Biology High School
Answers
Answer 1
Skin color is determined by three skin pigments: melanin, carotene, and hemoglobin. The first statement about skin pigment is true because it protects the skin from UV light. The second statement is also true because melanin is the most important skin pigment.
The third statement is true because hemoglobin passes from melanocytes to keratinocytes in the stratum basale. However, the fourth statement is incorrect because hemoglobin is not the skin pigment responsible for the pink color in light skin. Hemoglobin is a protein molecule found in red blood cells that carries oxygen from the lungs to other tissues and organs.
When hemoglobin circulates close to the surface of the skin, it can give a pink or reddish hue, especially in fair-skinned individuals. Therefore, the correct option is: Hemoglobin is not the skin pigment responsible for the pink in light skin.
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Related Questions
How are T-cells activated? What role do dendritic cells
and major histocompatibility complexes (MHCs) play in this
process?
Answers
T-cells are activated through antigen presentation on the major history compatibility complex (M H C) class II molecules, that activate an immune response
T-cells are white blood cells that play a vital role in the immune system's reaction to disease-causing pathogens such as bacteria, viruses, fungi, and parasites. They recognize and react to specific antigens, which are substances that activate an immune response such as the secretion of cytokines and cell proliferation.Cross-presentation.
To T-cells via their surface M H C class II molecules. This antigen presentation causes the T-cell to differentiate into cells and initiate an immune response, such as the secretion of cytokines and cell proliferation.Cross-presentation.hey recognize and react to specific antigens, which are substances that activate an immune response
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Which of the following muscles is NOT biarticular? O Long head of tricep Short head of bicep Long head of bicep Lateral head of tricep QUESTION 2 begins at the sternum and travels to the humerus. The Pectoralis minor O OOO O Pectoralis major Infraspinatus Subscapularis QUESTION 3 In general, muscles that are on the anterior forearm will cause which motion around the wrist? Flexion Extension оооо Rotation Circumduction QUESTION 4 The shoulder typically has the greatest range of motion (ROM) in which motion Internal rotation External rotation OOOO Extension Flexion
Answers
The muscle that is not biarticular among options provided is Lateral head of tricep.
2 The muscle that begins at sternum or travels to humerus is Pectoralis major.
3 muscles on the anterior forearm generally cause flexion around the wrist.
4 the shoulder typically has greatest range of motion (ROM) in external rotation.
Muscles are vital for movement and stability in the human body. They consist of contractile fibers that generate force and enable various actions, like walking, running, and lifting. Muscles are categorized as skeletal, smooth, or cardiac, depending on their location and function. Regular exercise helps strengthen muscles, improves overall health, and enhances endurance. Adequate nutrition, rest, and proper hydration are essential for maintaining healthy muscles.
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the
pkasma membrane may contain all except which proteins?
Answers
The plasma membrane may contain all types of proteins. There are no specific proteins that are completely excluded from the plasma membrane.
The plasma membrane, also known as the cell membrane, is a selectively permeable barrier that surrounds the cell and separates its internal environment from the external environment. It is composed of a phospholipid bilayer embedded with various types of proteins. These proteins play crucial roles in the functioning of the cell membrane, including the transport of molecules, cell signaling, and structural support.
There are different types of proteins found in the plasma membrane, such as integral membrane proteins, peripheral membrane proteins, receptor proteins, transport proteins, and enzymes. These proteins are essential for the proper functioning of the cell and its interactions with the external environment.
Therefore, it can be concluded that all types of proteins can be present in the plasma membrane, and there are no specific proteins that are excluded from it.
