Answer:
Cluster of treats that define how people normally react in the situation are called disposition.
Explanation:
Disposition refers to mood swing and the attitude towards life which surrounds one person. It is also a description of ones temperament. it mean that the qualities establishing the person or a group. The persuade habit of one person and the natural mind set or an emotional outlook is disposition. It is a 'state of mind' regarding something.Evaluate one strategy that manufactures can use to reduce the human costs of extracting minerals for use in cell phones. In a paragraph, explain how the strategy can help resolve the problem citing evidence to support your claim.
Answer:
Manufacturers could promote the recycling of cell phones.
Explanation:
Mining for the manufacture of cell phones and other electronic devices has caused at least six million refugees in the Democratic Republic of Congo in the last two decades. The high demand for minerals important for mobile technology causes humanitarian and environmental conflicts in the Democratic Republic of Congo. This is a serious issue that should be discussed among telephone companies to establish measures that will lower the human costs of depleting these ores.
One way to alleviate these problems would be through the recycling of disused electronic devices. Telephone companies could set up telephone collection points that are not in use and thus recover the ores needed to make new devices.
Minerals such as coltan, tungsten and cassiterite are widely used components in the manufacture of cell phones and electronic devices, many of which could be recovered and reused, but there is little awareness about cell phone recycling around the world.
Below is a hypothetical developmental pathway. This pathway leads to the expression of a black pigment in certain cells of the ectoderm (the spot cells) which results in the ladybug's spotted appearance. Cells of the ectoderm where black pigment is not produced are red.
L ---| D → Y ---| B → black
Based on the pathway above, will the spot cells in ladybug mutant lacking the D gene be able to make black pigment?
Select one:
O not enough information
O yes
O no
No, in a ladybug mutant lacking the D gene, the spot cells would not be able to make black pigment.
Why is No the best choiceThe developmental pathway indicates that the expression of black pigment in spot cells (ectoderm cells) depends on the presence of the D gene. Without the D gene, the pathway is interrupted, and the subsequent steps (Y → B) that lead to the production of black pigment cannot occur.
Therefore, in a ladybug mutant lacking the D gene, the spot cells would not be able to make black pigment.
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In mountain Koalas, brown coat color is recessive to gray coat color. A gray female rabbit gives birth to four gray-coated and three brown-coated baby rabbits. What can be deduced about the genotype of the baby rabbits’ father?
Answer:
gg
Explanation:
It is given that allele for brown colored coat is recessive to gray colored coat. Let the allele for brown color coat be represented by “g” and gray color coat be represented by “G”
Since brown color is a recessive trait. It will be present only in offspring having genotype “gg”. This means that genotype of both the parents will have “g” allele.
Since it is given that female is gray colored thus genotype of female would be “Gg”
So the father would have genotype “gg”
Suppose you are investigating an autosomal recessive disease known as "studius toxicosis" which occurs at a rate in the American college student population of 1 in 16 individuals. Students who inherited two recessive alleles (tt) have the disease. If we assume Hardy-Weinberg equilibrium in a large population of college students, what is the percentage of heterozygous carriers of the studius toxicosis allele in the population? That is, individuals who have the genotype Tt. Refer to the Hardy-Weinberg equation. Be sure to answer in a percentage, not decimal, but do not include the % sign. Example: answer 45, not 0.45.
Answer:
37.5
Explanation:
Acording to the the Hardy-Weinberg equation:
[tex]p^2 +pq+q^2=1[/tex]
[tex]p+q = 1[/tex]
Where p is the frequency of the dominant allele "T", and q is the frequency of the recessive allele "t".
Therefore, p^2 is the frequency of the dominant homozygous genotype "TT", q^2 is the frequency of the recessive homozygous genotype "tt" and
2pq is the frequency of the heterozygous genotypes "Tt" and "tT".
