Answer:
B. PP×PP, PP×Pp, or PP x pp
Explanation:
The allele for the purple flower (P) is dominant over the one for the white flowers (p). To be a purple-flowered plant, the progeny must have homozygous dominant (PP) or heterozygous dominant (Pp). A cross between two homozygous dominant purple-flowered parents (PP x PP) would produce all the homozygous dominant progeny (PP) having purple flowers.
A cross between a homozygous dominant (PP) and a heterozygous dominant (Pp) parent plant would produce homozygous and heterozygous dominant progeny in a 1:1 ratio (PP x Pp = 1 PP: 1 Pp). Similarly, a cross between a homozygous dominant (PP) and homozygous recessive (pp) parent plant would produce all heterozygous dominant (Pp) progeny with purple flowers.
Answer:
pp
Explanation:
A mutation occurs in an operon that prevents the transcription factor from binding to its recognition site on the DNA. In which type(s) of gene regulation would this mutation result in constitutive expression of the structural gene?
Answer:Negative inducible and Negative Repressible
Explanation:
Negative control of operon includes Negative inducible and Negative repressible both prevent transcription, but are different in there mode of operation.
Negative inducible operon: This is usually a process by which an active repressor regulatory protein binds to the DNA thereby inhibiting RNA polymerase from transcribing gene.
Negative repressible operons: This is also a process when an negative inducer binds to the operator to which prevent DNA transcription
Answer:
Explanation:
In the negative inducible and negative repressible regulations of gene.
Coat color is determined by two loci in large cats. Two pink panthers fall in love and produce a large litter of baby panthers with the following phenotypic ratios: 12/16 pink; 3/16 black; and 1/16 white. What is the genotype of the white progeny?A) A_ B_B) A_ bbC) aa B_D) aa bb *Correct answerE) A_ B_ and A_ bb
Answer:
D) aa bb
Explanation:
Whenever there are two loci involved in a certain trait and after a cross the phenotypic ratio of the offspring is 12:3:1 (instead of the expected 9:3:3:1) there is dominant epistasis, where the presence of a dominant allele in one locus masks the expression of the other locus. In this case, for loci A and B, we could say that the offspring have the following genotypes:
9 A_B_ : pink3 A_bb : pink3 aaB_: black1 aabb: whiteThe presence of the dominant allele A in the A/a locus gives pink coat color, regardless of the alleles in the B/b locus.
Final answer:
The genotype of the white progeny in large cats given the phenotypic ratio is aa bb, indicating complete dominance when dominant alleles are present, and an autosomal epistatic interaction determines the white coat color.
Explanation:
When considering the phenotypic ratio of the large cats' offspring, which are 12/16 pink, 3/16 black, and 1/16 white, and given that both parents are pink panthers, we can deduce the inheritance pattern. The progeny that are white must exhibit the recessive phenotype at both loci, indicating a genotype of aa bb. This genotype suggests that the A and B genes are showing complete dominance when present in the dominant form. For the white progeny to be white, they must lack both dominant alleles responsible for the pigmentation.
In terms of coat color inheritance in cats, we know that it is usually related to genes located on the X chromosome. However, the white coloring mentioned in the phenotypic ratio suggests an autosomal epistatic interaction where a second gene can override the expression of the first gene, determining the coat color.
The question on the phenotypes of offspring from a cross between a yellow and a black cat needs further information, particularly the specific genotype of the parents, as multiple outcomes could occur based on X-linked inheritance and potential for codominance or incomplete dominance.
DNA polymerases are capable of editing and error correction, whereas the capacity for error correction in RNA polymerases appears to be quite limited. Approximately one error occurs in every 104 to 105 nucleotide incorporated in RNA. Given that a single base error in either replication or transcription can lead to an error in the protein coded by the gene or mRNA. Please suggest a possible explanation for this striking difference.
Explanation:
DNA polymerase replicates the DNA supplied to all new cells produced while RNA polymerase drives DNA copy RNA synthesis. Unless corrected, error in DNA replication could result in the transmission of the error DNA to all next-generation cells.
Protein synthesis error will cause faulty copies of RNA and degraded proteins. To order to ensure the transfer of key genetic information to future generations of cells, failure to DNA replication must be corrected.
DNA polymerases, involved in DNA replication, have robust error correction mechanisms. This contrasts with RNA polymerases which do not have the same level of error correction. Additionally, DNA repair mechanisms correct errors and play a vital role in preventing mutations.