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Complete a flow chart of an immune response beginning with entrance of antigen. Start with the response of the innate immune system, describe antigen processing, cell interaction, involvement of cytokines, and the end results for B and T cells
Answers
Here is a flow chart outlining the immune response starting from the entrance of an antigen:
Entrance of Antigen
↓
Recognition by Pattern Recognition Receptors (PRRs) of Innate Immune Cells
↓
Activation of Innate Immune Response
- Release of Cytokines (e.g., Interleukins, Interferons)
- Recruitment of Phagocytes (Macrophages, Neutrophils) to the Site of Infection
- Phagocytosis of Pathogens by Phagocytes
↓
Antigen Processing and Presentation
- Phagocytes engulf and degrade antigens
- Antigen fragments are presented on the surface of Antigen-Presenting Cells (APCs)
↓
Interaction with Helper T Cells
- Antigen presentation by APCs to Helper T Cells
- Binding of T Cell Receptor (TCR) on Helper T Cells to antigen-Major Histocompatibility Complex (MHC) complex on APCs
- Co-stimulatory signals between APCs and Helper T Cells
↓
Activation of Helper T Cells
- Release of Cytokines by Helper T Cells
- Stimulation of B Cells and Cytotoxic T Cells
↓
Activation of B Cells
- Binding of Antigen to B Cell Receptor (BCR)
- Co-stimulatory signals from Helper T Cells
- Differentiation into Plasma Cells
- Production and Secretion of Antibodies specific to the antigen
↓
Activation of Cytotoxic T Cells
- Recognition of Antigen-MHC complex on Infected Cells
- Binding of T Cell Receptor (TCR) on Cytotoxic T Cells to antigen-MHC complex
- Co-stimulatory signals from Helper T Cells
- Killing of Infected Cells through release of cytotoxic molecules (e.g., Perforin, Granzymes)
↓
Effector Phase
- Antibodies and Cytotoxic T Cells eliminate pathogens or infected cells
↓
Resolution of Infection
- Decrease in pathogen load
- Return to homeostasis
It's important to note that this flow chart provides a simplified overview of the immune response and does not include all the intricacies and details of each step. Additionally, the immune response can vary depending on the specific antigen, pathogen, and individual's immune system.
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Which ovarian hormone is secreted during the follicular phase of the ovarian cycle? progesterone LH GnRH FSH estrogen
Answers
Estrogen is the ovarian hormone predominantly secreted during the follicular phase of the ovarian cycle. Its production and actions play a vital role in preparing the reproductive system for potential fertilization and implantation.
The ovarian hormone that is primarily secreted during the follicular phase of the ovarian cycle is estrogen. The ovarian cycle consists of two main phases: the follicular phase and the luteal phase. The follicular phase begins on the first day of menstruation and lasts until ovulation occurs. During the follicular phase, the anterior pituitary gland releases follicle-stimulating hormone (FSH), which stimulates the growth and development of follicles in the ovary. Within the developing follicles, granulosa cells produce estrogen. As the follicles grow, they secrete increasing amounts of estrogen into the bloodstream. Estrogen plays a crucial role in the follicular phase. It promotes the thickening of the endometrium (the lining of the uterus) in preparation for potential implantation of a fertilized egg. Estrogen also stimulates the growth of blood vessels in the uterus and promotes the development of secondary sexual characteristics. Additionally, estrogen exerts negative feedback on the anterior pituitary, suppressing the secretion of FSH and luteinizing hormone (LH) during most of the follicular phase. This helps to prevent the development of multiple follicles and allows one dominant follicle to mature.
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True or False
48. Phenotypic variations in quantitative traits is the results of
genetic variation variation, interactions of heredity and the
environment
49. In animal breeding programs, the average performance of
selected parents is always lower than that of the population from
which they were selected
50. The chicken can have a colored plumage only when the two
epistatic genes, dominant and recessive white exist in the
genotypes iiCC or iiCc
51. The genotype P-R in chickens produces a comb type called
walnut comb
52. Genes that are responsible for quantitative traits do not follow
the mendelian inheritance
53. In poultry, the female is homogenetic and the male is
heterogenetic
54. Epistasis could be used to explain the rise in performance in
hybrid individuals above the average of their parents
55. In additive gene action, the genotype reflects the phenotype
56. Feed conversion ratio is a trait that shows discrete variation
57. In quantitative traits, the offspring inherits 50% of superiority
of genes above the average of the population
58. Bodyweight, egg numbers and polydactyl are all examples of
quantitative traits
59. Animal breeding deals with application of genetic principles
and statistics for the improvement of farm animals
60. The gene mf masks the expression of the gene F which is
responsible for the frizzling
Answers
True. The gene mf masks the expression of the gene F, which is responsible for the frizzling trait in chickens.
True. Phenotypic variations in quantitative traits are the result of genetic variation and interactions between genetics and the environment.
False. The average performance of selected parents in animal breeding programs is generally higher than that of the population from which they were selected. This is because the purpose of breeding programs is to improve traits and select individuals with desirable characteristics.
