The correct answer is A. Permian.
Explanation
The synapsids were terrestrial vertebrate mammals or animals related to them rather than to reptiles or birds that lived during the Permian (a period that began about 299 million years ago and ended about 251 million years ago) and were the dominant terrestrial vertebrates during it. These animals survived until the Triassic; however, due to the Permian-Triassic extinction, they were not the predominant vertebrate as in the previous period. Later, during the Cretatic and Jurassic periods, their development was minimal and did not have great importance as reptiles were dominant in this period. So, the correct answer is A. Permian.
Cocaine acts in the nervous system by blocking
a. the dopamine transporter.
b. the serotonin transporter.
c. the norepinephrine transporter.
d. voltage-gated sodium channels in axons.
e. All of the above
Answer:
The correct option is: e. All of the above
Explanation:
Cocaine is a habit-forming illegal recreational drug and a very strong nervous system stimulant. The effects of this drug include euphoria, hallucinations, paranoid delusions, increased heart rate, large pupils, perspiration, itching and high body temperature.
Cocaine acts on the nervous systems of the human beings by blocking the reuptake of dopamine, serotonin, and norepinephrine neurotransmitters in the brain.
It also interferes with the action potential propagation by blocking the voltage-gated sodium channels.
Why does it make sense that the sensors and controller of body temperature reside in the brain?
A. Because the brain enables the body's response to temperature changes
B. Because deep inside the brain is considered the body's core
C. Because temperature changes happen most quickly in the brain
D. Because the brain works best at the correct temperature
E. Because temperature changes affect the brain first
The following enzymes or processes are known to regulate eukaryotic gene expression. In the blank provided by each enzyme or process, enter the number 1 if the enzyme or process increases gene expression or the number 2 if the enzyme or process decreases gene expression.
(a) histone acetyltransferases (HATs) ____
(b) histone deacetylases (HDACs) ____
(c) acetylation ____
(d) deacetylation ____
(e) decreased attaraction between DNA and histones ____
(f) increased attraction between DNA and histones ____
Answer:
a) 1 b) 2 c) 1 d) 2 e) 1 f) 2
Explanation:
a) A coactivator CBP contains a subunit that has histone acetyltransferase (HAT) activity. These enzymes transfer acetyl groups from an acetyl CoA donor to the amino groups of specific lysine residues on histone proteins.
-Gradually this leads to activation of several other coactivators and as a result initiation of transcription takes place.
b) HDACs are associated with transcriptional repression. HDACs are present as subunits of larger complexes described as corepressors.
-Corepressors are similar to coactivators, except that they are recruited to specific genetic loci by transcriptional factors (repressors) that cause the targeted gene to be silenced rather than activated.
c) Acetylation of histone residues is thought to prevent chromatin fibers from folding into compact structures, which helps to maintain active, euchromatic regions.
-On a finer scale, histone acetylation increases access of specific regions of the DNA template to interacting proteins, which promotes transcriptional activation.
d) Deacetylation of the the histone tails makes the DNA more tightly wrapped around the histone cores, making it difficult for transcription factors to bind to the DNA. This leads to decreased levels of gene expression.
e) Decreased attraction is directly proportional to loosening of the DNA from histone core hence accessible to the transcriptional factors which further leads to transcription and gene expression
f) Increased attraction means DNA gets tightly wound around the histone core and inaccessible to the transcription factor hence decreased gene expression.
If an enzyme in solution is saturated with substrate, the most effective way to obtain a faster yield of products is to
a. add more of the enzyme.
b. heat the solution to 90°C.
c. add more substrate.
d. add a noncompetitive inhibitor.
Answer:
a. add more of the enzyme.
Explanation:
Enzymes speed up the rate of reaction by lowering the activation energy for the reaction. They are not used themselves in the reaction. They are specific to substrate molecule. Substrate molecule binds to enzyme's active site and they undergo the reaction to form the product and release back the enzyme.