Explanation:The difference in error correction capacity between DNA polymerases and RNA polymerases can be attributed to their different roles and structures. DNA polymerases, involved in DNA replication, have a robust proofreading mechanism. As they add nucleotides, they read each newly added base. If an incorrect base has been added, they cut the phosphodiester bond and release the wrong nucleotide, replacing it with the correct one.
This is contrasted with RNA polymerases that do not have the same level of error correction. This may be due to the more 'temporary' nature of RNA in the cell; the errors in RNA do not have as long-lasting effects as those in DNA because RNA is not permanently stored in the cell.
DNA repair mechanisms are another aspect involved in maintaining the integrity of the genetic code. These mechanisms - including mismatch repair and nucleotide excision repair - excise and correct incorrectly incorporated bases, and play a significantly important role in preventing mutations and serious consequences like cancer.
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Which of the following statements is correct concerning the spinal cord?A. The white matter contains cell bodies for spinal nuclei.B. Damage to sensory tracts in the spinal cord leads to paralysis.C. Just like the cerebrum, the gray matter is found on the superficial surfaces.D. Spinal nerves have mixed motor and sensory function.
Answer:
D. Spinal nerves have mixed motor and sensory function.
Explanation:
spinal's nerves are mixed nerves, which means that they carry sensory and motor information. They carry sensory information to the central nervous system to give a motor answer that will travel through the spinal nerves to specific muscles. The White matter, which is peripheral in the spinal cord, contains axons while the gray matter, which is central in the spinal cord, contains cell bodies.
Spinal nerves have mixed motor and sensory function is the statements is correct concerning the spinal cord. Thus, option (d) is correct.
The spinal cord is essential for the movement of nerve impulses from the brain to the rest of the body. It transmits impulses for pressure, temperature, movement, feeling, and pain.
Gray matter and white matter both make up the spinal cord, with the gray matter being central and the white matter being on either side. While nerve fibres that carry messages are found in the white matter, the gray matter houses the neuronal cell bodies.
As a result, the significance of the concerning the spinal cord are the aforementioned. Therefore, option (d) is correct.
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Why is conserving biodiversity important? Choose all that apply.
protecting endangered species
protecting habitat diversity
maintaining genetic diversity
preserving heirloom varieties of food crops
Answer:
The correct options are :
protecting endangered species protecting habitat diversitymaintaining genetic diversityExplanation:
Biodiversity can be described as the different varieties of plants and animals which are present in a habitat. If biodiversity wasn't conserved then survival of most plants and animals species would become difficult whenever changes occurred. Without biodiversity, degradation of a habitat would occur. Hence, the conservation of biodiversity is very important for protecting the wild-life animals and plants. Biodiversity boosts up an ecosystem. No matter how small an organism is, their conservation is important for the maintenance of biodiversity.
Conserving biodiversity is important for protecting endangered species, maintaining ecosystems, ensuring the adaptability of species through genetic diversity, and ensuring food security through preservation of heirloom crop varieties.
Conserving biodiversity is critical for several intertwined reasons. Protecting endangered species ensures the survival of organisms that may have unique ecological roles or genetic traits beneficial to the ecosystem or human society. Protecting habitat diversity helps maintain the health and integrity of ecosystems, which in turn supports a variety of life forms. Maintaining genetic diversity within species is crucial for their adaptability and survival in the face of environmental changes, including climate change. Lastly, preserving heirloom varieties of food crops is important for food security and agricultural sustainability, as these varieties can be more resilient to pests, diseases, and changing climate conditions.
In peas, tall plants (T) are completely dominant over short plants (t). You cross two heterozygous pea plants Use a Punnett square to determine the expected genotypic and phenotypic ratios of the offspring.What is the male pea plant's genotype?What is the female pea plant's genotype?What are the male pea plant's gametes?What are the female pea plant's gametes?What is the expected genotypic ratio of the offspring?What is the expected phenotypic ratio of the offspring?
Answers and Explanation:
Find enclosed the Punnet square.
T= dominant allele
t= recessive allele
Tt= heterozygous genotype
TT and tt= homozygous genotype
- The male pea plant's genotype is Tt (heterozygous)
- The female pea plant's genotype is Tt (heterozygous)
- Male gametes: T and t
- Female gametes: T and t
- The expected genotypic ratio of the offpring will be:
TT: Tt: tt ⇒ 1: 2: 1
1/4 of the offpring will be TT genotype (dominant homozygous), 1/2 of the offpring will be Tt (heterozygous) genotype and 1/4 of the offpring will be tt genotype (recessive homozygous).