Since 1 in 16 individuals have inherited two recessive alleles (tt) and have the disease:
[tex]q^2=\frac{1}{16} \\q=\sqrt{\frac{1}{16}} \\q= 0.25[/tex]
Thus, the fraction of heterozygous carriers of the studius toxicosis allele in the population is given by:
[tex]p=1-q\\p = 1 - 0.25 = 0.75\\2pq=2*(0.75)*(0.25)\\2pq= 0.375[/tex]
37.5 percent of the population are heterozygous carriers
Growth factors and cytokines both lead to tyrosine phosphorylation through receptors, but they do so through different mechanisms. What is the key difference between the receptors for these two classes of ligands in terms of their tyrosine kinase activity?
Answer:
The growth factor receptors have a kinase domain while the Cytokines receptors do not contain a kinase domain as part of their structure.
Explanation:
The two are signaling molecules that control cell activities in some manners, such paracrine, endocrine and autocrine manners.
The receptor kinase domain can be specific for substrate sites in which phosphorylation occurs.
Which of the following is an example of exploitation?
A. ants eating trash
B. humans that create a beehive
C. flowers blooming
D. rabbits that feed on a homeowner's garden
Answer:
B.
Explanation:
The answer is B, because bee farmers create false bee hives for h.bees to start a hive. They also can purchase a queen, (that reproduce more bees) that is her only job. The worker bees collect pollen and mix it with the bee's saliva to store honey for the hive, which also is food 4 the bees.
Suppose Alia recently learned that she inherited a mutant RB1 allele from her mother, who had retinoblastoma. RB1 is a tumor suppressor gene that is related to retinoblastoma. Why would Alia be at higher risk for getting retinoblastoma at an earlier age than her sister, Francine, who inherited a normal RB1 allele from their mother?
Answer:
Tumour suppressor genes can readily convert into tumour causing genes if mutations arise in them.
As Alia has one of the alleles that is a mutant, she will have more chances of getting retinoblastoma if the other allele gets mutated. A person who has inherited two functioning alleles of a tumour suppressor gene will require both the alleles to get mutated for the development of the tumour, hence her sister, Francine, who inherited a normal RB1 allele will have lesser chances.
Answer:
A person who inherited two functioning alleles of a tumor suppressor gene needs both alleles to mutate for tumors to develop
Explanation:
In a long bone, the osteons are:
A. lined up in the same direction as the diaphysis of the bone
B. lined up perpendicular to the long axis of the bone, in the direction of perforating canals.
C. arranged in an irregular pattern
D. are separated by medullary spaces
E. are lacking in the diaphysis of the bone
Answer:
B. lined up perpendicular to the long axis of the bone, in the direction of perforating canals.
Explanation:
Osteons are formations of concentric bone layers (lamellae) that tend to run parallel to the long axis of a bone. They sorround the Haversian canal, a canal that contains blood vessels that supply the osteocytes and are the structural unit of compact bone.
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Suppose that RNA polymerase was transcribing a eukaryotic gene with several introns. In what order would the RNA polymerase encounter the following elements in the DNA sequence of the gene?A-stop codonB-translation initiation codonC-TATA box elementD-3' UTRE-5' UTRF-splice acceptor site
RNA polymerase, during the transcription of a eukaryotic gene with introns, would encounter DNA elements in the order of TATA box, 5' UTR, translation initiation codon, splice acceptor site, stop codon, and 3' UTR. Transcription occurs in the 5' to 3' direction, beginning with the promoter region and moving downstream.
If RNA polymerase were transcribing a eukaryotic gene with several introns, the order in which it would encounter the elements in the DNA sequence of the gene would be as follows:
C-TATA box element - This is a component of the promoter region where transcription factors assemble and RNA polymerase II binds to start transcription.E-5' UTR (Untranslated Region) - The segment of mRNA that is transcribed first but is not translated into protein.B-translation initiation codon - This is typically the AUG codon, where the ribosome will begin the translation process.F-splice acceptor site - The location within the intron where splicing will occur during mRNA processing.A-stop codon - Signals the end of the protein-coding sequence within the mRNA.D-3' UTR (Untranslated Region) - The segment of mRNA that follows the coding sequence and is involved in regulation and stability of mRNA.It's important to note that RNA polymerase II transcribes the gene from the 5' to 3' direction, starting with the promoter region and moving downstream through the gene.
