Answers
Answer:
The female (XcX) is carrier for color blindness while the male (XY) is normal. The carrier female transmits its one X chromosome with gene for color blindness to one of her sons making him color blind.
Explanation:
Color blindness is X linked recessive disorder and followed the criss-cross inheritance wherein the color blind sons get the gene from their carrier/affected mother while the color blind father would transmit the gene to his daughters along with X chromosome.
Since both the parents are normal, the mother (XcX) should be carrier for the disease to get the color blind son. A cross between carrier mother (XcX) and normal father (XY) would make 50% of their sons color blind.
Related Questions
Identify all the amino acid-specifying codons where a point mutation (a single base change) could generate a nonsense codon.
Answers
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.
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.
Answers
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.
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
Answers
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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Unusual nucleosides that are found in mature tRNA molecules are added during:
a. transcription initiation
b. transcription elongation
c. transcription termination
d. post transcriptional modification
e. it is not currently known when the unusual nucleosides are added
Answers
Answer:
The correct answer will be option-D.
Explanation:
tRNA or transfer RNA is an adaptor molecule used during the translation process to add specific amino acids to the polypeptide.
The tRNA structure contains modified bases like inosine which is the first nucleotide of the anticodon loop and has the ability to form a hydrogen bond with more than one base of the codon. Inosine is derived from the adenine base during post-transcriptional modification of mRNA by methylation.
Thus, option-D is the correct answer.
The reason that there are Okazaki fragments during DNA replication is ultimately because:
A) Some parts of the chromosome are made of RNA and different polymerases have to be used
B) Polymerases can only synthesize in the 5' to 3' direction
C) There are only enough dNTP's to support quick replication on one strand, so the other strand lags behind it
D) It is necessary to "slow down" one of the strands so that replication does not occur before cells have a chance to grow
Answers
Answer:
B) Polymerases can only synthesize in the 5' to 3' direction
Explanation:
The leading strand's directionality is 3' to 5', so polymerase has no problem with replicating this one. But the lagging strand has the opposite directionality, so the polymerase must work in the opposite direction of the replication fork.In consequence, the replication process undergoes periodic breaks, and the enzymes have to stop and start again while helicase separates both strands, resulting in the polymerization of okazaki fragments.
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
Answers
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.
Which of the following is NOT a stage of translation?
a. Initiation
b. Expression
c. Elongation
d. Termination
Answers
Answer:
b. Expression is the correct answer.
Explanation:
Translation is a process through which protein is produced from the messenger RNA.
Three stages of translation are
initiation: during this process, tRNA gets attach at the codon called start codon.elongation: during this process amino acids are added continuously, which result in the formation of long-chain joined together through a peptide bondtermination: In this translation gets stop when the stop codon enters the ribosome.Expression is not a stage of translation. The actual stages are initiation, wherein the ribosome assembles around the mRNA; elongation, during which the mRNA is read by the ribosome; and termination, where the process ends and the polypeptide chain is released.
Explanation:In the process of translation, which is a part of protein synthesis, there are three main stages: Initiation, Elongation, and Termination. Therefore, Expression is not a stage of translation. The steps are:
Initiation: the ribosome assembles around the mRNA to be read and the first tRNA is attached at the start codon. Elongation: the ribosome translocates along the mRNA chain and synthesizes the polypeptide chain. Termination: the process ends when a stop codon in the mRNA is reached and the polypeptide chain is released. Learn more about Translation here:https://brainly.com/question/23277238
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What type of bond occurs between the nucleotide bases in the codon and the anticodon?
a. ionic
b. covalent
c. hydrogen
d. disulfide bridges
e. peptide
Answers
Answer:
C. Hydrogen bonds
Explanation:
Anticodon refers to the set of three nucleotides present in tRNA. The anticodon is complementary to the codon of mRNA. The nucleotide bases of anticodon and mRNA codons are paired by hydrogen bonds.
Here, the adenine of anticodon makes the hydrogen bond with the uracil base of codon while the guanine base of anticodon forms the hydrogen bond with the cytosine base of the codon.
There is a specific tRNA with an anticodon complementary to the mRNA codon for each amino acid. For example, the tRNA for phenylalanine has an anticodon 3' AAG 5' and binds to the complementary mRNA codon base via hydrogen bonds.