- The expected phenotypic ratio of the offpring will be:
tall : short ⇒ 3 : 1
3/4 of the offpring will be tall (TT and Tt) and the restant 1/4 will be short (tt).
Cancer is characterized by uncontrolled cell division. Cell division is preceded by DNA replication. Several proteins of the DNA-replication machinery can be targets for chemotherapeutic agents. One such protein is topoisomerase. Select the effects that result from the inhibition of topoisomerase. Supercoils accumulate, resulting in chromosomal instability.
Inhibition of topoisomerase can result in the accumulation of supercoils, leading to chromosomal instability.
Explanation:
Cancer is characterized by uncontrolled cell division, which is preceded by DNA replication. Inhibiting the protein topoisomerase, which plays a role in DNA replication, can have effects such as the accumulation of supercoils. Supercoils are twists or turns in the DNA structure, and their accumulation could lead to chromosomal instability.
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- Proper aseptic technique is crucial to ensuring growth of pure bacterial colonies. What is one experimental way you can test your practices to confirm that you are using proper aseptic technique? (5 points)- In a laboratory setting, what are three ways you can properly sterilize culturing equipment?
- For each inoculation tool, give one scenario in which use of that tool would be appropriate.
- Do some research and describe two or three scenarios in which it would be preferable to use a stab tube vs. a growth plate.
(Hint: What do bacteria use to help them move? Can motility be used to help identify many medically important pathogenic bacteria such as the Enterobacteriaceae?)
Answer:
1. Aseptic techniques can be tested for their suitability by flaming the loop or by using agar slants.
2. Culturing equipment can be sterilized by three methods and these include wet heat, dry heat and filtration.
3. Inoculating needles are used for transferring bacterial culture to a soft agar medium. Inoculating loops are used to transfer culture into liquid medium or plate.
4. Growth plates or agar plates, Liquid media and Stab-tube media are used for growth of bacterial colonies.
Explanation:
1. A-septic techniques can be tested for their suitability by flaming the loop or by using agar slants. The loop is flamed to an extent that it becomes red-hot and is then cooled before picking up the organisms from the bacterial culture to be transferred. The hot loop is kept into the tube for a few seconds before removing the inoculum so that a bacterial aerosol is not produced.
2. Culturing equipment can be sterilized by three methods and these include wet heat, dry heat and filtration. Wet heat method is the most preferred method for sterilizing culture material using an autoclave. The material is heated using pressurized steam in an autoclave; which is an effective procedure for killing microbes, spores and viruses at 100◦C. However, it is interesting to know that pressurized steam has 7 times more heat than water at 100◦C which allows rapid delivery of heat and good penetration for sterilizing dense materials. High pressured steam hydrolyzes bacterial protein and removes any chances of contamination.
Dry heat method kill microbes by oxidation of cells thus requires more energy and is conducted at a higher temperature of 121◦C for efficient sterilization.
Filtration is another method of sterilizing liquids without heating and is performed by passing the solution through a filter with a pore diameter of 0.2 µm. The filter material used for sterilization can be heat-fused glass funnels or cellulose membranes. However, viruses can pass through these filters and filtration is not a preferred sterilization method.
3. Inoculating needles are used for transferring bacterial culture to a soft agar medium because needles supports appropriate spreading of culture and produces growth along the stab line also.
Inoculating loops are used to transfer culture into liquid medium or plate because it is free following and adequate spreading of culture is not required.
4.Growth plates or agar plates are the best bet when morphology of the species is to be identified or a bacterial colony needs to be isolated. In this scenario, growth plates allow the separation of a single clone and aids microscopic analysis.
In order to expand bacterial culture for the purpose of isolating DNA plasmid, liquid media is used. These liquid cultures allow easy inoculation of growth parameters such as antibiotics, and control of temperature and air pressure.
Stab tube media is prepared by filling test tube with soft agar and is specifically designed for growth of bacteria in low oxygen conditions. The stab tube allows immersion of inoculation loop into the agar depth where oxygen is extremely low and bacterial colonies can be produced without any shaking of media.
Final answer:
Testing aseptic technique can be done using control cultures, and equipment sterilization can be achieved through autoclaving, dry heat, or chemical methods. Inoculation tools like loops, needles, and pipettes are used for specific purposes. Stab tubes are preferable for studying anaerobic bacteria and bacterial motility, crucial for identifying certain pathogens.