The law of independent assortment allows for genetic recombination. The following equation can be used to determine the total number of possible genotype combinations for any particular number of genes: 2g = Number of possible genotype combinations (where g is the number of genes)1 gene: 21 = 2 genotypes2 genes: 22 = 4 genotypes3 genes: 23 = 8 genotypesa) Consider the following genotype: YySsTt. How many different gamete combinations can be produced?
Answer:
Gametes that will be produced are:
YST, YSt, YsT, Yst, yST, ySt, ysT, yst
Explanation:
From YySsTt, it is obvious that as there 3 type of genes.
The number of gametes = [tex]2^{3}[/tex] = 8
5 genes = [tex]2^{5}[/tex] = 32 genotypes
10 genes = [tex]2^{10}[/tex] = 1024 genotypes
20 genes = [tex]2^{20}[/tex] = 1048576 genotypes
In the Cori cycle, when glucose is degraded by glycolysis to lactate in muscle, the lactate is excreted into the blood and returns to the liver. In the liver, lactate is converted back into glucose by gluconeogenesis. For each given enzyme, identify whether it is involved in the glycolysis pathyway, gluconeogenesis pathway, both pathways, or neither pathway.1. glyceraldehyde 3-phosphate dehydrogenase O glycolysis O gluconeogenesis O both O neither 2. glucose-6-phosphatase O glycolysis O gluconeogenesis both O neither 3. alcohol dehydrogenase O glycolysis O gluconeogenesis both neither 4. phosphoenolpyruvate carboxykinase O glycolysis O gluconeogenesis O both 5. phosphofructokinase-1 O glycolysis O gluconeogenesis both O neither 6. phosphoglycerate mutase O glycolysis gluconeogenesis O both neither 7. hexokinase O glycolysis O gluconeogenesis. O both neither 8. pyruvate dehydrogenase O glycolysis gluconeogenesis
Answer:
Explanation:
1. glyceraldehyde 3-phosphate dehydrogenase: O both
2. glucose-6-phosphatase: O gluconeogenesis
3. alcohol dehydrogenase: O neither
4. phosphoenolpyruvate carboxykinase: O gluconeogenesis
5. phosphofructokinase-1: O glycolysis
6. phosphoglycerate mutase: O both
7. hexokinase: O glycolysis
8. pyruvate dehydrogenase: O neither
Which characteristic would not be useful in a vector? A) A recognition sequence for a restriction enzyme B) Large size relative to the host chromosomes C) The ability to replicate independently inside the host cell D) A reporter gene E) An origin of replication
Answer:
The correct answer is B) Large size relative to the host chromosomes
Explanation:
A vector for clonation or expression must contain sequence elements which allow it independent replication and identification of host cells which have incorporated the vector. Among these elements are:
- Replication origin (option E): allows it to replicate independently of the host genome. Option B is the same as it.
- Restriction enzyme sites (option A): in order to cut the vector and to inser the desired gene into the vector.
- Reporter gene (option E): it allows to visualize the host cells which are expresing the inserted gene of the vector. For example, a reporter gene can codificate a fluorescent protein, thus the host cells which have expresion of the vector are fluorescent.
The only option which is unuseful is option B because if the vector has too large size, the transformation efficiency will be low. That means that fewer host cells will contain the vector after transformation.
A female rabbit heterozygous at two genes (named A and B) is test-crossed with a male rabbit. Their progeny include 442 A/a;B/b, 458 a/a;b/b, 46 A/a;b/b, and 54 a/a;B/b. Explain these results. Identify which alleles are recessive, whether genes A and B are linked, and, if so, how far apart they are on the chromosome
Answer:
The genes are linked and 10 mu apart.