False. The presence of two epistatic genes, dominant and recessive white, in the genotypes iiCC or iiCc does not guarantee colored plumage in chickens. The expression of plumage color is influenced by multiple genetic factors and interactions.
False. The genotype P-R in chickens does not produce a comb type called walnut comb. The specific genetic combinations determine the comb type in chickens.
True. Genes responsible for quantitative traits often do not follow simple Mendelian inheritance patterns. They can be influenced by multiple genes and environmental factors. False. In poultry, the male is homogenetic, meaning it carries two identical sex chromosomes (ZZ), while the female is heterogenetic, meaning it carries two different sex chromosomes (ZW). True. Epistasis, which refers to gene interactions, can contribute to the rise in performance in hybrid individuals above the average of their parents. True. In additive gene action, the phenotype reflects the cumulative effect of multiple genes in an additive manner. False. Feed conversion ratio is a trait that shows continuous variation rather than discrete variation.
False. The inheritance of superiority of genes in offspring is not fixed at 50% above the average of the population. The degree of inheritance depends on the specific genetic architecture and inheritance patterns of the traits.
True. Bodyweight, egg numbers, and polydactyl (extra digits) are all examples of quantitative traits, which show continuous variation.
True. Animal breeding involves the application of genetic principles and statistical methods to improve the characteristics of farm animals.
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The wet mount and Gram stain of vaginal secretions from a patient who has bacterial vaginosis caused by Gardnerella vaginalis will reveal large:
squamous epithelial cells with numerous attached pleomorphic gram-variable or gram-negative coccobacilli and rods.
numbers of large gram-positive rods.
numbers of polymorphonuclear leukocytes.
pleomorphic gram-negative rods with many leukocytes.
Answers
The wet mount and Gram stain of vaginal secretions from a patient who has bacterial vaginosis caused by Gardnerella vaginalis will reveal large squamous epithelial cells with numerous attached pleomorphic gram-variable or gram-negative coccobacilli and rods. There may also be small numbers of large gram-positive rods and polymorphonuclear leukocytes.
Bacterial vaginosis caused by Gardnerella vaginalis is a common vaginal infection that may or may not cause signs and symptoms in women. The wet mount and Gram stain of vaginal secretions from a patient who has bacterial vaginosis caused by Gardnerella vaginalis will reveal large squamous epithelial cells with numerous attached pleomorphic gram-variable or gram-negative coccobacilli and rods.
The wet mount of the vaginal secretions may also show clue cells. These are cells that are covered with bacteria.The Gram stain can show small gram-negative or gram-variable rods called Gardnerella vaginalis. There may be small numbers of large gram-positive rods such as lactobacilli.
A culture can be done to confirm the presence of Gardnerella vaginalis and to rule out other bacteria that can cause similar symptoms.
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All of the following are distinctions between protostome and deuterostomes except:
A) cleavage type
B) cell determination
C) presence of a cuticle
D) position of blastopore and anu s
E) organ of coelom
Answers
All of the following are distinctions between protostome and deuterostomes except C) presence of a cuticle
A cuticle is a non-cellular layer that covers the external body surface of nematodes, arthropods, and several other invertebrates. Protostomes and Deuterostomes are two of the major groups of animals. The differences between these groups are significant and are the result of divergence that has occurred over hundreds of millions of years. These differences include their cleavage type, cell determination, position of blastopore and anus, and the origin of their coelom.
Each of the above features distinguishes protostome from deuterostomes except the presence of a cuticle. The cuticle is a non-cellular layer that covers the external body surface of nematodes, arthropods, and several other invertebrates. It serves to protect the animal from physical damage and helps it retain moisture. The cuticle is not a distinction between protostome and deuterostomes. So the correct answer is C) presence of a cuticle.
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7) Which statement is true?
A) Kinases phosphorylate proteins.
B) Kinases can only activate a protein or enzyme.
C) Kinases can only deactivate a protein or enzyme.
D) Kinases can activate some proteins and enzymes while deactivating other proteins and enzymes.
E) Both A and B are correct.
F) Both A and D are correct.
8) Which statement is true?
A) Kinases phosphorylate proteins.
B) Kinases can only activate a protein or enzyme.
C) Kinases can only deactivate a protein or enzyme.