Explanation:
Proper aseptic technique is crucial in microbiology to prevent contamination and ensure the growth of pure bacterial cultures. An experimental way to test your aseptic technique is by performing a control culture without introducing bacteria, allowing it to grow in the same conditions as your experimental cultures. If no growth is observed, it suggests that the aseptic technique was successful.
There are several methods for sterilizing culturing equipment, including:
Autoclaving, which uses pressurized steam to sterilize.Dry heat sterilization, where equipment is heated in an oven.Chemical sterilization, using disinfectants or antiseptics for items that cannot be exposed to high temperatures.Each inoculation tool, such as loops, needles, and pipettes, has its specific use. Loops are typically used for streaking bacteria on agar plates to isolate single colonies. Needles are for transferring bacteria to slants or stab cultures. Pipettes are used to transfer liquids, essential for creating dilutions or adding bacteria to broth cultures.
Stab tube cultures are used in scenarios where it's important to study anaerobic growth or bacterial motility. A stab tube provides an environment with reduced oxygen exposure, ideal for growing anaerobic bacteria. Moreover, the stabbing action can demonstrate motility, which is important for identifying pathogenic bacteria such as those in the Enterobacteriaceae family.
Which of the following statements about valves in the heart is/are true? A. A valve enables blood to only flow in one direction. B. A valve keeps blood from flowing in a reverse direction. C. All of the valves open when the blood pressure on one side of the valve is higher than on the other side of the valve. D. Statements A and B are correct. Statement C is incorrect. E. Statements A, B, and C are correct.
Answer: Option D is the correct answer.
The correct statements about valves are;The heart valves enable blood to flow in one direction.The valves keep blood in flowing in reverse direction.
Explanation:
Heart valves are thin tissue paper membrane that is attached to the heart. The heart valves and chambers are lined with endocardium.The heart valves is divided into four which are;
1. The two atrioventricular valves , the biscupid and triscupid valves which are between the upper chambers of the atria and the lower chambers of the ventricles
2. The two semilunar valves, the aortic and the pulmonary valves which are in the arteries leaving the heart. The biscupid and the aortic valves are in the left heart while the pulmonary and triscupid valves are in the right heart. During ventriculsr systole when pressure rises in the left ventricle and it is greater in the aorta, the aortic valves then open and allow blood to leave in through the left ventricle in the aorta.
The statement which is true about valves in the heart is: Choice D.
A valve enables blood to only flow in one directionA valve keeps blood from flowing in a reverse direction.Discussion:
Heart valves by definition are thin tissue membranes attached to the heart. The heart valves and chambers are lined with endocardium.
The heart valves are divided into four which are;
The two atrioventricular valves , the biscupid and triscupid valves.The two semilunar valves, the aortic and the pulmonary valves located in the arteries leaving the heart.All of which are characterized by enabling blood flow only in one direction and keeping blood from flowing in the reverse direction.
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Part B - Diet and Lifestyle Choices During Pregnancy
Pregnancy comes with a unique set of diet and lifestyle guidelines. Some substances are particularly beneficial for the mother or baby, others are harmful, and some should be used with caution.
Drag the appropriate items to their respective bins.
1- green leafy vegetables
2- sashimi and sushi made with
raw fish
3-alcoholic beverages
4-unpasteurized milk
5-cooked fish and seafood
6-sodas and sports drinks
7-caffeinated beverages
8-low-fat sources of calcium
and vitamin D
Avoid During Pregnancy
Consume During Pregnancy
Consume with Caution During Pregnancy
Answer:
Explanation:
Avoid During Pregnancy
3-alcoholic beverages
7-caffeinated beverages
4-unpasteurized milk
2.sashimi and sushi made with
raw fish
Consume During Pregnancy
1- green leafy vegetables
8. low-fat sources of calcium
and vitamin D
Consume with Caution During Pregnancy
6-sodas and sports drinks
5-cooked fish and seafood
soda and sports drink should be used with Doctor recommendation during pregnancy. A high amount of cooked fish is also not advisable due to its adverse effects on the pregnant women but a daily dose recommended is good for body protein needs.
green leafy vegetables and low-fat sources of calcium and vitamin D are highly recommended during pregnancy
alcohol and caffeine are not good
while raw fish is also not good during pregnancy because they have high mercury, bacteria and viruses which are more harmful during pregnancy.