Explanation:
A female AaBb rabbit is test crossed with a male rabbit (aabb). The male can only produce ab gametes (all the progeny will have ab on one of the homologous chromosomes).
If the genes assorted independently, the female would produce 4 types of gametes with the same frequency: 1/4 AB, 1/4 Ab, 1/4 aB and 1/4 ab.
However, the observed AB and ab gametes were much more frequent than Ab and aB, which means that the genes are linked and alleles on the same chromosome do not assort independently during meiosis.
Recombination is a rare event, so the most abundant gametes are the parentals. That is how we know that the mother had the AB/ab genotype. The recombinant gametes therefore are Ab and aB.
Distance (mu) = # Recombinants × 100/ Total progeny
Distance = (54 + 46) × 100/ 1000
Distance = 100 × 100/1000
Distance = 10 mu
Certain species of frogs in the genus Phyllobates have a powerful defensive adaptation−their skin can secrete a milky fluid that contains an extremely toxic compound called batrachotoxin (BTX). These frogs, which are found in Colombia, are known as poison dart frogs because some indigenous Colombian hunters coat the tips of their blowgun darts with the frogs' skin secretions. An animal hit by one of these darts dies quickly. What is the mechanism of action of BTX?
Answer:
Batrachotoxin is one of the most potent biotoxins, more toxic than strychnine (250x approx.), causing ventricular fibrillation and neuromuscular toxicity. Its action mechanism is producing a conformational change that allows a permanent state of the channel opened that increases the resting sodium permeability. This generates action potentials in excitable membranes of nerve and muscle; causing the blocking of neuromuscular transmission and evoking muscular contracture. Death results from respiratory paralysis. In addition, BTX also causes arrhythmias, ventricular tachycardia, and fibrillation.
Explanation:
Batrachotixin (BTX, PubChem CID: 6324647), obtained from frogs of the genus Phyllobates of the Choco rain forest of Colombia, activate permanently the voltage-gated sodium channels. Previous work demonstrated that a phenylalanine residue approximately halfway through pore-lining transmembrane segment IVS6 is a critical determinant of channel sensitivity to BTX. In addition, Li 2002 demonstrated an electrostatic ligand-receptor interaction at this site, possibly involving a charged tertiary amine on BTX.
- Cataldi, M. (2016). Batrachotoxin.
- Li, H. L., Hadid, D., & Ragsdale, D. S. (2002). The batrachotoxin receptor on the voltage-gated sodium channel is guarded by the channel activation gate. Molecular pharmacology, 61(4), 905-912.
- National Center for Biotechnology Information. PubChem Database. Batrachotoxin, CID=6324647, https://pubchem.ncbi.nlm.nih.gov/compound/Batrachotoxin (accessed on Dec. 4, 2019).
Dr. Smith's parents have normal hearing. However, Dr. Smith has an inherited form of deafness. Deafness is a recessive trait that is associated with the abnormal allele d. The normal allele at this locus, associated with normal hearing, is D. Dr. Smith's parents could have which of the following genotypes?A) DD and ddB) dd and ddC) Dd and DdD) Dd and DD
Answer:
The correct option is C) Dd and Dd
Explanation:
To explain the answer, lets make a punnet square with a cross between Dd and Dd.
D d
D DD Dd
d Dd dd
As deafness is a recessive trait, this means that both the alleles in a person should be same and recessive for that trait to occur in the phenotype. A cross between heterozygous carriers show that there will be a 25% probability that the offsprings produced will get the recessive alleles and hence they will be deaf. 50% of the offsprings will have the chance to be carriers like their parents. And there willbe a 25% chance that they will have normal homozygous alleles.