D) Kinases can activate some proteins and enzymes while deactivating other proteins and enzymes.
E) Both A and B are correct.
F) Both A and D are correct.
Answers
Statement that is true is Kinases phosphorylate proteins, while the statement that is true regarding Kinases is that Kinases can activate some proteins and enzymes while deactivating other proteins and enzymes.
Kinases are enzymes that phosphorylate various molecules such as proteins, nucleotides, and lipids. Kinases are a type of enzyme that catalyzes the transfer of a phosphate group from an ATP molecule to another molecule, resulting in the phosphorylation of that molecule.
The addition of a phosphate group to a protein can activate or deactivate it, depending on the protein, while the phosphorylation of nucleotides and lipids serves other purposes.Statement that is true
Kinases phosphorylate proteins. It is true because Kinases are enzymes that catalyze the transfer of a phosphate group to another molecule, resulting in the phosphorylation of that molecule. The addition of a phosphate group to a protein can activate or deactivate it, depending on the protein.
Kinases are a type of enzyme that catalyzes the transfer of a phosphate group from an ATP molecule to another molecule.Statement that is true regarding KinasesKinases can activate some proteins and enzymes while deactivating other proteins and enzymes.
This is true because the addition of a phosphate group to a protein can activate or deactivate it, depending on the protein. The transfer of a phosphate group from an ATP molecule to another molecule is catalyzed by Kinases.
Therefore, Kinases can activate some proteins and enzymes while deactivating other proteins and enzymes as they are enzymes that phosphorylate various molecules such as proteins, nucleotides, and lipids.
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Constructing a phylogenetic tree using parsimony requires you to:
choose the tree with the fewest branching points.
choose the tree that represents the fewest evolutionary changes, either in DNA sequences or morphology.
build the phylogeny using only the fossil record, as this provides the simplest explanation for evolution.
choose the tree that assumes all evolutionary changes are equally probable.
choose the tree in which the branching points are based on as many shared derived characters as possible.
Answers
Constructing a phylogenetic tree using parsimony requires you to choose the tree that represents the fewest evolutionary changes, either in DNA sequences or morphology.
Phylogenetic trees using parsimony
Parsimony assumes that the simplest explanation is the most likely, so the tree with the fewest evolutionary changes is preferred.
This approach aims to minimize the number of evolutionary events required to explain the observed data. It does not assume equal probabilities for all evolutionary changes but rather focuses on minimizing the total number of changes by selecting the tree with the fewest branching points and utilizing shared derived characters as much as possible.
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Seek out information on what types of roles our gut flora or gut microbes play regarding our health and well-being.
Answers
Our gut flora or gut microbes play an important role in our overall health and well-being. These microbes, which are found in our digestive system, help break down the food we eat and support the functioning of our immune system, among other things. In this answer, I will discuss the roles that gut flora plays in our health in more detail.
One of the key roles of gut flora is to support our digestion. These microbes help break down complex carbohydrates, proteins, and fats into smaller, more easily digestible molecules. They also produce enzymes that we need to digest certain types of food, such as lactose in dairy products.
Another important function of gut flora is to support our immune system. These microbes help train our immune system to recognize and respond to harmful pathogens. They also produce molecules that help regulate inflammation in the body, which is important for maintaining good health.
Gut flora has also been linked to a number of chronic diseases, including obesity, type 2 diabetes, and heart disease. Research has shown that imbalances in gut flora can lead to inflammation, insulin resistance, and other metabolic problems that can contribute to these conditions.
In addition to these health benefits, gut flora has also been shown to play a role in our mental health. Research has linked imbalances in gut flora to a number of mental health disorders, including depression and anxiety.
Overall, gut flora plays a critical role in our health and well-being. By supporting our digestion, immune system, and mental health, these microbes help keep us healthy and strong. If you want to maintain good gut health, it is important to eat a healthy diet that is rich in fiber and fermented foods, avoid unnecessary antibiotics, and seek out other ways to support your gut health, such as probiotic supplements.
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Hello,
I was hoping someone might be able to help me out with this for
funeral services or even another class? I am highly confused and
have tried multiple times. Thanks.
