbCc is: [tex]\frac{1}{2}[/tex]
Recall:
The Principle of Independent Assortment of genes holds that genes will separate from each one another independently during development of reproductive cell.A Punnett Square can be used to show the offspring that will be produced during a cross between two parents.Thus, the images below shows the various outcome of each cross.
a. Figure A shows the cross between:
AABBCC × aabbcc
The genotype of all offspring produced are only AaBbCc.
Therefore, the probability for a cross of AABBCC × aabbcc to produce AaBbCc is: 1.
b. Figure B shows the cross between:
AABbCc × AaBbCc
From the cross, there are only 2 AAbbCC out of 64 offspring produced.
That is: [tex]\frac{2}{64} = \frac{1}{32}[/tex]Therefore, the probability for a cross of AABbCc × AaBbCc to produce AAbbCC is: [tex]\frac{1}{32}[/tex]
c. Figure C shows the cross between:
AaBbCc × AaBbCc
From the cross, there are only 8 AaBbCc out of 64 offspring produced.
That is: [tex]\frac{8}{64} = \frac{1}{8}[/tex]Therefore, the probability for a cross of AaBbCc × AaBbCc to produce AaBbCc is: [tex]\frac{1}{8}[/tex]
d. Figure D shows the cross between:
aaBbCC × AABbcc
From the cross, there are only 32 AaBbCc out of 64 offspring produced.
That is: [tex]\frac{32}{64} = \frac{1}{2}[/tex]Therefore, the probability for a cross of aaBbCC × AABbcc to produce AaBbCc is: [tex]\frac{1}{2}[/tex]
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Which of the following statements concerning chromosomal distribution is INCORRECT?
A. All human somatic cells contain 23 chromosomal pairs for a total diploid number of 46 chromosomes.
B. Each gamete contains 23 chromosomes, one member of each chromosomal pair.
C. During meiotic division, the members of the chromosome pairs regroup themselves into the original combinations derived from the individual's mothers and father for separation into haploid gametes.
D. Sex determination depends on the combination of sex chromosomes, an XY combination being a genetic male, XX being a genetic female.
E. The sex chromosome content of the fertilizing sperm determines the sex of the offspring.
Answer:
C. During meiotic division, the members of the chromosome pairs regroup themselves into the original combinations derived from the individual's mothers and father for separation into haploid gametes.
Explanation:
During meiosis, there is a random segregation of chromosomes. Metaphase-I of meiosis-I includes alignment of homologous pairs of chromosomes at the cell's equator.
During anaphase I, the homologous chromosomes are separated from each other and move towards the opposite poles. This segregation of homologous chromosomes to the opposite poles is a random event and creates unique combinations of maternal and paternal chromosomes at each pole and finally in each gamete.
Those parts of a new mRNA transcript that get spliced out and don't wind up getting translated into proteins are called:
a. exons
b. spliceosomes
c. introns
d. protons
Answer:
The correct answer is c. introns
Explanation:
Newly transcribed mRNA contains coding and non coding sequences. Coding sequences are called exons and non coding sequences are called introns. Introns do not code for proteins and hinder translation so they are removed from the mRNA in a process called splicing.
After splicing of introns from mRNA only exons are left in the mRNA which contains coding region for protein synthesis and are translated into functional proteins.
So introns are parts of a new mRNA transcript that get spliced out and don't wind up getting translated into proteins. Sometimes introns join together to form their own proteins.
Thus, the correct answer is c. introns.
What are potential mechanisms that can lead to endocrine dysfunctions?
Answer:
Hypothalamus-pituitary dysfunction
Adrenal disorders
Endocrinopathies of the reproductive system
Endocrinopathies related to the parathyroid glands
Thyroid pathologies
Endocrine Pancreas Disorders
Explanation:
Hypothalamus-pituitary dysfunction
Diseases of the anterior pituitary: Pituitary hypofunction may be due to a disease of the pituitary itself or the hypothalamus. In any case there is a decreased secretion of pituitary hormones with subsequent effects on the function of the rest of the body. Thus the TSH deficit produces hyperthyroidism without goiter; the deficit of LH and FSH causes hypogonadism; ACTH deficiency results in hypoadrenalism and poor skin color; Prl deficiency causes postpartum breastfeeding failure and GH deficiency causes short stature (dwarfism), facial wrinkles and occasionally fasting blood glucose in children.