Securin is a cytoplasmic protein that binds to separase, an enzyme that degrades cohesin. When separase is bound to securin, separase is inactive. When separase is released, it immediately becomes active. Securin and separase remain attached to each other as long as securin is phosphorylated. Based on the role of securin, it must become dephosphorylated just prior to
metaphase.
anaphase.
prometaphase.
prophase.
telophase.
cytokinesis.
Answer:
Anaphase.
Explanation:
Two main process of cell division are mitosis and meiosis. The regulation of the cell cycle is maintained by the regulation of cyclins, cdks and important protein factors.
Securin and separase are important proteins that allow the transition of one phase of the cell cycle to the another phase. The securin helps in the separation of the chromosome and sister chromatid that occurs in anaphase. Hence, dephosphorylation must occur prior to anaphase.
Thus, the correct answer is option (2).
Interferons are cytokines produced by host cells in response to intracellular infection. There are two types of interferons, type I and type II. Which of the following is a difference between these two types? Choose one:
A. Type I is part of innate, nonspecific immunity, while type II is part of adaptive, specific immunity.
B. Type I binds to extracellular pathogens, while type II binds to intracellular pathogens.
C. Type I includes interferon-alpha and interferon-beta, while type II includes interferon-omega.
D. Type I interferes with viral replication, while type II activates white blood cells.
Answer:
A. Type I is part of innate, nonspecific immunity, while type II is part of adaptive, specific immunity.
Explanation:
Type I interferons: Are produced early on during infection and are responsible for activation of the innate immune response, e.g Natural Killer cells.
Type II interferons: Are produced as part of the innate immune response and act as a link between innate immune response and activation of the adaptive immune response.
Interferon type I is part of innate immunity, while type II is part of adaptive immunity.
Explanation:The correct answer is option A: Type I interferons are part of the innate, nonspecific immunity, while type II interferons are part of the adaptive, specific immunity. Type I interferons, including interferon-alpha and interferon-beta, are produced by most cells in response to viral infection and have antiviral effects. Type II interferon, also known as interferon-gamma, is mainly produced by activated T cells and natural killer cells to activate white blood cells and enhance immune response.
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The sodium-potassium ion exchange pump_____________.
A) must reestablish ion concentrations after each action potential.
B) transports sodium ions into the cell during depolarization.
C) transports potassium ions out of the cell during repolarization.
D) moves sodium and potassium opposite to the direction of their electrochemical gradients.
E) depends on a hydrogen gradient for energy.
Answer: option D
The sodium potassium ion exchange move sodium and potassium in opposite direction in electrochemical gradients.
Explanation:
The sodium potassium pump is found in many animals plasma membranes and its moves the sodium and potassium ion in opposite direction across the plasma membrane with the hydrolysis of ATP(adenosine triphosphate) to supply the needed energy. It is an active transport process.
Final answer:
The sodium-potassium pump moves sodium ions out and potassium ions into the cell against their concentration gradients, using ATP for energy, to maintain the correct ionic balances and the neuron's resting membrane potential.(Option D)
Explanation:
The sodium-potassium ion exchange pump plays a critical role in maintaining the electrochemical gradients necessary for the proper functioning of neurons. This pump actively transports three sodium ions out of the cell and two potassium ions into the cell, against their respective concentration gradients.
It uses energy in the form of ATP to carry out this process, which is essential for keeping the ionic concentrations at proper levels inside and outside the cell. By moving ions against their electrochemical gradients, the pump maintains the negative charge inside the cell, which is crucial for the transmission of nerve impulses.
In MendAliens, normal head top is dominant to Bart head top, in which the top of the head is jagged. Suppose a MendAlien homozygous for normal head shape is crossed with one homozygous for Bart head shape (see picture above). What are the genotypes of the two parents? What are the genotypes in the F1 generation? What are the phenotypes present in the F1 generation?