Theresa
31. Assuming a funeral director has signed a variable workweek agreement form, and is salaried at $500 per week, what is the gross pay (pay prior to taxes) for the following week? 56 hours the first w
Answers
To calculate the gross pay of the funeral director, we can use the formula given below;
Gross pay = Base salary + Overtime pay
Where,Base salary = $500
Overtime pay = (Total number of hours worked in excess of 40 hours) x (Hourly rate of pay x 1.5)
Hourly rate of pay = Base salary ÷ 40
Assuming a funeral director has signed a variable workweek agreement form, and is salaried at $500 per week, the hourly rate of pay would be;$500 ÷ 40 = $12.50 per hour.The funeral director has worked for 56 hours the first week, we can find the overtime pay for this week using the formula given above.
Overtime pay = (Total number of hours worked in excess of 40 hours) x (Hourly rate of pay x 1.5)= (56 - 40) x ($12.50 x 1.5)= 16 x $18.75= $300Gross pay = Base salary + Overtime pay= $500 + $300= $800Therefore, the gross pay for the funeral director for the given week is $800 prior to taxes.
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1. Which of the following characteristics is not shared by ALL cells?
A) Have internal, membrane-bound compartments.
B) Capable of transcription and translation.
C) Capable of respiration.
D) Utilize enzyme driven reactions.
2. Which of the following structures contains the thylakoid membranes?
A) The Nucleus.
B) Lysosomes
C) Mitochondria
D) Chloroplasts
Answers
1. The characteristic that is not shared by ALL cells is "Have internal, membrane-bound compartments." .2. The structure that contains the thylakoid membranes is Chloroplasts
Explanation:Cells come in all shapes and sizes, but they all have a few things in common. For example, they all have a cell membrane that separates them from the surrounding environment. They also have cytoplasm, which is a fluid-like substance that fills the inside of the cell.
Additionally, they all have DNA, the genetic material that controls the cell's activities. While all cells share these characteristics, they do not all have internal, membrane-bound compartments. For example, bacteria are a type of cell that lacks these compartments.
2. The structure that contains the thylakoid membranes is Chloroplasts. Explanation: Chloroplasts are organelles found in plant cells that are responsible for photosynthesis. They contain a complex system of membranes called thylakoids, which are arranged in stacks called grana.
These membranes are the site of photosynthesis, the process by which plants convert sunlight into energy. The other options in the question do not contain thylakoid membranes.
The nucleus contains DNA, lysosomes contain enzymes that break down waste products, and mitochondria are the site of cellular respiration.
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17) Whan Mendel cross-bred paternal pes plants with yellow seeds with maternal pes plants with gem seeds and the dominant that was yellow send, he found
Q) all of the Fs progeny had green soods
R)all of the Fi progeny had yellow seeds
s) half of the F progeny had grem seeds
T) 75% of the F progeny had yellow seeds
18) question 7, the maternal pes places with the green sent were
U) recessive homotypes
V) dominant homerygtes
X) a mixture of recessive and dominam bomo
19) If a test cross is conducted using a purple flowered pes plane of unknown genitype herd with a white flowered pea plant, and the progeny all have purple flowers, that means
A) he unknown genotype is PP (homozygous dominant)
B) he unknown gmotype is pp (honorygnus recessive)
C) he unknown genotype is Pp heterozygote
D) the unknown genotype in node of the above
20) If we have epistasis in mice, and we have grees for black far (8), brown for (0) colored fur (C) and white fur (c), what will be the color of a mouse with the protype
A) back
B) brown
C) white
D) mixed white and black
21) If the disease phenylketonuria follows the same rules as Mendel's pes plast, and la baly a bor bor so 2 parts that cach are benirygous for the bad (dacase) gme the chances that the haby
sonnal (not died)
A) 25%
B) 75%
C) 100%
D) 0%
22) If the Hung's des gene is dont and your father has me bad mother har no bod genes, your chances of getting i
A)100%
B) 0%
C) 25%
D) 50 %
Answers
When Mendel cross-bred paternal pea plants with yellow seeds and maternal pea plants with green seeds, he found that all of the F1 progeny had green seeds.
The maternal pea plants with green seeds were likely recessive homozygotes.
If a test cross is conducted using a purple-flowered pea plant of unknown genotype and all the progeny have purple flowers, it suggests that the unknown genotype is Pp heterozygous.
If epistasis is present in mice with genes for black fur, brown fur, colored fur, and white fur, the color of a mouse with the genotype would be white.