Rate of reaction depends on both substrate and enzyme concentration. Maximum rate of reaction is reached when all the active sites of enzyme molecules have been occupied by the substrate molecules which means that they are saturated. If more substrate is added at this point it wont have any effect on rate of reaction since there are no free active sites. Hence more enzyme is required to be added so that extra substrate can be utilized and rate of reaction can further be increased.
What fibers surround the xylem and phloem? Give their functions.
Answer:
The xylem fiber surround xylem and phloem fiber surround the phloem.
Explanation:
The fibers surround the xylem are xylem fibers. This is a complex plant tissue. These cells are the dead cells.
Some of the xylem fibers have cross walls and some don't have. The xylem fibers have a protoplast in it. These fibers have thick cell walls. The xylem fibers provide mechanical support to the plant.
The phloem is surrounded by phloem fibers. The phloem fibers are associated with sclerenchyma fibers. These have thick cell walls. They become dead at maturity.
These fibers also provide mechanical support to the plants and protect the phloem. Because phloem tissues have thin walls. The phloem fibers of jute, flax plants used for making ropes.
Bark fibers surround the xylem and phloem and provide structural support and protection to the plant. They are part of the scleraenchym cells and are primarily composed of cellulose, hemicellulose, and lignin.
Explanation:The fibers that surround the xylem and phloem are known as bark fibers. They essentially form a part of the peripheral bark of the tree and fall into the category of sclerenchyma cells, meaning they are hardened and thickened. These fibers, being composed primarily of cellulose, hemicellulose, and lignin, provide critical structural support to the plant. In addition to providing support, bark fibers also play a role in protecting the plant from physical damage and invasion by pathogens.
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From the bacteria's perspective, why is it helpful that it produce diarrhea in people?
A. Because it gets the bacteria out of the person and, likely, into the next one
B. It's not helpful really. That's just what that toxin causes.
C. Because that quickly kills the person
D. Because it makes the patient too unpleasant to be around
E. Because there is no real treatment for that
Answer:
A
Explanation:
The vast majority of bacteria that produce gastrointestinal symptoms are transmitted via fecal-oral transmission. This means that when sub-optimal hygiene is present, bacteria from the hands of the infected person that got there from contact with its own feces (after going to the bathroom, for example) is passed to the next person. This happens a lot in the food business, and it's the mode of transmission of the much famous Salmonella.
Mutations that result in no gene product or totally nunfunctional gene products are called:
a. isomers
b. isoalleles
c. null alleles
d. lethal alleles
e. forward alleles
Answer:
c. null alleles
Explanation:
Genetic mutation can produce a non functional allele called as null allele. There might not be any product from this allele or a defective product might be produced. If the mutant allele does not even produce a RNA transcript, it is called as RNA null. If it is not able to produce the final protein product, it is called a protein null.
An example of null allele is The O blood group. Due to a mutation in the gene producing A antigen, an enzymatically inactive protein is produced by O blood group people. Both the copies of null allele O are required for O phenotype to be present.
Which DNA segment is not required specifically for an expression vector (for use in E. coli)?
a. origin of replication
b. operator
c. ribosome binding site
d. promoter
e. transcription termination sequence
Answer:
b. operator
Explanation:
An expression vector is usually a plasmid or a virus designed for gene expression in cells, it's designed to act as an enhancer and promoter to efficient transcription of the gene that it carries, meaning it is used for protein production.
As any other vector expression vector may have an origin of replication, a selectable marker, and multiple cloning sites, it also needs elements that are necessary for gene expression, a promoter, a ribosomal binding site, and a start, terminal codon and a transcription termination sequence.
An operator contains the code to begin the process of transcribing the DNA message of one or more structural genes into mRNA, it works as a repressor to regulate the gene expression, since the function of an expression vector is the protein production an operator is not specifically required.