Answer:
The genotypes of the two parents: normal head shape (HH) and Bart head shape (hh)
The genotypes in the F1 generation= Hh
The phenotypes present in the F1 generation= All normal head shape
Explanation:
Let's assume that the allele "H" is responsible for the normal head shape while the allele "h" gives Bart head shape. According to the given information, both the parents are homozygous. The genotype of the parent with normal head shape would be "HH" while that of the one with Bart head shape would be "hh". Since the normal head shape is dominant, all the F1 hybrid progeny would exhibit "normal head shape".
The genotypes of the two MendAlien parents are homozygous dominant (HH) and homozygous recessive (hh), respectively. The F1 generation will all have heterozygous genotypes (Hh) and exhibit the dominant phenotype of a normal head shape.
The MendAlien scenario described involves a monohybrid cross, which is a basic genetic cross considering one trait at a time. We are told that a homozygous dominant individual (with a normal head shape) is crossed with a homozygous recessive individual (with a Bart head shape).
The genotype of the homozygous dominant parent would be HH (where 'H' represents the allele for a normal head shape) and the genotype of the homozygous recessive parent would be hh (where 'h' represents the allele for a Bart head shape).
When these individuals are crossed, the F1 generation will all be heterozygous (Hh), receiving one dominant allele for a normal head shape from the dominant parent and one recessive allele for a Bart head shape from the other parent. As a result, all F1 offspring will display the phenotype of the dominant trait - normal head shape.
How often do human skin cells divide? Why might that be? Compare this rate to how frequently human neurons divide. What do you notice?
Answer:
explained
Explanation:
The skin cells divide every day, as the epithelial layer is the outer protective layer, it needs to be replaced from time to time, which helps in healing wear and tear. Our skin cells divide once in each day.
In contrast, the nerve cells do not divide at all. That is their frequency of cell division is zero. They cannot divide because they lack centriole in them.
Human skin cells undergo division approximately every 27 days. This rapid turnover is necessary to replenish and repair the skin, which is exposed to constant wear and tear, environmental factors, and potential damage.
On the other hand, human neurons generally do not undergo division once they are fully developed. Unlike skin cells, neurons are post-mitotic, meaning they have exited the cell cycle and have limited or no ability to divide further.
The notable difference in the division rate between skin cells and neurons highlights the contrasting needs and functions of these cell types.
Thus, skin cells require frequent division to maintain the integrity and functionality of the skin, while neurons focus on specialized functions and long-term stability rather than continuous replication.
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___________ is a measure of the degree to which the phenotypic variation of a given trait is due to genetic factors.
Metabolism can be bisected into two subcategories:
(A) catabolism and anabolism.
(B) takes complex organic molecules and breaks them down into simpler molecules; this is often accompanied by the of energy.
(C) builds up biomolecules from simpler substances; this is often accompanied by the of energy.
Answer: A (catabolism and Anabolism)
Explanation:
Metabolic pathway are basically divided into two categories.
1. Catabolic
2. Anabolic
Catabolic (degradation) pathways, where energy rich complex macramolecules are degraded into smaller molecules. Energy released during this is trapped as chemical energy, usually as ATP.
Anabolic (biosynthesis) pathways. The cells synthesize complex molecules from simple precursors. This needs energy.
Describe the similarities and differences between voltaic and electrolytic cells. Be sure to address the following items in your description: spontaneous reduction or oxidation reactions and which electrode(s) they occur at charges on electrodes .
Answer:
Both cell are opposite to each other in function and make
Explanation:
a) A voltaic cell works opposite to that of electrolytic cell. It produces chemical energy from electrical energy while the later one produces electrical energy from chemical energy.
b) The chemical reactions in voltaic cell are spontaneous while chemical reactions in electrolytic cells are non –spontaneous.
c) The anode and cathode carry opposite charges in the two cells. In case of voltaic cell, anode carries negative charge while cathode carries positive charge.However, in case of electrolytic cell both cathode and anode bear opposite charges wrt voltaic cell
Which of the following is NOT part of genetic "linkage mapping"?