Vasopressin disorders: (SIADH) is characterized by objectifying an excess of ADH, hyponatremia and water intoxication, all in the absence of hypovolemia, hypotension, heart failure, hypothyroidism or corticosuprarenal insufficiency
Adrenal disorders
The adrenal glands are responsible for the synthesis of various hormones. In the cortical zone the following hormones are synthesized: the mineralcorticoids whose production is related to the glomerular zone, the glucocorticoids whose secretion is attributed to the fasciculate zone and that of androgens with the reticular zone. Although it is clear that in the glomerular zone only the synthesis of aldosterone occurs, because it lacks 17-a-hydroxylase that incapacitates it to secrete cortisol and androgens . Includes:
Adrenal pathology with hyperfunction: Mineralcorticoid hyperfunction, Glucocorticoid hyperfunction or Cushing syndrome, Androgenic hyperfunction, Adrenal medulla hyperfunction
Adrenal pathology with hypofunction: Chronic primary adrenal corticosteroid hypofunction or Addison's disease, Acute corticosuprarenal hypofunction, Secondary adrenal corticosteroid hypofunction, Selective hypocorticisms.
Endocrinopathies of the reproductive system
Ovarian hyperfunction: Ovarian hyperfunction refers to the excessive production of androgens or estrogens by the ovary, possibly due to a primary tumor of the ovary or a gonadotropodependent ovarian hypoplasia
Ovarian hypofunction: Ovarian hypofunction may be primary or secondary, as due to disorders in the ovary itself or as a result of extragonadal disorders. The most common disorders in primary hypofunction are sexual infantilism and short stature, accompanied by a series of manifestations such as low implantation ears, short neck, chest chest, shortening of the 4th and 5th metacarpal and metatarsal
Disorders of the male reproductive system: The testicles fulfill two functions: hormonal production and spermatogenesis. Male reproductive disorders are grouped into hypogonadism, infertility, varicocele and gynecomastia
Endocrinopathies related to the parathyroid glands
The regulation of calcium and phosphate metabolism is very complex. The concentration of both remains constant in the blood although its administration varies considerably
Includes: Hyperparathyroidism, Hypoparathyroidism
Thyroid pathologies
Thyroid disorders include a series of syndromes that include the effects of a hypofunction of the gland or a hyperfunction of the gland. The different types of thyroiditis include a set of inflammatory disorders of diverse etiology that have in common the destruction of the thyroid follicle.
Includes: Hyperthyroidism, Hypothyroidism
Endocrine Pancreas Disorders
It is widely known that the pancreas in addition to its digestive functions, is responsible for the secretion of the hormones insulin and glucagon whose functions are closely related to the regulation of the metabolism of lipids, proteins and mainly carbohydrates. In both cases it is a small protein.
Includes: Diabetes Mellitus
Compared with 31P, the radioactive isotope 32P has a.
a. different atomic number.
b. One more proton.
c. One more electron.
d. One more neutron.
Answer:
d. One more neutron
Explanation:
Phosphorus-32 is known as a radioactive isotope of the phosphorus element. The nucleus of this isotope has 15 protons and 17 neutrons.
There is also Phosphorus-31 which is the most common isotope of phosphorus. This isotope has at its core 15 protons and 16 neutrons.
Therefore, the difference between them is one neutron more than the other.
The radioactive isotope 32P has one more neutron compared to 31P; the number of protons and the atomic number remain unchanged.