A group of paleontologists have discovered a new fossil animal resembling a horse with intact DNA in the fossil. The fossil animal is theorized to have a phylogenetic relationship with a horse. What is the most appropriate method for determining if a close relation existed between these two species? A : Morphological studies B : Biochemical estimation C : Genetic analysis D : Physiological studies
Answers
The correct answer is C. Genetic analysis
Explanation:
In biology, two organisms have a phylogenetic relationship if they share a common ancestor and therefore have genetic similarities although in most cases there are also similarities in morphology, physiology, etc. However, two organisms might have similarities in morphology and physiology without genetic similarities. Due to this, if you need to determine whether two species are related or not the best method is a genetic analysis as only those organisms that share a common ancestor (phylogenetic relationship) are genetically similar.
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
Answers
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.
Which (if any) of the following statements regarding chickenpox, smallpox, and syphilis is correct?
a. all can be transmitted through contact with the rash/lesion
b. all can cause congenital infection
c. all can be prevented by vaccination
d. all are caused by viruses
Answers
Answer:
a. all can be transmitted through contact with the rash/lesion
Explanation:
All the three diseases i.e. chickenpox, smallpox, and syphilis can be transmitted through contact with the rash/lesion.
Chicken pox: Chickenpox is a viral infection in which red blisters appear on the skin. It is caused by Varicella-zoster virus.
Smallpox: Smallpox was certified the global eradication by WHO in the year of 1980. Smallpox is a viral infection caused by the Variola virus. Variola viruses are of two types named Variola major and Variola minor.
Syphilis: Syphilis is a bacterial disease which is the most dangerous infection spread through sexual contact.
Bacterial transformation and bacteriophage labeling experiments proved that DNA was the hereditary material in bacteria and in DNA-containing viruses. Some viruses do not contain DNA but have RNA inside the phage particle. An example is the tobacco mosaic virus (TMV) that infects tobacco plants, causing lesions in the leaves. Two different variants of T MV exist that have different forms of a particular protein in the virus particle that can be distinguished. It is possible to reconstitute T MV in vitro (in the test tube) by mixing purified proteins and RNA. The reconstituted virus can then be used to infect the host plant cells and produce a new generation of viruses. Design an experiment to show that RNA acts as the hereditary material in TMV.
Answers
Answer:
Mix RNA from virus type I with the protein from virus type II to reconstitute a hybrid virus. In a parallel experiment, mix protein from virus type I with the RNA from virus type II. After that infect the cells with each of these reconstituted hybrid viruses distinctly, and assess the protein in the progeny viruses, which originates from each of the infections.
One will see that the progeny viruses in each case exhibit the protein, which matches the type of RNA in the parent hybrid virus. The protein in the progeny did not match with the protein in the parent hybrid virus.
Final answer:
To demonstrate RNA's role as hereditary material in TMV, an experiment involving modifying and tracking RNA in recombinant viruses, which are then used to infect plants, could be conducted. Progeny exhibiting the modified RNA's characteristics would confirm RNA as the hereditary material.
Explanation:
Experiment to Show RNA as Hereditary Material in TMV
To design an experiment proving that RNA is the hereditary material in Tobacco Mosaic Virus (TMV), a researcher could follow the precedent set by the Hershey and Chase experiments. First, one would need to acquire two strains of TMV with distinct protein coats but identical RNA. The RNA of one TMV strain should be modified with a mutagenic agent to introduce a detectable change, while keeping the protein coat unaltered. Next, the modified RNA and protein coats from both strains should be reconstituted to form new TMV particles. These recombinant viruses would be used to infect tobacco plants. If RNA is indeed the hereditary material, the progeny viruses isolated from the plants should exhibit the introduced change from the modified RNA irrespective of the protein coat it was packaged with.
Additionally, one could label the RNA with a radioactive or fluorescent marker to track its inclusion into the host cells and the manufacturing of new virus particles. If the marker is found in the progeny viruses, it would provide further evidence that RNA is passed onto the next generation, confirming it as the hereditary material of TMV.
To cause a human pandemic, the H5N1 avian flu virus would have to
a. spread to primates such as chimpanzees.
b. develop into a virus with a different host range.
c. become capable of human-to-human transmission.
d. become much more pathogenic.
Answers
Answer:
c. become capable of human-to-human transmission.