_________ account for the hereditary influence on the phenotype in a cumulative way.
Answer:
Additive alleles
Explanation:
Additive alleles account for the hereditary influence on the phenotype in a cumulative way.
Alleles are nothing but variations of a gene. that simply gives variation to the traits.
Whereas, Additive alleles are alleles that contribute to most observable traits, such as height, weight, hair color, eye color,and complexion. Therefore, they account for the hereditary influence on the phenotype.
What is the energy requirement in kJ/mol to transport a proton across the mitochondrial inner membrane in plant cells at night when the outside temperature is 15ºC, the pH differential across this membrane is 1.4 pH units, and the membrane potential of 160 mV. Choose the ONE correct answer.
A. +21.8 kJ/mol
B. +75.4 kJ/mol
C. +23.1 kJ/mol
D. +46.2 kJ/mol
E. -21.8 kJ/mol
F. -23.1 kJ/mol
G. +19.0 kJ/mol
H. +44.6 kJ/mol
Answer:
C. +23.1 kJ/mol
Explanation:
the formula to use to calculate the energy requirement in kJ/mol to transport a proton across the mitochondrial inner membrane in plant cells is:
ΔGt = RTIn [tex]\frac{C_{2} }{C_1}[/tex] + ZFΔV
let's list the values of the data we are being given in the question to make it easier when solving it.
Z= 1
F= 96500C (faraday's constant)
ΔV= 160mV = 0.160V
R= 8.314( constant)
T= 15ºC ( converting our degree Celsius into kelvin, we will have 273.15k+ 15 = 288.15K)
∴ T= 288.15K
Putting it all together in the formula, we have:
ΔGt = 8.314 × 288.15 × 2.303 log [tex]\frac{C_{2} }{C_1}[/tex] + 1 × 96500 × 0.160
ΔGt = 5517.25 [tex][ -(log(H^{+}_{out} ) + log(H^{+}_{in} )_{in}[/tex] +15440
ΔGt = 5517.25 [tex](pH_{out} - pH_{in}[/tex] +15440
Given that the pH differential gradient across the membrane is 1.4pH units. It implies that;
ΔGt = 5517.25 × 1.4 + 15440
= 7724.15 +15440
= 23164.15 Joules/moles
= +23.1 KJ/mole
Individual cells do not survive for the entire lifetime of an organism because a. the enzyme telomerase is readily destroyed by the environment, resulting in cell death. b. DNA replication is subject to errors that cause cell death. c. Okazaki fragments disrupt protein synthesis, resulting in cell death. d. the repeating telomeric sequence of TTAGGG interferes with normal DNA replication and leads to cell death.
Answer:
b. DNA replication is subject to errors that cause cell death.
Explanation:
Answer : none
Explanation:the removal of the RNA primer following dna replication leads to a shortening of the chromosome and eventual cell death
Tasting involves many different cell-signaling processes that ultimately generate nerve signals transduced by membrane depolarization. Sweet tastes result in PIP2 hydrolysis, while salty tastes allow sodium ions to directly alter the membrane potential. What can you deduce about the signaling mechanisms for sweet and salty?
Answer:
We can deduce that:
1. Sodium ions directly enters the cells, indicating the signal is transduced by an ion channel.
2. Sweet utilizes the GRCP signaling pathway, activating phospholipase C.
Explanation:
IN SALTY, the Sodium ion passes through dynamic various chambers that can be located on the apical membrane of the taste buds.
Sweet utilizes GRCP signaling pathway va activation of phospholipase C in order to produce IP3 and DAG.
_____ arise from a single fertilized egg and are thus genetically almost identical.
Answer:
Monozygotic twins
Explanation:
Monozygotic or identical twins are those that originate from a single ovary_ and a single sperm and therefore share the same genetic load.
What happens in these cases is that the embryo splits in two, and depending on the moment in which this happens, different configurations can be expected in the development of the placenta.
When the embryo is cleaved within the first four days after fertilization it results in a bicorial-biamnotic pregnancy, that means, two sacs and two independent chorions will be visible (exactly the same as in the case of dizygotic twins from two different embryos).
Answer:
Zygote
Explanation:As it only organism after fertilization formed
What is the most common type of fossils? Group of answer choices:
O Trilobite (class Trilobita)
O Snail (class Gastropoda)
O Bryozoan (phylum Bryozoa)
O Foram (order Foraminiferida)
O Cephalopod (class Cephalopoda)
Answer:
Snail (class Gastropoda)
Explanation:
The most common body fossils (here we exclude ichnofossils, that are fossils of footprints, trails and other activities of an organism) in number of occurrences in the collections are of snails of the genus Turitella, which are present in the fossil record from the Cretaceous to the recent periods.