Compared with 31P, the radioactive isotope 32P has one more neutron. Isotopes are atoms that have the same number of protons but a different number of neutrons. The atomic number, which represents the number of protons, remains the same across isotopes of the same element. Therefore, option a, different atomic number, and option b, one more proton, are incorrect because the atomic number for phosphorus is 15 for both isotopes, and therefore, the proton count is also the same. Option c, one more electron, is also incorrect because the number of electrons in a neutral atom matches the number of protons, which has not changed between these isotopes. 32P has one more neutron than 31P, making option d the correct answer.
Fill in the Blank: Transcription factors (proteins) help our cells to promote gene expression (e.g., the production of insulin) by binding directly to DNA and by assisting ______________________ to initiate transcription.
a. actin monomers
b. tight junctions
c. RNA polymerases
d. muscle fibrils
Answer:
The correct answer will be option-C.
Explanation:
The RNA polymerase is the enzyme which synthesizes the mRNA molecules using a single strand of DNA. RNA polymerase has the ability to bind nucleotide at 3' end of the strand thus proceeding the strand in 5' to 3' direction.
In the given question, gene expression of the insulin-producing gene has been discussed which uses transcription factors. The transcription factors assist RNA polymerase enzyme to attach to the promoter sequence and start synthesizing RNA molecule.
Thus, Option-C is the correct answer.
During aerobic respiration, electrons travel downhill in which sequence?
A) food --> Krebs cycle --> ATP -->NAD+
B) food --> NADH --> electron transport chain --> oxygen
C) glucose --> ATP --> oxygen
D) glucose --> ATP --> electron transport chain --> NADH
E) food --> glycolysis --> Krebs cycle --> NADH --> ATP
Answer:
The correct answer will be option-B.
Explanation:
Cellular respiration is a slow process which produces energy from the oxidation of food components which could be glucose, fatty acids and proteins.
The process proceeds in four steps: glycolysis, link reaction, Krebs cycle and electron transport chain.
The energy molecules are produced when a high energy electron gets reduced and the released energy is stored in the form of energy molecules which could be ATP or energy equivalents like NADH and FADH₂.
These reducing equivalents donate its electrons during electron transport chain where the electrons flow to the last electron acceptor called oxygen.
Thus, Option-B is the correct answer.
Option B is correct. The correct order of electron travel downhill is
[tex]\rm \bold {Food\rightarrow NADH \rightarrow \texttt{Electron transport chain} \rightarrow Oxygen}[/tex]
Aerobic Respiration is the process by which food convert to energy.
It completes in few steps
The food digested and convert to Glucose.Glucose broken down in Glycolysis to form pyruvate.The third step is Krebs cycle where NADH is generated.NADH is used in Electron transport chain to Form ATP.In the last electron transferred to Oxygen.So,option B is Correct. other sequences are incorrect.
Hence we can say that the correct order of electron Down hill is
[tex]\rm \bold {Food\rightarrow NADH \rightarrow \texttt{Electron transport chain} \rightarrow Oxygen}[/tex]
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The fuel usually used in cellular respiration is ___________.
a. galactose
b. fructose
c. chitin
d. glucose
Answer: d. glucose
Explanation:
Cellular respiration starts with a process called glycolysis. Glycolysis breaks down one molecule of glucose into two molecules of pyruvate. Then pyruvate oxidates and forms Acetyl CoA, which enters the citric acid cycle and ends up producing 36 ATP molecules that will be used as energy by the cell.
Other sugars can enter the cycle but must be turned into some intermediate molecule first. For example, fructose must be turned into fructose-6-phosphate, the third molecule used in glycolysis.
A cell in the body is recognized as "self" by its _________ and is therefore not targeted by the immune response for destruction.
a. particular region of genomic DNA
b. antibodies
c. major histocompatibility complex (MHC) membrane proteins
d. mRNA sequences
e. membrane phospholipids
Answer:
C
Explanation:
The major histocompatibility complex (MHC) is a group of genes whose function is to codify proteins that participate in the immune response, helping the system to recognize foreign substances to develop an immune response.