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Which of the following is true of spermatogenesis?
a. occurs before puberty
b. involves the differentiation of spermatids leading to the formation of mature spermatozoa
c. the process of spermatozoa production
d. results in the production of primary spermatocytes
e. involves the production and secretion of testosterone
Answer:
The correct answer is c. the process of spermatozoa production
Explanation:
Spermatogenesis is the process of formation of sperms in the seminiferous tubules present in the testes. The process of spermatogenesis starts when germ cells present in the seminiferous tubule starts dividing mitotically.
Some of these cells become primary spermatocytes after mitosis and enters meiosis I. After completion of meiosis I they become secondary spermatocytes.
Each secondary spermatocytes converts into two haploid spermatids after completion of meiosis II. Spermatids then differentiate into mature sperms called spermatozoa.
Thus, the correct answer is c. the process of spermatozoa production
Which of the following is a hydrophobic material?
a. paper c. wax
b. table salt d. sugar
Answer:
Wax.
Explanation:
Hydrophobic material may be defined as the material that contains the non polar molecule and hydrophobic interaction in their structure. The hydrophobic molecule are soluble in non polar solvents.
Wax is made of lipids. The lipids are non polar molecule and are hydrophobic in nature. Wax are made of large hydrocarbon part that are insoluble in water and can be easily soluble in non polar solvents.
Thus, the correct answer is option (c).
The term hydrophobic refers to substances that repel water. Out of the mentioned options, wax best fits this description as it doesn't mix with water. Paper, table salt, and sugar dissolve in or absorb water and are thus hydrophilic, or attracted to water.
Explanation:In the context of the question, which asks for the material that is hydrophobic, the correct answer is c. wax. Hydrophobic substances are those that don't mix with or repel water. For instance, wax is a hydrophobic substance you commonly encounter. When you pour water on wax, it simply rolls off or forms beads and doesn't mix, this is due to its hydrophobic characteristic. Materials like table salt, paper, and sugar, are all hydrophilic, they tend to dissolve in and interact with water.
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Describe the route blood takes as it flows through the body, starting with the heart receiving oxygen-poor blood from the body.
Answer:
the blood enters into the right atrium by inferior and superior vena cava--- left ventricle---pulmonary artery ---- lungs----left atrium----left ventricle---aorta----different body tissues.
Explanation:
The circulation in man is called the double circulation. Because in the heart the two types of blood circulate separately i.e. oxygenated blood/oxygen-poor blood and deoxygenated blood/oxygen-rich blood and they never mix.
The route of deoxygenated blood:
In the heart, the deoxygenated blood enters by the superior and inferior vena cava. Then this oxygen-poor blood stores in the right atrium and from there it passes into the left ventricle. The flow of blood from the right atrium to left ventricle occurs by the tricuspid valve.
When the blood enters into the right ventricle the valves closed to stop the backflow of the blood.
The blood from the right ventricle goes to the pulmonary artery by the pulmonary valve. From the pulmonary valve, the deoxygenated blood passes to the lungs. In the lungs, the deoxygenated blood/oxygen-poor blood gets purified.
The rout of oxygenated blood:
Now the blood becomes rich in oxygen and called oxygenated blood. This oxygenated blood enters into the left atrium by pulmonary vein. From the left atrium, the oxygen-rich blood moves into the left ventricle. Then the oxygenated blood reaches to the different parts of the body by the aorta.
In this way, the right half of the heart circulates the deoxygenated blood and the left half of the heart provides deoxygenated blood.
Applying the "If . . . then" logic of science (see Chapter 1), what are a few of the predictions that arise from the hypothesis that plants evolved from green algae? Put another way, how could you test this hypothesis?
Answer:
Systematics, molecular analysis, and cellular morphology offer enough proof that plants had originated from green algae. In fact, the green algae, red algae, and the land plants come under a similar eukaryotic group, that is, Archaeplastida. The group had descended from an ancient protist that engulfed a cyanobacterium.
If the plants had originated from green algae than this can be demonstrated by sequencing the genes of both the green algae and plants. If the sequence of gene of plants is closely associated with the gene sequence of green algae, then the hypothesis that the plants had originated from green algae would be certainly true.