Fill in the blanks.
Osteichthyans" commonly referred to as "fishes" have a ________________ endoskeleton.
Fishes control their buoyancy with an air sac known as a ________________.
Fishes breathe by drawing water over _______ in chambers covered by a bony flap called the ________________.
Answer:
bony
swim bladder
mouth
operculum
Explanation:
Fill in the blanks.
Osteichthyans" commonly referred to as "fishes" have a _____Bony___________ endoskeleton.
Fishes control their buoyancy with an air sac known as a _____swim bladder_
Fishes breathe by drawing water over ___mouth ____ in chambers covered by a bony flap called the __operculum.______________.
Which of the following statements is/are true of corals?
Choose all that apply.
A. Corals are animals.
B. Corals are benthic organisms.
C. Corals secrete silica to build the coral reef.
D. Corals are solitary.
E. Corals live in tropical water.
Answer:
The correct answers are option A. "Corals are animals". B. "Corals are benthic organisms"., and E. "Corals live in tropical water".
Explanation:
Corals are animals, what we know as a coral, is in fact a group of small animals called polyps that need food to survive. Corals are benthic organisms because they live at the bottom of the sea. The subclass of benthic organisms that corals belong is called Macrobenthos, for being large enough to be seen at the naked eye. Corals live mostly in tropical waters, because they do not tolerate waters with a temperature below 18 Celsius.
Answer:
a
Explanation:
The genetic code is said to be "degenerate" becauseA. A single tRNA may couple a single mRNA codon to multiple amino acids. B. A single codon in an mRNA can couple to the anticodons of many different types of tRNAs. C. A single tRNA specific for a particular amino acid may respond to multiple codons in an mRNA D. More than one type of tRNA can be charged with a particular amino acid.
Answer:
The correct answer is option C. "A single tRNA specific for a particular amino acid may respond to multiple codons in an mRNA".
Explanation:
Genetic code is said to be "degenerate" because there are more codons than amino acids, which is possible because a single tRNA specific for a particular amino acid may respond to multiple codons in an mRNA. tRNA recognize the codons of the mRNA by a specific nucleotide sequence called anticodon, however the nucleotide sequence of the anticodon can recognize more than one codon sequence. This phenomenon takes place at the third position of the codon and is known as "wobble".
Both prokaryotes and eukaryotes use the information encoded on the genes in their DNA to synthesize proteins. How does the process of translation in prokaryotes differ from translation in eukaryotes?
Translation in prokaryotes differs from eukaryotes in that it occurs simultaneously with transcription and may involve polycistronic mRNA.
Explanation:The process of translation in prokaryotes differs from translation in eukaryotes in several ways. In prokaryotes, translation occurs simultaneously with transcription, meaning the mRNA is translated as it is being transcribed. This is because prokaryotes lack a nucleus and do not have separate compartments for transcription and translation. In contrast, in eukaryotes, transcription occurs in the nucleus and the mRNA must undergo several processing steps before it can be transported to the cytoplasm for translation. Additionally, prokaryotes often have polycistronic mRNA, meaning multiple genes are translated from the same mRNA molecule, whereas eukaryotes generally have monocistronic mRNA, with each mRNA molecule encoding only one gene.
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Ribosomes are the RNA-protein enzyme complexes that synthesize proteins. There are no ribosomes inside the lumen of the secretory pathway. Instead, secreted and membrane-bound proteins are synthesized in the cytosol, but have short stretches of amino acids called "signal sequences" that target the newly-forming proteins to a special channel in the membrane of the ER that enables the movement of the newly-forming proteins into the ER. This "translocation" ceases if a protein sequence is reached that will be embedded in the plasma membrane. Protein synthesis will proceed, and the membrane-spanning sequence will eventually be embedded in the ER membrane. Which of the following features would such a transmembrane amino acid sequence have?
A. The amino acids would have mostly hydrophobic side-chains.
B. The amino acids would have mostly hydrophilic side-chains.
C.There's no way to predict the relative number of hydrophobic and hydrophilic side-chains in transmembrane amino acid sequences.