If the disease phenylketonuria follows Mendelian rules and both parents are heterozygous carriers, the chances of the baby being unaffected (not having the disease) are 25%.
If the Huntington's disease gene is dominant and the father has the gene while the mother has no bad genes, the chances of the individual getting the gene are 50%.
When Mendel cross-bred paternal pea plants with yellow seeds and maternal pea plants with green seeds, all of the F1 progeny had green seeds. This suggests that the green seed trait is dominant over the yellow seed trait.
If the maternal pea plants with green seeds produced green-seeded offspring when crossed with another plant, it indicates that the maternal plants were likely recessive homozygotes for the green seed trait.
In a test cross between a purple-flowered pea plant of unknown genotype and a white-flowered pea plant, if all the progeny have purple flowers, it suggests that the unknown genotype is Pp heterozygous. The dominant purple flower trait is expressed in the presence of either homozygous dominant (PP) or heterozygous (Pp) genotypes.
Epistasis refers to the interaction between genes where one gene masks or modifies the expression of another gene. If in mice the genes for black fur (B), brown fur (b), colored fur (C), and white fur (c) are involved in an epistatic interaction, and the genotype is "ccBB," the mouse will have white fur, as the "c" gene masks the expression of fur color genes.
If phenylketonuria follows Mendelian rules, and both parents are carriers (heterozygous for the bad gene), the chances of the baby being unaffected (not having the disease) are 25%. This is because there is a 25% chance for the baby to inherit two normal alleles (non-carrier).
If the Huntington's disease gene is dominant, and the father has the gene while the mother has no bad genes (homozygous normal), the chances of the individual getting the gene are 50%. In this case, the individual has a 50% chance of inheriting the dominant gene from the father and a 50% chance of inheriting the normal gene from the mother.
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Design a messenger RNA transcript with the necessary prokaryotic
control sites that codes for the octapeptide
Lys-Pro-Ala-Gly-Thr-Glu-Asn-Ser.
Answers
A designed mRNA transcript for the octapeptide Lys-Pro-Ala-Gly-Thr-Glu-Asn-Ser require a promoter sequence, a Shine-Dalgarno sequence, a start codon, a coding region for the peptide, and a stop codon.
To design an mRNA transcript for the octapeptide Lys-Pro-Ala-Gly-Thr-Glu-Asn-Ser in a prokaryotic system, several key elements need to be included.
First, a promoter sequence is necessary to initiate transcription. The promoter sequence is recognized by RNA polymerase and helps to position it correctly on the DNA template.
Next, a Shine-Dalgarno sequence is required. This sequence, typically located upstream of the start codon, interacts with the ribosome and facilitates translation initiation.
Following the Shine-Dalgarno sequence, a start codon, such as AUG, is needed to indicate the beginning of the coding region for the octapeptide.
The coding region itself will consist of the corresponding nucleotide sequence for the octapeptide Lys-Pro-Ala-Gly-Thr-Glu-Asn-Ser. Each amino acid is encoded by a three-nucleotide codon.
Finally, a stop codon, such as UAA, UAG, or UGA, is required to signal the termination of translation.
By incorporating these elements into the mRNA transcript, the prokaryotic system will be able to transcribe and translate the genetic information to produce the desired octapeptide.
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Insects are also known as the _________. They have ____ walking
legs and ____ pairs of wings. A. Myriapoda, 10, 0 B. Arachnida, 4,
2 C. Hexapoda, 3, 2 D. Uniramia, 2, 0
Answers
Insects are also known as Hexapoda. They have 3 pairs of walking legs and 2 pairs of wings.
The correct answer is option C. Insects are commonly referred to as Hexapoda, which means "six legs." This is because they possess six legs, which are divided into three pairs. The legs are used for walking, running, jumping, and various other forms of locomotion.
Additionally, insects have two pairs of wings, which are typically used for flight. The wings are often thin, membranous structures that allow insects to maneuver and travel through the air. However, it is important to note that not all insects have functional wings. Some insect species, such as ants or fleas, may have reduced or absent wings.
Options A, B, and D are incorrect. Myriapoda (option A) refers to a different group of arthropods that includes centipedes and millipedes, which have more than six legs. Arachnida (option B) includes spiders, scorpions, and ticks, which have eight legs and do not possess wings. Uniramia (option D) is not a commonly used term for insects and does not accurately describe their leg or wing structure.