Thus, the mentioned hypothesis can be tested with the help of gene sequencing.
The hypothesis that plants evolved from green algae can be tested by comparing DNA sequences, looking for shared structural and biochemical traits, and identifying adaptations for terrestrial living such as the presence of a cuticle or vascular tissue.
Explanation:When applying the "If... then" logic of science to the hypothesis that plants evolved from green algae, several predictions can be made to test this hypothesis. One prediction might be that if plants evolved from green algae, then they should share common ancestral traits with green algae. Some of these ancestral traits can include the presence of cellulose in their cell walls and the use of similar pigments for photosynthesis.
To test this hypothesis, scientists could compare the DNA sequences of plants and green algae, expecting to find significant similarities. Additionally, they would look for structural and biochemical characteristics that plants and green algae have in common, such as the types of chlorophyll used in photosynthesis or the presence of similar reproductive strategies like the alternation of generations observed in some marine algae species.
Moreover, if plants have adapted from green algae, which are aquatic, we might expect early plants to have features that allowed them to cope with the challenges of transitioning to a terrestrial environment. Such adaptations could include the development of embryo retention, a protective cuticle, stomata for gas exchange, and vascular tissue for transport of water and nutrients. Confirmation of these predicted features would further strengthen the hypothesis that plants are indeed evolved from green algae.
During the light dependent reaction, the hydrogen ions go down their concentration from __________ producing ATP in the process.
a. The stroma into the thylakoid lumen
b. The thylakoid lumen into the stroma
c. Across the inner membrane into the stroma
d. The stroma across the inner membrane
Answer:
b. The thylakoid lumen into the stroma
Explanation:
During light reaction of photosynthesis, when electrons which have high energy travel across electron transport chain some energy is released by them. This energy causes some H+ from stroma to move into thylakoid lumen. Due to this, H+ concentration becomes more in lumen as compared to stroma. Soon after that because of concentration gradient, these H+ then start moving from high concentration to low concentration i.e. from thylakoid lumen to stroma. There are some reverse pumps i.e. ATP synthase located in the thylakoid membrane which are responsible for generating energy in the form of ATP from the movement of H+. From the movement of 3 H+ , 1 ATP is generated.
How many different proteins composed of 100 amino acids could possibly exist?
The number of possible proteins in humans is twenty.
The number of amino acids you can use it's called n, and the number of possible proteins then will be P=20^n.
In this case, you have 100 amino acids if you replace the n in the equation P=20^100 possible proteins.
Match the function or location to the correct connective tissue. Allows the growth of long bones Intervertebral discs, pubic symphysis External ear, epiglottis, and auditory tubes Outer portion of all bones Inside skull bones, vertebrae, and sternum Transports oxygen, carbon dioxide, and other substances Produces new blood cells and stores lipids Capable of strength with stretching and recoil in several directions Tensile strength capable of withstanding stretch in all directions Vocal folds and ligaments between vertebrae Tendons and ligaments Provides superstructure for lymphatic tissues Energy storage Epithelial basement membrane sits on this Precursor to adult connective tissues Umbilical cord of the newborn
1. Allows the growth of long bones - Periosteum
2. Intervertebral discs, pubic symphysis - Fibrocartilage
3. External ear, epiglottis, and auditory tubes - Elastic cartilage
4. Outer portion of all bones - Dense (compact) bone
5. Inside skull bones, vertebrae, and sternum - Spongy (cancellous) bone
6. Transports oxygen, carbon dioxide, and other substances - Blood
7. Produces new blood cells and stores lipids - Bone marrow
8. Capable of strength with stretching and recoil in several directions - Elastic connective tissue