Explanation:
The H5N1 flu virus is a virus that is capable of causing infectious respiratory disease in birds and hence, it is an avian influenza virus and the infection caused by it famously known as bird flu. When it comes to humans, the humans may acquire the infection from consuming or being in close contact with infected birds however, the incidences of the human pandemic are rare. The reason for fewer chances of a human pandemic is because for such a pandemic to occur person to person spread is needed which does not happen in case of bird flu.What are potential mechanisms that can lead to endocrine dysfunctions?
Answers
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
Taste receptors on the tongue are not related to smell receptors of the nose.
a. True
b. False
Answers
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.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
Answers
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.
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.
Answers
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.
Some birds can detect much lower frequency sounds that humans can. These sounds are called infrasounds. What is the leading hypothesis for why they are able to detect these low frequnecy sounds?
a. to be able to make quick flight changes to avoid predators that produce infrasounds
b. to detect the orientation of the sun even at night so that they can continue to fly
c. to detect pheromones to identify intraspecific members of the opposite sex during mating
d. to orient themselves during migration by detecting waterfalls, volcanoes and thunderstorms
e. to echolocate prey by producing these infrasounds that then are reflected back to their ears
Answers
Answer:
d. To orient themselves during migration by detecting waterfalls, volcanoes and thunderstorms.
Explanation:
It's known that natural events such as earthquakes, waterfalls/ocean waves, volcanoes and severe storms generate infrasound; the primary hypothesis about why some birds can detect infrasounds is to be able to orient themselves during their migration and avoid such natural events.
Describe the most common molecular mechanism for recessively inherited human genetic diseases such as cystic fibrosis.
Answers
Answer:
Explanation:
cystic fibrosis is an autosomal recessive disorder. When the child receives the defective gene from both of his parents, he suffers from cystic fibrosis. Because his parents are carriers. In recessive genetic disorder, the genes will be expressed when both recessive genes are present in one person. The person suffering from this disease have a lung infection and pancreatic dysfunction.
In this cystic fibrosis, genes are located in chromosome 7. The effective gene is the CFTR gene. The CFTR gene is present in the DNA and by transcription, this forms CFTR protein. This is a channel protein and transports chloride ion.
This CFTR protein transports chloride ions and it makes a balance in the cell membrane. These genes are commonly present in the epithelial cells. Outside the epithelial mucus is present to keep the cells moist.
The epithelium gets a lack of water and chloride due to the defect. Therefore cells need CFTR proteins also. This causes lung infection and pancreatic disorder.
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?
Answers
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.
Which of the following factors would tend to increase membrane fluidity?
a. a greater proportion of unsaturated phospholipids
b. a greater proportion of saturated phospholipids
c. a lower temperature
d. a relatively high protein content in the membrane
Answers
Answer:
The correct answer is A.
Explanation:
Membrane fluidity can be increased by lipid chains with carbon-carbon double bonds (unsaturated phospholipids). They are more fluid than lipids that are saturated because unsaturated double bonds make it harder for the lipids to pack together.
Also higher temperatures can increase membrane fluidity because lipids acquire thermal energy, this results in many arranges and rearranges of the membrane and therefore makes it more fluid.
The fluidity of cell membranes is generally increased by a higher proportion of unsaturated phospholipids and decreased by a higher proportion of saturated phospholipids and lower temperature. The protein content does not necessarily affect fluidity.
Explanation:The fluidity of cell membranes is affected by the proportion of unsaturated phospholipids and saturated phospholipids, temperature, and the protein content. Among the options listed:
A greater proportion of unsaturated phospholipids would increase membrane fluidity. Unsaturated phospholipids have kinks in their fatty acid tails due to double bonds, which prevent tight packing and hence increase fluidity.A greater proportion of saturated phospholipids would decrease membrane fluidity – the straight fatty acid tails of saturated phospholipids can pack closely together, reducing fluidity.A lower temperature generally decreases membrane fluidity, as it allows phospholipids to pack more closely together.A relatively high protein content in the membrane does not necessarily affect fluidity. It depends on the type of proteins and their interaction with the phospholipids.Learn more about Membrane Fluidity here:
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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.
Answers
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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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
Answers
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.