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How can the antiviral state be propagated in the absence of immune cells?
1) Type I IFNs are produced by infected epithelial cells; this induces anti-viral biochemical changes in the same cell and adjacent cells.
2) Viral nucleic acids are shuttled between cells through connexins (proteins that connect cells).
3) Extracellular TLRs expressed on epithelial cells directly recognize viral capsid proteins.
4) Dendritic cells produce type I IFNs
Answers
The antiviral state can be propagated in the absence of immune cells by Type I IFNs that are produced by infected epithelial cells which induces anti-viral biochemical changes in the same cell and adjacent cells.
The correct option is-1
Type I IFNs are produced by infected epithelial cells; this induces anti-viral biochemical changes in the same cell and adjacent cells.How can the antiviral state be propagated in the absence of immune cells?The immune system is composed of a network of cells, tissues, and organs that work together to identify and destroy foreign invaders such as viruses, bacteria, fungi, and parasites.
When an immune response is triggered, immune cells detect viral components such as nucleic acids or viral capsid proteins, and they respond by producing antiviral molecules such as interferons (IFNs) to limit viral replication and spread.However, there are situations when the immune system is not fully functional or absent. In these cases, the antiviral state can still be propagated through alternative pathways. For example, infected cells can produce Type I IFNs that induce antiviral biochemical changes in the same cell and neighboring cells. This leads to the production of antiviral proteins that inhibit viral replication and trigger cell death (apoptosis) to prevent the virus from spreading further. Therefore, the antiviral state can be propagated in the absence of immune cells by local production of antiviral molecules such as Type I IFNs.
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Module 4 (Chapter 14) Case Study A 51-year old healthy male received a minor abrasion at a local physical fitness center that resulted in a raised hard lesion on his thigh. He visited his primary care physician, who drained the lesion and prescribed an oral first-generation cephalosporin commonly used for skin infections and lesions. The patient was asked to drain the lesion daily and wipe the affected area with disposable cindamycin medicated pads. He was instructed to keep the infected area covered with a clean dry bandage and to no participate in any athletic activity unless he could keep the wound dry and covered. He was also told to practice good personal hygiene after cleaning the wound and to avoid shared items. A culture was performed, and catalsse, coagulase gram cocci were isolated Antimicrobial susceptibility testing showed the isolate was resistant to penicilin, oxacillin, and erythromycin and sensitive to dindamycin. Further testing by a double disk diffusion showed the isolate was positive for inducible clindamycin resistance. Word Documen: 120 Saved Home insert Layout TU B Lapr 19 Casqu 1. Indicate patient history facts that influence your diagnosis References *** p Share 28 Which tests were conducted on the bacteria cultured from the lesion? Descobe how each test mentioned in the case study fanctions and what it tells you 2b. Of the three species of Staphylococcus, we discussed in the lecture, which is consistent with the observed results of these tests?
2e Indicate the test results you would expect for the other two species we discussed in lecture 3. Indicate two additional tests or bacterial characteristics you could conduct or observe that would be consistent with your etiological diagnosis agent. Tell me the result you would expect to se
4. Indicate the mode of action of the antihistics used to treat this parent. First generation cephalosporin b. Clindamycin 5. What is the double disk diffusion test? Discuss how this text works and what is going on with the
Answers
The patient's history includes receiving a minor abrasion at a physical fitness center, resulting in a raised hard lesion on his thigh. The culture performed on the lesion identified catalase, coagulase gram cocci, which was found to be resistant to penicillin, oxacillin, and erythromycin, but sensitive to clindamycin.
The isolate also showed inducible clindamycin resistance. Two tests mentioned in the case study were the culture and antimicrobial susceptibility testing. Staphylococcus species are consistent with the observed results, and the expected results for the other two species (Staphylococcus aureus and Staphylococcus epidermidis) were not mentioned. Additional tests or observations that could support the etiological diagnosis include Gram staining and identification of specific virulence factors. The mode of action of the antibiotics used, first-generation cephalosporin and clindamycin, should be mentioned. The double disk diffusion test is a method used to determine inducible clindamycin resistance by placing two antibiotic disks on the same agar plate and observing the inhibition zones around them.