9. Tensile strength capable of withstanding stretch in all directions - Areolar (loose) connective tissue
10. Vocal folds and ligaments between vertebrae - Dense (regular) connective tissue
11. Tendons and ligaments - Dense (regular) connective tissue
12. Provides superstructure for lymphatic tissues - Reticular connective tissue
13. Energy storage - Adipose tissue
14. Epithelial basement membrane sits on this - Areolar (loose) connective tissue
15. Precursor to adult connective tissues - Mesenchyme
16. Umbilical cord of the newborn - Wharton's jelly
Here's the matching of functions or locations to the correct connective tissues:
1. **Allows the growth of long bones:**
- *Connective Tissue:* **Hyaline Cartilage**
2. **Intervertebral discs, pubic symphysis:**
- *Connective Tissue:* **Fibrocartilage**
3. **External ear, epiglottis, and auditory tubes:**
- *Connective Tissue:* **Elastic Cartilage**
4. **Outer portion of all bones:**
- *Connective Tissue:* **Periosteum**
5. **Inside skull bones, vertebrae, and sternum:**
- *Connective Tissue:* **Endosteum**
6. **Transports oxygen, carbon dioxide, and other substances:**
- *Connective Tissue:* **Blood (Connective Tissue Proper)**
7. **Produces new blood cells and stores lipids:**
- *Connective Tissue:* **Bone Marrow**
8. **Capable of strength with stretching and recoil in several directions:**
- *Connective Tissue:* **Elastic Connective Tissue**
9. **Tensile strength capable of withstanding stretch in all directions:**
- *Connective Tissue:* **Collagenous (Dense) Connective Tissue**
10. **Vocal folds and ligaments between vertebrae:**
- *Connective Tissue:* **Elastic Ligaments**
11. **Tendons and ligaments:**
- *Connective Tissue:* **Dense Regular Connective Tissue**
12. **Provides superstructure for lymphatic tissues:**
- *Connective Tissue:* **Reticular Connective Tissue**
13. **Energy storage:**
- *Connective Tissue:* **Adipose Tissue**
14. **Epithelial basement membrane sits on this:**
- *Connective Tissue:* **Basement Membrane (a form of Areolar Connective Tissue)**
15. **Precursor to adult connective tissues:**
- *Connective Tissue:* **Mesenchyme**
16. **Umbilical cord of the newborn:**
- *Connective Tissue:* **Wharton's Jelly**
Each of these connective tissues serves specific functions or is located in particular areas of the body to support and maintain the structure and function of various organs and tissues.
The largest mass of nervous tissues is found in __________
a. stomach
b. brain and spinal cord
c. pancreas
d. liver
The correct answer is B. Brain and spinal cord
Explanation:
In biology, the nervous tissue is one of the main tissues in the nervous system that is responsible for movement, communication between different parts of the body and perception/processing of information. This tissue can be found mainly in the brain, the spinal cord and the nerves which are also the most important structures in the Nervous system. Also. this tissue includes different types of specialized cells such as the neuroglia cells and neurons. According to this, the largest mass of nervous tissues is found in the brain and spinal cord, because it is in these structures most nervous tissues can be found.
Answer:
Largest mass of Nervous tissue is found in the brain, spinal cord, and nerves.
Explanation:
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Which of the following tools of DNA technology is incorrectly paired with its use?
a. electrophoresis—separation of DNA fragments
b. DNA ligase—cutting DNA, creating sticky ends of restriction fragments
c. DNA polymerase—polymerase chain reaction to amplify sections of DNA
d. reverse transcriptase—production of cDNA from mRNA
The one that is incorrectly paired is DNA ligase—cutting DNA, creating sticky ends of restriction fragments. The correct option is B.
What is DNA ligase?DNA ligase is a ligase enzyme that facilitates the joining of DNA strands by catalyzing the formation of a phosphodiester bond.
It aids living life forms repair single-strand cuts in duplex DNA, but some forms may specifically repair double-strand breaks.
By connecting cuts in the phosphodiester backbone of DNA that occur during replication and recombination, as well as as a result of DNA damage and repair, DNA ligases play an important role in maintaining genomic integrity.