Histocompatibility or compatibility of tissue is given by self-identifications molecules (antigens) located on the surface of cells, membrane, these molecules are almost unique to each person, letting the body to distinguish self from non-self.
Which of the following is an example of qualitative data?
a. the fish swam in a zigzag motion.
b. The contents of the stomach are mixed every 20 seconds.
c. The temperature decreased from 20oC to 15oC.
d. The six pairs of robins hatched an average of three chicks each.
Answer:
a. the fish swam in a zigzag motion.
Explanation:
When someone quantifies (quantity) usually assigns a certain number (1,2,3 etc.) quality is a subjective measure that depends on the point of view of the observer like color, behaviour.
cAMP activates cAMP-dependent protein kinase by
bindingregulatory subunits and inducing their release from the
catalyticsubunits.
stimulating itsphosphorylation.
stimulating thedimerization of kinase subunits.
stimulating therelease of a translational inhibitory protein
bound to itsmRNA.
Answer:
binding regulatory subunits and inducing their release from the catalytic subunits
Explanation:
cAMP molecules diffuse into the cytoplasm where they bind to an allosteric site on a regulatory subunit of a cAMP-dependent protein kinase ( protein kinase A, PKA).
-In its inactive form, PKA is a heterotetramer comprised of two subunits namely, regulatory (R) and two catalytic (C) subunits.
-The regulatory subunits normally inhibit the catalytic activity of the enzyme. cAMP binding causes the dissociation of the regulatory subunits, thereby releasing the active catalytic subunits of PKA.
-cAMP stimulates glucose mobilization by activating a protein kinase that adds a phosphate group onto a specific serine residue of the glycogen phosphorylase polypeptide.
Familial hypercholesterolemia (FH) is an inherited trait in humans that results in higher than normal serum cholesterol levels (measured in milligrams of cholesterol per deciliter of blood (mg/dl)). People with serum cholesterol levels that are roughly twice normal have a 25 times higher frequency of heart attacks than unaffected individuals. People with serum cholesterol levels three or more times higher than normal have severely blocked arteries and almost always die before they reach the age of 20. The pedigrees above show the occurrence of FH in four Japanese families: a. What is the most likely mode of inheritance of FH based on this data? Are there any individuals in any of these pedigrees who do not fit your hypothesis? What special conditions might account for such individuals? b. Why do individuals in the same phenotypic class (unfilled, yellow, or orange symbols) show such variation in their levels of serum cholesterol?
Answer:
Thanks for you question. Your hypothesis suggests a linear relationship between serum Cholesterol levels and MI. This hypothesis seems to ignore the difference in the prevalence and effectiveness of LDL receptors in the FH patient.
FH patients who have inherited the mutation from both parents have very few LDL receptors in their blood and therefore almost no ability to pass the unused Cholesterol through the liver. FH patients who are heterozygous will have more LDL receptors although both will find Cholesterol removal problematic without the addition of a PCSK9 inhibitor.
In short, your hypothesis need to account for other factors that are in play.
Explanation:
Consider my case. I am a 64 year old male who has Heterozygous Familial Hypercholesterolemia. Before treatment at age 12 my Total cholesterol was 510 mg/dl. My genetic testing shows two mutations to the LDL Receptor gene with only one mutation being pathogenic. My first heart attack was at 47 and first stroke at 62. My current LDL is too low to detect with the use of a PCSK9 inhibitor (Repatha®).
What is happening during the plateau phase of
cardiacdepolarization?
A) Sodium ions block voltage-gated calcium ion channels.
B) Acetylcholine is hyperpolarizing the SA node
C) Nothing - all ion gates and valves are closed.
D) voltage-gated potassium ion channels open to allowpotassium
ions to diffuse out.
E) Calcium ions enter the cystol of cardiac myofibers
fromsarcoplasmic reticulum and through voltage-gated
slowchannels.
Answer:
Option (3).