How are restriction enzymes and ligase used in biotechnology?
a. restriction enzymes cut DNA at specific locations, producing ends that can be ligated back together with ligase
b. only restriction enzymes that produce blunt ends after cutting DNA can be ligated with ligase
c. only restriction enzymes that produce sticky ends on the DNA can be ligated with ligase
d. restriction enzymes can both cut DNA at specific sites and ligate them back together
e. restriction enzymes randomly cut DNA, and the cut fragments can be ligated back together with ligase
Answers
Answer:
A. Restriction enzymes cut DNA at specific locations, producing ends that can be ligated back together with ligase.
Explanation:
Restriction enzymes are one of the endonucleases that cut the DNA at specific base sequences. The base sequences recognized and cut by the restriction enzymes are known as restriction sites.
Restriction enzymes are used in recombinant DNA technology to cut the DNA at specific sites. Restriction enzymes can produce DNA fragments with sticky ends or blunt ends. These DNA fragments are joined together by DNA ligase enzyme.
For example, the donor DNA and the vector DNA are cut at specific sites using a particular restriction enzyme. The resultant DNA fragments have complementary ends that are ligated together by the action of a DNA ligase enzyme. The result is the insertion of a gene of interest into the vector DNA.
Describe why oxygen moves from the air to the blood in the alveolar capillaries and again why oxygen moves from the blood to the interstitial fluid in the body tissues. Why is the partial pressure (concentration) of oxygen so low in the interstitial fluid?
Answers
Answer:
Higher partial pressure of oxygen in alveolar air than that of alveolar capillaries drives the diffusion of oxygen gas from air to the blood.
Higher partial pressure of oxygen in blood at tissue level as compared to the interstitial fluid drives diffusion of oxygen from blood to the tissue fluid.
Consumption of oxygen for cellular respiration in cells reduces the its concentration in the interstitial fluid.
Explanation:
Gaseous exchange in the body occurs through the process of diffusion wherein respiratory gases are exchanges down the concentration gradient.
Since the alveolar air has a higher partial pressure of oxygen than that of blood present in the surrounding blood capillaries, oxygen diffuses from the alveolar air to the blood making the blood oxygen-rich.
As the oxygen-rich blood reaches the body tissues, oxygen is diffused from the blood into the interstitial fluid since the later has lower oxygen concentration.
The supply of oxygen from the blood to the body tissues is required as the cells consume the oxygen to perform aerobic respiration and retrieve energy from the nutrients.
Oxygen moves from the air to the blood and from the blood to the interstitial fluid due to differences in partial pressure, with oxygen continuously diffusing from areas of higher to lower concentration.
Explanation:Oxygen Movement from Air to Blood and Tissues
The movement of oxygen from the air to the blood in the alveolar capillaries and from the blood to the interstitial fluid in the body tissues is driven by a gradient in partial pressure. Gases like oxygen diffuse from areas of high concentration to areas of low concentration according to Henry's Law. In the lungs, the partial pressure of oxygen is high in the alveoli (around 100 mm Hg) and low in the blood of the pulmonary capillaries (around 40 mm Hg). As a result, there is a strong gradient that drives oxygen diffusion across the respiratory membrane from the alveoli into the blood.
Once in the blood, oxygen binds to hemoglobin in red blood cells (RBCs), which transport it to the tissues. In the body tissues, the partial pressure of oxygen in the blood is higher than that in the interstitial fluid, causing oxygen to diffuse into the interstitial fluid and then into the cells of the tissues where it is used for cellular respiration. The partial pressure of oxygen is so low in the interstitial fluid because oxygen is continuously being consumed by the cells, thereby maintaining the pressure gradient necessary for diffusion.
Light-dependent repair corrects which of the following DNA alterations?
a. methylation
b. thymine dimers
c. mismatched basepairing
d. hydroxylation
e. inversions
Answers
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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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.
Answers
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).
Explain how we know that DNA breaks and rejoins during recombination.
Answers
Answer:
It occurs through homologous recombination
Explanation:
GENERAL RECOMBINATION OR HOMOLOGIST
Previously we defined its general characteristics. We will now describe a molecular model of this recombination, based on the classic Meselson and Radding, modified with the latest advances. Do not forget that we are facing a model, that is, a hypothetical proposal to explain a set of experimental data. Not all points of this model are fully clarified or demonstrated:
Suppose we have an exogenote and an endogenote, both consisting of double helices. In recombination models, the exogenote is usually referred to as donor DNA, and the endogenote as recipient DNA.