Electrophoresis is used to separate DNA fragments, DNA polymerase is used to amplify sections of DNA, and reverse transcriptase is used to produce cDNA from mRNA.
Thus, the correct option is B.
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Compare the chemical structures of DNA and proteins.
Answer:
DNA:
Dexoyribonucleic acid is present as a genetic material in all organism except some virus. DNA is made up of polymers of nucleotides. They are made up of nitrogenous base (adenine, guanine, thymine and cytosine) , deoxyribose pentose sugar and phosphate group. The nitrogenous bases are linked with each other with hydrogen bonds and nucleotides are linked together with phosphodiester bond.
Protein:
Proteins are the main building block of the body. Proteins are made of the polymers of amino acids. The amino acid consist of amine group, carboxylic group, hydrogen and R group attached with the carbon. The amino acids are linked together through the peptide bond.
Cellular Respiration (there is more than one correct answer)
a. produces oxygen
b. uses carbon dioxide
c. uses glucose
d. produces carbon dioxide
Answer: Uses glucose ;produces carbon dioxide
Explanation:
Cellular respiration can be defined as the process by which the chemical energy in the body is transferred into ATP and some of it is released in the form of heat.
During this process the glucose that is broken down and the energy released is converted into ATP.
Carbon dioxide is released as a by-product of this process.
Hence, the correct answer is option C and D
A fruit fly population has a gene with two alleles, A1 and A2. Tests show that 70% of the gametes produced in the population contain the A1 allele. If the population is in Hardy-Weinberg equilibrium, what proportion of the flies carry both A1 and A2?
a. 0.7 b. 0.49 c. 0.42 d. 0.21
Answer:
Option C
Explanation:
Given ,
A1 allele is carried by [tex]70[/tex] % people
Let us assume A1 s dominant genotype
This means [tex]p= \frac{70}{100} = 0.7[/tex]
Thus, frequency of allele in the given population is [tex]0.7[/tex]
It is also given that the population is in Hardy-Weinberg equilibrium thus
[tex]p+q=1\\0.7+q=1\\q = 1-0.7\\q= 0.3[/tex]
Frequency of fruit fly with genotype A2A2 will be
[tex]q^2\\= (0.3)^2\\= 0.09[/tex]
As per Hardy-Weinberg's second equation of equilibrium
[tex]p^{2} + q^{2} + 2pq = 1\\0.49+0.09+2pq = 1\\2pq = 1-0.49-0.09\\2pq= 0.42[/tex]
Hence, option C is correct
Which of the following statements about excision repair is correct?
a. base excision repair is initiated by DNA glycosylases that recognize abnormal deoxyriboses in DNA.
b. nucleotide excision repair removes large regions of DNA via an excinuclease which cuts on either side of the damaged bases.
c. e. coli exonuclease activity is carried out by uvrA, uvrB and uvrC, and the resulting gap is filled in by DNA Pol III
d. DNA glycosylases cleave the altered nucleoside (base and sugar) from the DNA backbone creating an apurinic or apyrimidinic site
e. in humans, a mechanism similar to E.coli is carried out where the protein XPA acts as the exonuclease.
Answer:
a. True, b. False, c.True, d. True
Explanation:
a. Base excision repair is started by a DNA glycosylase that recognizes the changes and removes the altered base by cleavage of the glycosidic bond binding the base and the deoxyribose sugar together.
b. Nucleotide excision repair works by a cut-and patch mechanism that removes their heavy lesions, including pyrimidine dimers and nucleotides . Endonucleases are responsible for the lesion of the damaged strand.
c. Nucleotide excision repair is initiated by the proteins namely UvrA, UvrC, and UvrB in Escherichia coli.
-UvrD (helicase II) later removes the damaged strand
-DNA polymerase I (PolI) fills in the resulting gap.
d. DNA glycolases removes the damaged nitrogenous base.
-It leaves the sugar-phosphate backbone intact and thus creating an apurinic/apyrimidinic site, which is commonly referred to as an AP site.
e. Xeroderma pigmentosum complementation group A(XPA)
-This is an essential protein in the nucleotide excision repair pathway.