Explanation:
Heart contains the cardiac myocetes cells that are surrounded by the sarcolemma. Heart has the ability to conduct the electrical impulse same as the muscle cell have.
The plateau phase is the second phase of the cardiac action potential. During this phase, the calcium ions moves out of the cell from the sarcoplasmic reticulum to the cytosol through voltage gated sodium calcium channels. Sodium ions will move inside the cell during this phase.
Thus, the correct answer is option (3).
Identify all the amino acid-specifying codons where a point mutation (a single base change) could generate a nonsense codon.
Answer:
Glutamic Acid (Glu): codons GAA, GAG
Glutamine (Gln): codons CAA, CAG.
Lysine (Lys): codons AAA, AAG, UCG
Serine (Ser): codons UCA, UCG
Leucine (Leu): codons UUG, UUA
Tyrosine (Tyr): codons UAC, UAU
Tryoptophan (Trp): codon UGG
Arginine (Arg): codons CGA, AGA
Glycine (Gly): codon GGA
Cysteine (Cys): codon UGU
Explanation:
Non sense codons are: UAA, UAG, UGA. Then following the genetic code (see attached file) a single base substitution in any of the codons indicated in the answer could generate a stop codon. This single base substitution might happen in the first, second or third base of the codon.
Final answer:
A nonsense mutation occurs when a point mutation changes an amino acid codon to a stop codon, prematurely terminating protein translation and potentially resulting in a nonfunctional protein.
Explanation:
A nonsense mutation is a type of point mutation that converts a codon encoding an amino acid (a sense codon) into a stop codon (a nonsense codon), such as TAA, TAG, or TGA. When a nonsense mutation occurs, translation of the mRNA will stop prematurely, leading to a potentially truncated and nonfunctional protein.
To identify all amino acid-specifying codons where a single base change could generate a nonsense codon, one would examine each codon in a gene sequence and determine if mutating one of its bases could result in TAA, TAG, or TGA. For example, changing CAA (which encodes glutamine) to UAA creates a stop codon, resulting in premature termination of the protein during translation and potentially rendering it nonfunctional. Other examples might include altering one base in CAG or CGA codons to similarly create stop codons.
Your girlfriend seems quite thin and is a cheerleader so she gets plenty of exercise. She hardly seems to eat when you go out though, and in addition to team practices, she spends a great deal of time exercising and expresses concern about her weight. Which eating disorder might you suspect?
A. Anorexia nervosa
B. Binge-eating disorder
C. Obesity
D. Bulimia nervosa
E. Not so much an eating disorder as just an obsession
Answer:A. Anorexia nervosa
Explanation: this shows signs of anorexia because she doesnt eat much. Anorexia is when people deorive themselfs of food and try to become as skinny as possible. A lot of people do this because they feel uncomfortable with their body, or if their going through a hard time in their life and doing this help them cope....it then becomes an addiction. Some people think this is signs of bulimia...but bulimia is actually somone eating but purging themselfs after they eat.
Arrange the following list of eukaryotic gene elements in the order they would appear in the genome and in the direction traveled by RNA polymerase along the gene. Assume the gene's single intron interrupts the open reading frame. Note that some of these names are abbreviated and thus do not distinguish between elements in DNA versus RNA. For example, "splice-donor site" is an abbreviation for "DNA sequences transcribed into the splice-donor site" because splicing takes place on the gene's RNA transcript, not on the gene itself. Geneticists often use this kind of shorthand for simplicity, even though it is imprecise. (a) splice-donor site; (b) 3' UT R; (c) promoter; (d) stop codon; (e) nucleotide to which methylated cap is added; (f) initiation codon; (g) transcription terminator; (h) splice-acceptor site; (i) 5' UT R; (j) poly-A addition site; (k) splice branch site.