Answer:
mostly hydrophobic
Explanation:
A membrane is made of amphipathic molecules with the hydrophobic part pointing to the lumen and cytosol and the hydrophobic part (much bigger makes up most of the membrane) in the middle, the protein chains that are embedded in the membrane are thus mostly hydrophobic because the membrane is mostly hydrophobic and like likes like
The correlation between the area of an island and the number of species found there is referred to as:
A. species-area relationship.
B. latitudinal gradient.
C.theory of island biogeography.
D. immigration and extinction.
Answer: Option A.
It is called specie- area relationship.
Explanation:
Specie- area relationship in ecology describe the relationship between the area or habitat and the total number of species present in the area. It is also called species-area curve.
A larger area will contain a larger number of species.
A lot of factors determine the slope and elevation of specie-area relationship and they are, immigration and extinction rate, predator- prey dynamics, rate of disturbances between large and small area e.t.c.
Many promoter regions in genes transcribed by eukaryotic RNA polymerase II do not contain a TATA box sequence. However, all three eukaryotic polymerases require TBP for transcription initiation regardless of the presence of a TATA box in the promoter region. What hypothesis might explain why TBP is necessary for transcription from TATA-less promoters
A. The DNA-bending function of TBP cannot be performed by another protein in the general transcription factor complex.
B. TBP is still required for initial recognition of promoter sequences. This protein has an equal affinity for DNA sequences with and without a TATA box.
C. TBP is required for stabilizing the complex as it assembles on the promoter.
D. TBP is required for the catalytic activity of the RNA polymerase.
Answer:
A. The DNA-bending function of TBP cannot be performed by another protein in the general transcription factor complex.
Explanation:
TATA box sequence binding protein binds to aspecific region on DNA known as TATA Sequence. TATA binding protein binding to the DNA Sequence is necessary because it generates bending of DNA till 80° angle.
BINDING of all transcription factors depends solely upon the binding of TBP directly, that is why TATA binding proteins bonding to DNA is very important for initiation of transcription. Any other protein does not have the capacity of band formation with DNA. Hence, A. the DNA-bending function of TBP cannot be performed by another protein in the general transcription factor complex.
In an organism's DNA, what makes one nucleotide different from another?
A. The sugar
B. The phosphate
C. The base
D. Only some nucleotides have carbon atoms
E. Only some nucleotides have nitrogen atoms
Answer:
C. The base
Explanation:
The nitrogenous bases, Adenine, Guanine, Thymine and Cytosine in DNA are the variable elements in the nucleotides. Nucleotides are constituted by three elements: The sugar (deoxyribose) and the phosphate group constitutes the backbone of the double helix and are constant in all nucleotides. Also, all nucleotides have carbon and nitrogen atoms.
C) The base is what makes one nucleotide different from another in an organism's DNA.
Explanation:In an organism's DNA, what makes one nucleotide different from another is the base. Nucleotides are composed of three components: a sugar molecule, a phosphate group, and a nitrogenous base. The base is the variable part of the nucleotide and can be adenine (A), thymine (T), cytosine (C), or guanine (G). The sequence of bases in a DNA molecule is what determines the genetic code or instructions for building and functioning of an organism.
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Two nutrient solutions are maintained at the same pH. Actively respiring mitochondria are isolated and placed into each of the two solutions. Oxygen gas is bubbled into one solution. The other solution is depleted of available oxygen. Which of the following best explains why ATP production is greater in the tube with oxygen than in the tube without oxygen?
A. The rate of proton pumping across the inner mitochondrial membrane is lower in the sample without oxygen. B. Electron transport is reduced in the absence of a plasma membrane. C. In the absence of oxygen, oxidative phosphorylation produces more ATP than does fermentation. D. In the presence of oxygen, glycolysis produces more ATP than in the absence of oxygen.
Was told the answer is A. but i need help justifying that and why it is not D
Answer:
Glycolysis is not an oxygen dependent process and the rate of glycolysis is not affected by the amount of oxygen present. The amount of oxygen present affects the rate of oxidative phosphorylation in the mitochondria. Hence with more oxygen present, there is increased oxidative phosphorylation in the presence of high amounts of final electron acceptor O2 which pumps more protons across the membrane and results in a steeper proton gradient across the mitochondrial membrane thus increasing the rate of ATP production. Thus the answer is A as the rate of proton pumping is increased with higher rate of oxidative phosphorylation due to the abundance of final electron acceptor O2.
Oxygen doesn't affect glycolysis' rate. Oxygen affects oxidative phosphorylation. With more oxygen, oxidative phosphorylation rises, pumping more protons across the mitochondrial membrane and increasing ATP synthesis.