- It helps to make a pre-incision complex along with other proteins.
Fats are broken down into ________________
The P53 protein normally promotes
A) DNA repair.
B) tumor formation.
C) cell division.
D) apoptosis.
Answer:
A. DNA repair
Explanation:
The p53 protein acts as a tumor remover by controlling a set of genes required for the cell division. This protein protects the genome by restricting abnormal cells to proliferate avoiding the replication of damaged DNA. The first resource the protein uses and prometes is DNA repair and if the DNA has irreparable damage, the last resource used by this protein is apoptosis.
Explain generaly how Cyclin and Cyclin Dependent Kinase (CDK) work to control the cell cycle.
Answer:
Various cyclins are the regulatory proteins that associate with respective cyclin dependent kinases. The cyclin-cdk complex activate or inhibit the proteins required for specific stage of cell cycle by phosphorylation.
Explanation:
Cyclin-dependent kinases (Cdks) are the protein kinases that are involved in the regulation of the cell cycle. Cyclins are the regulatory proteins whose levels change during the cell cycle. Cyclins associate with their respective cyclin-dependent kinases to activate them. The cyclin-Cdk complex phosphorylates the enzymes and other proteins required for a particular stage of the cell cycle.
For example, the G1-cdk (the cyclin- G1 specific Cdk complex) allows the cell to enter from G1 to S phase while the G1/S-Cdk complex makes the cell to commit for DNA replication while the S-Cdk complex serves in the initiation of DNA replication. The M-Cdk complex makes the cell to enter the mitosis by phosphorylation of proteins required for various stages of mitosis. Cdks are not activated when not associated with cyclins.
When the cell is not prepared for cell division, phosphorylation of protein p27 is followed by degradation of the [proteins required for cell division. The protein p27 is one of the inhibitors of cell division.
The nebular hypothesis suggests that our solar system evolved from a huge _____.
rotating cloud
asteroid collision
volcanic eruption
star explosion
Answer:
rotating cloud
Explanation:
The nebular hypothesis -
According to this hypothesis , the stars are made up of dense and massive clouds of hydrogen gas , called the giant molecular clouds .
The clouds are gravitationally not stable .
And the formation of the planetary system is a natural result of the star formation .
The solar system evolved a huge rotating cloud .
A woman, Penelope, has a sister with polycystic kidney disease (ARPKD), which is inherited in an autosomal recessive mode. Penelope does not have the disease, and both of her parents do not have ARPKD. Penelope marries a man from Europe who does not have ARPKD. There is no information about whether this disease runs in his family, but 96% of the population does not have ARPKD (assume the disease is in Hardy-Weinberg equilibrium).
What is the probability that they will have a child with ARPKD?
A) 0.25
B) 0.15
C) 0.05
D) 0.04
E) there is not enough information to answer this question
Answer:
D) 0.04
Explanation:
Assuming that a is the recessive allele when homozygous that causes the polycystic kidney disease and A is the dominant one:
Since Penelope's sisters has the disease, she must be homozygous recessive (aa). Her parents must both be Aa (otherwise her sister wouldn't have inherit the disease). So, the probability of Penelope of being heterozygous (Aa) is 0.5.
On the other hand, the probability of having the disease in the population is 0.04 (1 minus the probability of not having the disease which is 0.96). This, according the Hardy–Weinberg principle, in the population would represent the genotype frequency [tex]q^{2}[/tex]. So, the allele a would have a frequency [tex]\sqrt{q}[/tex] = [tex]\sqrt{0.04}[/tex] = 0.2. Since the gene has only two alleles, all alleles must be either A or a , therefore p + q = 1. So, the A population's frequency is p = 1 - q = 1 - 0.2 = 0.8.
The European man's probability of being Aa (doesn't have the disease but can carry the disease allele) is 2pq = 2 x 0.8 x 0.2 = 0.32.