1) Start of recombination: Homologous recombination begins with an endonucleotide incision in one of the donor double helix chains. Responsible for this process is the nuclease RecBCD (= nuclease V), which acts as follows: it is randomly attached to the donor's DNA, and moves along the double helix until it finds a characteristic sequence called c
Once the sequence is recognized, the RecBCD nuclease cuts to 4-6 bases to the right (3 'side) of the upper chain (as we have written above). Then, this same protein, acting now as a helicase, unrolls the cut chain, causing a zone of single-stranded DNA (c.s. DNA) to move with its 3 ’free end
2) The gap left by the displaced portion of the donor cut chain is filled by reparative DNA synthesis.
3) The displaced single chain zone of the donor DNA is coated by subunits of the RecA protein (at the rate of one RecA monomer per 5-10 bases). Thus, that simple chain adopts an extended helical configuration.
4) Assimilation or synapse: This is the key moment of action of RecA. Somehow, the DNA-bound RecA c.s. The donor facilitates the encounter of the latter with the complementary double helix part of the recipient, so that in principle a triple helix is formed. Then, with the hydrolysis of ATP, RecA facilitates that the donor chain moves to the homologous chain of the receptor, and therefore matches the complementary one of that receptor. In this process, the chain portion of the donor's homologous receptor is displaced, causing the so-called "D-structure".
It is important to highlight that this process promoted by RecA depends on the donor and the recipient having great sequence homology (from 100 to 95%), and that these homology segments are more than 100 bases in length.
Note that this synapse involves the formation of a portion of heteroduplex in the double receptor helix: there is an area where each chain comes from a DNA c.d. different parental (donor and recipient).
5) It is assumed that the newly displaced chain of the recipient DNA (D-structure) is digested by nucleases.
6) Covalent union of the ends originating in the two homologous chains. This results in a simple cross-linking whereby the two double helices are "tied." The resulting global structure is called the Holliday structure or joint.
7) Migration of the branches: a complex formed by the RuvA and RuvB proteins is attached to the crossing point of the Holliday structure, which with ATP hydrolysis achieve the displacement of the Hollyday crossing point: in this way the portion of heteroduplex in both double helices.
8) Isomerization: to easily visualize it, imagine that we rotate the two segments of one of the DNA c.d. 180o with respect to the cross-linking point, to generate a flat structure that is isomeric from the previous one ("X structure").
9) Resolution of this structure: this step is catalyzed by the RuvC protein, which cuts and splices two of the chains cross-linked at the Hollyday junction. The result of the resolution may vary depending on whether the chains that were not previously involved in the cross-linking are cut and spliced, or that they are again involved in this second cutting and sealing operation:
a) If the cuts and splices affect the DNA chains that were not previously involved in the cross-linking, the result will be two reciprocal recombinant molecules, where each of the 4 chains are recombinant (there has been an exchange of markers between donor and recipient)
b) If the cuts and splices affect the same chains that had already participated in the first cross-linking, the result will consist of two double helices that present only two portions of heteroduplex DNA.
Final answer:
DNA breaks and rejoins during recombination to exchange genetic material between chromosomes and create genetic diversity in species.
Explanation:
During recombination, DNA breaks and rejoins to exchange genetic material between homologous chromosomes. This process occurs during meiosis and involves the breakage and rejoining of parental chromosomes. It leads to the generation of novel chromosomes that share DNA from both parents, resulting in genetic diversity in species. The repair of DNA breaks during recombination is essential for accurate DNA repair and the survival of species.
why is it important to stain youngcultures of bacteria with
the grain stain?
Answers
Answer:
Because older cultures of gram-positive bacteria tend to lose their ability to retain crystal-violet in the peptidoglycan of their cell walls and can be confused with gram-negative bacteria.
Explanation:
Gram staining is used to differentiate between two major groups of bacteria. Gram-positive and gram-negative, these bacteria differ in the amount of peptidoglycan in their cell walls. Gram-positive bacteria have a higher amount of peptidoglycan, which absorbs the violet crystal complex used in gram staining, staining them purple/violet. Old cultures of gram-positive bacteria tend to lose the ability to retain the violet crystal and are stained by safranine, staining them red/pink and appear to be gram-negative.
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.
Answers
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.