What is oxidative phosphorylation?During the process of oxidative phosphorylation, electrons that were obtained from NADH and FADH2 mix with oxygen (O2). The energy that is released as a result of these oxidation and reduction events is then used to fuel the synthesis of ATP from ADP.
Glycolysis does not require oxygen, and the presence of oxygen has no effect on the rate at which it occurs. Oxygen has an effect on the oxidative phosphorylation that occurs in mitochondria. In the presence of high amounts of final electron acceptor O2, the process of oxidative phosphorylation speeds up when there is more oxygen present. This results in the transfer of more protons across the mitochondrial membrane, which boosts the production of ATP.
Therefore, option (A) is correct and (D) can't be true.
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What effect does histone acetylation have on transcription? Group of answer choices Extremely decrease expression Moderately decrease expression No change in expression Measurably increase expression Either increase or decrease expression
Answer:
Measurably increase expression
Explanation:
Acetylation of histones, or more precisely their amino acids lysine, causes their positive charge to be neutralized, and weakening of the interaction of the histone tail with negatively charged local DNA induces local opening of chromatin structures. Thus, local DNA is exposed, increasing access to transcription factors and promoting significant increases in gene transcription.
In other words, histone acetylation results in the decompression of chromatin, which measurably increases expression.
The authors of a cross-sectional study hypothesized that lack of regular exercise is associated with obesity in children. Their study of 12 children in Michigan, however, failed to show a statistically significant association between exercise habits and obesity (OR = 1.9, p = 0.11). Is this A) Selection Bias, B) Confounding, C) Information Bias, D) Random Error
Answer:
A) Selection Bias
Explanation:as it depends on sampling method of data collection
In the scheme of feeding bacteria to worms to induce an RNAi response in worms, explain the purpose of the following components: lac promoter incorporated into the genome of E. coli, T7 gene incorporated into the genome of E. coli, and T7 promoter on the plasmid introduced into E. coli.
Answer:
lac promoter incorporated into the genome of E. coli, = for the repression of the gene
T7 gene incorporated into the genome of E. coli,+ for the formation of T7 gene
and T7 promoter on the plasmid introduced into E. coli= for the attachment of Polymerase to transcriped RNA from 5 to 3 direction
Explanation:
Answer:
For repression of gene the lac parameters must be incorporated into genome of E.coli
Where for the formation T7 gene,T7 gene must be incorporated into genome of E.coli
For attachment of polymerase to transcript RNA from 5 to 3 direction the T7 on the placimid must be introduced into E.coli
Which of the following statements about interphase is FALSE?
A. Most cells in adult human bodies are in interphase (including G0)
B. DNA is replicated in S phase
C. Once a cell enters interphase, it always moves on to M phase again
D. Cells grow in G1 and G2
E. Cells can stop the cell cycle by going to G0
Answer:
C. Once a cell enters interphase, it always moves on to M phase again
Explanation:
Interphase is the first stage of cell cycle and is followed by M phase. Interphase includes two growth phases called G1 and G2. The S phase of interphase is marked by DNA replication. The G2 phase of interphase is followed by M phase. However, not all the cells that have entered the interphase ends up in M phase.
A cell can be withdrawn from the cell cycle if the checkpoints are not passed through. Many cells halt the process of cell division for some time or even longer and may or may not resume the process of cell division again. For instance, if replication fork is stalled or there is error in DNA replication, the cell will be withdrawn from cell cycle at the G2 checkpoint.
n mice, black fur (B) is dominant to white (b) . At different locus, a dominant allele (A) produces a band of yellow just below the tip of each hair in mice that have black fur, which gives a frosted appearance that is termed agouti. Expression of the recessive (a) allele results in a solid coat color. If double heterozygous agouti parents are crossed, what color will most of the offspring be ?
Answer:
Black fur with solid coat
Explanation:
Given -
Black fur allele is represented by "B"
and white fur allele sis represented by "b"
Yellow band allele is represented by "A" and
solid coat color allele is represented by "a"
Genotype of double heterozygous agouti parents will be
"BbAa"
Crossing BbAa we get -
BA Ba bA ba
BA BBAA BBAa BbAA BbAa
Ba BBAa BBaa BbAa Bbaa
bA BbAA BbAa bbAA bbAa
ba BbAa Bbaa aaAa bbaa
Most of the offsrping will be Black fur with solid coat