The probability of Penelope and the European man of having an ill kid (assuming they both are Aa) is 0.25.
Finally, given the Penelope's probability of being Aa (0.5), the European man's 's probability of being Aa (0.32) and the probability of having a kid homozygous recessive aa (0.25) = 0.5 x 0.32 x 0.25 = 0.4 is the probability that they will have a child with ARPKD.
Conventionally pasteurized milk is sterile.
a. True
b. False
Answer: False
Explanation:
Conventionally pasteurized milk is not sterile it gets sterile after it is homogenized. The process of pasteurization involves sudden heating and sudden cowling of milk which does not kills all the microorganism but is intended to kill some of the bacteria and inactivates some enzyme.
Homogenization is a completely separate process which involve mechanical breakdown of fat molecules in the milk which increases the shelf life of milk by preventing the cream from rising at the top.
Hence, the given statement is False.
List a few terms that are unique for regulation of transcription in eukaryotic cells?
Answer:
Eukaryotic transcription refers to an elaborate procedure, which is used by the eukaryotic cells to duplicate the genetic information present in DNA into the units of RNA replica. The transcription of the gene takes place in both prokaryotic and eukaryotic cells. The expression of eukaryotic genes is mainly monitored at the level of the beginning of transcription, though in certain cases, transcription may be monitored and attenuated at subsequent steps.
Like in bacteria, the transcription in eukaryotic cells is monitored by the proteins, which combines with the unique regulatory sequences and modulate the activity of RNA polymerase. The intricate task of monitoring the expression of gene in various kinds of cells of multicellular species is achieved mainly by the combined activities of various different transcriptional regulatory proteins.
In supplementation, the packaging of DNA into chromatin and its variation by methylation produce further levels of complexity to the control of eukaryotic gene expression. The transcription in bacteria is monitored by the combination of proteins to the cis-acting sequences, which regulate the transcription of adjacent genes.
The same kind of cis-acting sequence monitors the expression of eukaryotic genes. These sequences have been determined in mammalian cells mainly by the application of gene transfer assays to examine the activity of suspected regulatory regions of cloned genes.
Which position on the anticodon is the wobble position?
a. 5' end
b. 3' end
c. middle of the anticodon
d. 5' end or 3' end
e. any position can be the wobble position
Answer:
A. 5' end
Explanation:
Wobble base refers to the specific base of some tRNA anticodons that have the ability to pair with more than one kind of base in codon. There is a total of 61 codons for 20 amino acids but only 20 different tRNAs.
A wobble base allows a specific tRNA anticodon to recognize and pair with more than one mRNA codon. The 5' base of the anticodon (the third nucleotide in an anticodon) can form hydrogen bonds with more than one type of 3' base of mRNA codon.
For example, Inosinate present at the 5' end of anticodon can pair with as many as three different nucleotides (U, C, and A).
Between glycolysis and the citric acid cycle (also known as
theKrebs cycle), there are two molecules of ATP and two molecule
ofGTP generated directly. Most of the energy derived from
theoxidation of glucose comes from the oxidation of
FADH2generated from the citric acid cycle.
NADPH generatedfrom the citric acid cycle.
NADH generatedfrom the citric acid cycle.
NADH generatedfrom glycolysis.
Answer:
The correct answer will be option-NADH generated from the citric acid cycle.
Explanation:
During cellular respiration, few ATP molecules are synthesized by substrate-level phosphorylation in glycolysis and citric acid cycle.
The process also produces high energy equivalents which are NADH and FADH₂ during the citric acid cycle. Two pyruvate molecules generate 6 NADH and FADH₂ molecules. These molecules release electron and hydrogen during the electron transport chain which is utilized to generate the ATP.
Each NADH produces 3 ATP molecule whereas FADH₂ molecule produces 2 ATP molecules. Since 6 NADH (6x3=18 ATP) produced and 2 FADH₂ (2x2).
Thus, NADH provides the highest amount of ATP molecules.