Answer:
The alignment of the elements in the following sequence will take place in the eukaryotic genome:
a. Promoter
b. Nucleotide to which methylated cap is added
c. 5 prime UTR
d. Initiation codon
e. Splice donor
f. Splice branch site
g. Splice acceptor
h. Stop codon
i. 3 prime UTR
j. Transcription terminator
k. Poly A addition site
After the process of splicing, the ultimate transcript will comprise the elements b, c, d, h, i. In eukaryotes, the RNA polymerase begins the process of transcription after it crosses the promoter region, and ceases at the transcription terminator. At the time of RNA processing, a 5 prime cap is supplemented to the transcript, splicing occurs, and a poly-A tail is supplemented. The 5 prime UTR and 3 prime UTR regions are found in the final transcript, that is, the mature RNA, however, are not translated.
The eukaryotic gene elements, in the order of appearance and direction of RNA polymerase travel along the gene, would be promoter, nucleotide to which methylated cap is added, 5' UTR, initiation codon, splice-donor site, splice branch site, splice-acceptor site, stop codon, 3' UTR, poly-A addition site and finally transcription terminator.
Explanation:The order and direction of the eukaryotic gene elements, as the RNA polymerase travels along the gene, would be as follows:
Promoter (c)Nucleotide to which methylated cap is added (e)5' UTR (i)Initiation Codon (f)Splice-donor site (a)Splice branch site (k)Splice-acceptor site (h)Stop codon (d)3' UTR (b)Poly-A addition site (j)Transcription terminator (g)The process begins with the promoter, continues with the addition of the methylated cap, initiation codon then the intron interruption with the splice-donor, branch, then splice-acceptor sites. After the intron, the gene sequence continues until it hits the stop codon and the untranslated regions ending with the Poly-A addition site and finally, the transcription terminator.
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The dietary energy content of food is measured in Calories.
a. True
b. False
Answer:
A. True
Explanation:
I hope I was helpful!
When the bases are present in their rare imino or enol states, they can form which of the folowing pair combination?
a. A:T
b. C:G
c. A:C
d. A:T and C:G
e. C:G and A:C
Answer:
The correct answer is c. A:C
Explanation:
Thymine and guanine are most stable in their keto form and adenine and cytosine are most stable in their amino form. Sometimes these bases undergo tautomeric shift(proton shift) which converts them into rare less stable form called tautomeric form which are imino and enol form.
In this rare form, these bases can pair with the base which are not complementary to them. Like adenine base pairs with cytosine and guanine base pairs with thymine when present in rare form.
When these tautomeric forms of nucleotides incorporated in DNA they may cause mutation in the DNA.
Therefore the correct answer is c. A:C .
Taste receptors on the tongue are not related to smell receptors of the nose.
a. True
b. False
Answer:
b. False
Explanation:
As human has about 350 olfactory receptors and subtypes work in various conditions allowing us to sense about 10,000 doors. All senses of smell and tastes merge at the back of the throat. As you taste something before smelling it the smell lingers on inside in the nose which makes you smell it. As both the smell and taste are chemoreceptors which means both have chemically same sensing environments. In addition, the division of taste receptors within the nose coordinate with activities, although humans can distinguish between the tastes from the smells of the objects. Working together to create a perception of flavor through the nasal passage.(a) What cell structures best reveal evolutionary unity?
(b) Provide an example of diversity related to specialized cellular modifications.
Answer:
The correct answer will be-
1. Evolutionary unity- DNA
2. specialized cellular modifications- plant cells have chloroplasts and large central vacuole.
Explanation:
Case I
Evolutionary unity is a concept which explains that all the organisms possess some common structure or universal structure or molecules.
The best characterized evolutionary molecule is DNA molecule which took the role of genetic material in past and is passed on to the daughter cells and still is made up of the same components which are a five-carbon sugar, four types of nitrogenous bases and a phosphate group. This DNA molecule is present in almost every organism except for a few viruses.
Case II
The specialization of the cellular components depends on the function a cell has to perform which became established in the population. The plant cells depended on the sunlight to prepare food for themselves which was done by the chloroplast organelle which is absent in the animal kingdom.