A. Calculating recombination frequencies using a testcross
B. Assuming that recombination frequency is proportional to the distance
between two linked genes
C. Finding and studying alternative (e.g., mutant) phenotypes of a character
D. Using information on DNA sequences
E. Studying characters two at a time
Answer:
The correct answer is option D.
Explanation:
Linkage mapping is the mapping process in which the genes present on the chromosomes are mapped on the base of their linkage. The linkage mapping helps in calculating how frequent recombination occurs using testcross.
Linkage mapping does not utilize the information that is on DNA sequences, however, it helps in assuming the distance of two linked genes is proportional of the recombination frequency. By the recombination, frequency mutations can be found and study.
Thus, the correct answer is option D.
A male Flagus fly with the Barkus phenotype is crossed with a female who has the wild-type phenotype. A second cross is performed in which the female Flagus fly has the Barkus phenotype and the male has the wild-type phenotype. This second cross is known as a:
Answer:
Reciprocal cross
Explanation:
a reciprocal cross is a breeding experiment designed to test the role of parental sex on a given inheritance pattern. In this type of test cross, male of one genotype and female of different genotype are crossed and the opposite is true as well. ( i.e the female of one genotype and male of different genotype are crossed.)
As we can see in the male Flagus fly with the Barkus phenotype is being crossed with a female with the wild-type phenotype.
In the second test, we can see that the opposite is true of our first statement where the female Flagus fly has the Barkus phenotype and the male has the wild-type phenotype after being crossed again thereby reversing the genotype.
Why is a frameshift missense mutation more likely to have a severe effect on phenotype than a nucleotide-pair substitution missense mutation in the same protein? Why is a frameshift missense mutation more likely to have a severe effect on phenotype than a nucleotide-pair substitution missense mutation in the same protein? A frameshift missense will cause the codons to be out of order, but a substitution missense does not change the order of the codons. A substitution missense mutation causes the protein to be shorter and thus non-functional. A substitution missense affects only one codon, but a frameshift missense affects all codons downstream of the frameshift. A frameshift missense mutation will cause an early Stop codon, but a substitution missense might be silent.
A substitution missense affects only one codon, but a frameshift missense affects all codons downstream of the frameshift.
A frameshift mutation occurs during deletion of one of two nucleotides. The arrangement of codons downstream of the mutation will definitely change, hence it being called a frameshift mutation. On the other hand, a substitution mutation will only cause a change in one nucleotide of one codon.
Explanation:
A frameshift mutation causes new codons downstream of the mutation that will code for different amino acids hence changing the properties of the translated proteins. This will have great ramifications for the phenotype of the organism.
On the other hand, substitution mutation will only cause a change in the amino acid at that point of mutation during translation. This will not have a drastic change in the protein hence will not be as lethal as a frameshift mutation.
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A frameshift mutation typically has a more severe effect on protein function than a nucleotide-pair substitution missense mutation because it changes the reading frame, leading to widespread amino acid changes and potentially a nonfunctional protein, while a missense mutation usually alters a single amino acid and may still produce a functional protein.
Explanation:A frameshift mutation is caused by insertions or deletions of nucleotides that are not in multiples of three, leading to a major shift in the genetic reading frame. This results in a different set of amino acids being encoded from the mutation point onward, which often produces a nonfunctional protein. Contrarily, a nucleotide-pair substitution mutation, also known as a missense mutation, typically changes only a single amino acid in the protein. This type of mutation has a less profound effect compared to a frameshift mutation because it does not alter the reading frame for the rest of the coding sequence.
A missense mutation from a nucleotide-pair substitution may sometimes lead to a protein that retains functionality, especially if the new amino acid is chemically similar to the original. However, if that amino acid is in a critical part of the protein, such as the active site, then the consequences can be more severe. In contrast, a frameshift mutation can result in a completely altered protein with many incorrect amino acids, which is more likely to be nonfunctional and have a severe effect on the phenotype.