Cell structures like ribosomes, mitochondria, and nuclei reveal evolutionary unity. Specialized cellular modifications, such as the dendrites and axons in neurons, demonstrate diversity.
Explanation:The cell structures that best reveal evolutionary unity are the organelles that are universally present in cells across different species, such as ribosomes, mitochondria, and nuclei among others. These organelles play fundamental roles in cellular life processes, indicating that all life forms evolved from a common ancestor which also possessed these structures.
An example of diversity related to specialized cellular modifications can be seen in neuron cells. These cells have specialized structures known as dendrites and axons, which enable them to perform their specific function of transmitting electrical signals throughout the body. Other cells may also exhibit modifications in their structure, size, and shape according to their specific functions.
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Light-dependent repair corrects which of the following DNA alterations?
a. methylation
b. thymine dimers
c. mismatched basepairing
d. hydroxylation
e. inversions
Answer:
The correct answer will be option-B.
Explanation:
The light-dependent repair process is a process which repairs the pyrimidine dimers formed from the UV radiation. This process is also known as photo-reactivation which issued by the bacteria to repair the DNA.
The process requires specific enzymes like photolyase which binds to pyrimidine dimers usually thymine dimers formed and catalyze the reaction in the presence of visible light. The process returns the DNA state to its prior state before UV damage.
Thus, Option-B is the correct answer.
The DNA alterations that is corrected by light-dependent repair is: b. thymine dimers.
What is Light-dependent Repair?A light-dependent repair, which is also known as photo-reactivation, can be described as a process which repairs the pyrimidine dimers that are formed from ultra-violet radiation.
This process repairs thymine dimers formed, and returns the DNA to its initial state before being damaged.
Thus, the DNA alterations that is corrected by light-dependent repair is: b. thymine dimers.
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Many indicators show that temperature are increasing and climate patterns are changing on a global scale. What is NOT a likely result of these changes?
a. many species will become extinct because they won't be able to cope with the changes
b. many species will benefit from the changes
c. new species that are well adapted to new climatic conditions will evolve
d. the earth will become devoid of life due to the changes
e. nearly all species will be affected by the changes
Answer:
d. the earth will become devoid of life due to the changes
Explanation:
There is evidence of great catastrophes throughout Earth's history. There is also evidence of great mass extinctions events but life has always find ways to adapt and persist. Climate change could cause a mass extinction specially because climate is changing faster than species can adapt. Is difficult to think that all life will cease to exist at a planet level.
Pure-breeding sweet peas with purple flowers and round pollen are crossed with pure-breeding sweet peas with red flowers and long pollen. The resulting F1 plants all have purple flowers and long pollen. When one of these plants is test crossed, 20% of the resulting offspring have purple flowers and long pollen. By how many map units are the genes for flower color and pollen shape separated? A. 10 B. 20 C. 40 D. 60 E. None of the above
Answer:
C. 40
Explanation:
Pure-breeding means that the individuals are homozygous for the genes being analyzed.
From Mendel's Law of Dominance we know that the traits that appear in the F1 are the dominant ones.
I will call:
P_ = purple flowers
pp = red flowers
L_ = long pollen
ll = round pollen
Initial cross:
P Pl/Pl x pL/pL
F1 Pl/pL
Test cross (cross with a homozygous recessive individual):
Pl/pL x pl/pl
Expected progeny:
Pl/pl = Parental (purple flowers, round pollen)
pL/pl = Parental (red flowers, long pollen)
PL/pl = Recombinant (purple flowers, long pollen)
pl/pl = Recombinant (red flowers, round pollen)
20% of the offspring have purple flowers and long pollen (PL/pl).
Every time crossing over happens in the meiosis of the F1 individual, both a PL gamete and a pl gamete form. That means that 20% of the offspring will also be pl/pl, and the total proportion of the offspring that will be recombinants will be 40%.
A distance of 1 map unit corresponds to a recombinant frequency of 1%.
A recombinant frequency of 40% therefore means that 40 map units separate the glower color and pollen shape genes.