It is important to note that while a missense mutation affects only the specific codon where the substitution occurs, a frameshift mutation has the potential to affect all downstream codons after the mutation point. Additionally, a frameshift mutation can introduce an early stop codon, prematurely ending the protein, while a missense mutation does not introduce a stop codon and usually only impacts the incorporation of a single amino acid.
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Color-blindness is an X-linked recessive disorder. Mike is color-blind. His wife, Meg, is homozygous for normal color vision allele. Using Punnett squares, derive and compare the genotypic and phenotypic ratios for the offspring of this marriage.
a. What is Mikes's genotype?b. What is Meg's genotype?c. What is the genotypic ratio of offspring?d. What is the phenotypic ratio of offspring?e. If they have eight children, how many of them would you expect to be color-blind?
Mike's genotype is XcY, Meg's is XX. Using Punnett square, the genotypic ratio of their offspring is 1:1 and the phenotypic ratio is 100% normal vision. If they have eight children, none are expected to be color-blind.
Explanation:In terms of genetics, color blindness is an X-linked recessive disorder. Mike, being color-blind, must have the genotype XcY. His wife, Meg, has normal color vision and is homozygous dominant, so her genotype is XX.
When we use a Punnett Square to derive the genotypic ratio of their offspring, we see that all daughters will be heterozygous carriers (XcX) and all sons will have normal vision (XY). Thus, the genotypic ratio is 1:1 (XcX: XcY).
Regarding the phenotypic ratio, all offspring are expected to have normal vision because they will inherit at least one normal vision allele. Therefore, the phenotypic ratio is 100% normal vision.
If Mike and Meg have eight children, because female offspring will be carriers and male offspring will possess the normal vision allele, we would expect none of them to be color-blind.
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the vertices of a triangle are (2,2),(4,4),and (4,2). Reflect the triangle in the y-axis. Give the coordinates of the reflected triangle
Which of the following is an example of homologous structures?
A. Whale hip bones.
B. Embryos of fish and birds both make gill slits.
C. Mice and chicken differ by only 25 amino acids when they make hemoglobin protein.
D. Rabbits and birds have the same bones in the same order in their forelimbs, even though they use them for different purposes.
Final answer:
Homologous structures are those that have a similar structure and origin but may have different functions in various species. The example of rabbits and birds, which have the same bones in the same order in their forelimbs for different purposes, represents homologous structures.(Option D)
Explanation:
The student's question asks which of the following is an example of homologous structures. Homologous structures are those that are found in different species and share a common ancestral origin, despite possibly having different functions in the organisms where they're found.
Considering this, the correct answer is D: Rabbits and birds have the same bones in the same order in their forelimbs, even though they use them for different purposes. This resembles the forelimbs of mammals like humans, cats, and whales, as well as the wing bones of bats, which while serving different purposes, reflect a common structural plan pointing to a shared evolutionary history.
Which of the following takes place during translation?
a) DNA replication
b) The conversion of genetic information from DNA nucleotides into RNA nucleotides
c) The conversion of genetic information from the language of proteins to the language of enzymes
d) The conversion of genetic information from the language of nucleic acids to the language of proteins
Answer:
d) The conversion of genetic information from the language of nucleic acids to the language of proteins
Explanation:
Answer:
The answer is: d) The conversion of genetic information from the language of nucleic acids to the language of proteins.
Explanation:
This is what is known as the central dogma of biology. All genetic information is found in genes, which are segments of DNA that contain genetic information. The human genome contains approximately 30000 genes. However, only a small part is codifying.
For this information to happen, the first thing is that it must be copied (replication) and this process happens in the kernel. DNA is copied to messenger RNA (transcription). This information is then used for protein construction (translation), which occurs in the cytoplasm.
These three processes are known as the central dogma of biology, which is that information goes from DNA to RNA and from it to proteins.