1. The patient's history of a minor abrasion and the raised hard lesion on his thigh are important facts that influence the diagnosis. These symptoms suggest a localized skin infection or lesion.
2. The tests conducted on the bacteria cultured from the lesion include culture and antimicrobial susceptibility testing. Culture helps identify the type of bacteria present, while antimicrobial susceptibility testing determines the effectiveness of various antibiotics against the isolated bacteria.
2b. Staphylococcus aureus is consistent with the observed results in the case study. However, the results for Staphylococcus epidermidis and the other mentioned Staphylococcus species are not provided.
3. Two additional tests or bacterial characteristics that could be conducted or observed to support the etiological diagnosis include Gram staining and identification of specific virulence factors. Gram staining can help determine the bacterial morphology and arrangement, while identifying virulence factors can provide insights into the pathogenicity of the bacteria.
4. The mode of action of the antibiotics used are as follows:
a) First-generation cephalosporin: It inhibits bacterial cell wall synthesis, leading to cell lysis and death.
b) Clindamycin: It inhibits bacterial protein synthesis by binding to the 50S subunit of the bacterial ribosome, thereby preventing peptide bond formation.
5. The double disk diffusion test is a method used to detect inducible clindamycin resistance in bacteria. It involves placing clindamycin and erythromycin disks on the same agar plate and observing the inhibition zones around the disks. If the zone of inhibition around the erythromycin disk is larger than that around the clindamycin disk, it indicates the presence of inducible clindamycin resistance, where the bacteria can express resistance to clindamycin under certain conditions. This resistance mechanism can be due to the presence of the erm gene, which can be induced by exposure to erythromycin.
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1. The number of phosphate units in a phospholipid is a. 1 b. 2 c. 3 2. The number of ester linkages in a phospholipid is a. 1 b. 2 c. 3 d. 4 d. 4 3. The inner bilayer of the nuclear envelope is continuous with a. SER b. RER c. cell membrane 4. The lumen and the cytosol are separated by the a. SER b. RER c. ER 5. When a sugar attaches to a protein gets the name a. glycoprotein b. lipoprotein c. glycan 6. A vesicle released from the Golgi a. has double membrane b. can be considered an organelle d. is a lipoprotein c. is a glycoprotein d. none d. nuclear membrane d. sweet protein
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. The number of phosphate units in a phospholipid is b
. 2. Phospholipids consist of a glycerol molecule, two fatty acid chains, and a phosphate group.
2. The number of ester linkages in a phospholipid is d.
4. Esters are organic molecules that have the functional group -COO- with two alkyl or aryl groups attached.
3. The inner bilayer of the nuclear envelope is continuous with the b. RER (Rough Endoplasmic Reticulum).
4. The lumen and the cytosol are separated by the a. SER (Smooth Endoplasmic Reticulum).
5. When a sugar attaches to a protein gets the name a. glycoprotein. Glycoproteins are proteins that contain oligosaccharide chains (glycans) covalently attached to polypeptide side-chains.
6. A vesicle released from the Golgi can be considered an organelle. The Golgi Apparatus consists of flattened stacks of membranes or cisternae, and vesicles that transport and modify proteins and lipids.
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Approximately what percentage of species on the ESA have changed status improved, declined or gone extinct) since being added to the list? O 0.002% O 0.12 % O 2.0% O 12% O 20%
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Approximately 12% of the species on the ESA have changed status improved, declined or gone extinct since being added to the list.
The ESA, or Endangered Species Act, was established in 1973 with the goal of protecting and recovering imperiled species and the ecosystems they inhabit. When a species is added to the ESA list, it is given a status of either endangered or threatened.Over time, the status of many species on the ESA list has changed. Some have improved, some have declined, and unfortunately, some have gone extinct. Approximately 12% of the species on the ESA have changed status since being added to the list.
Since being added to the list, approximately 12% of the species on the ESA have changed status improved, declined, or gone extinct. This means that a little over one in ten species on the list have seen a change in their status. These changes can occur due to a variety of reasons, such as successful conservation efforts, habitat loss, invasive species, and climate change. It is important to continue monitoring the status of species on the ESA list to ensure their continued protection and recovery.
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43) Reindeer herders in Norway introduce a new population of reindeer from Russia to breed with their existing herd in order to increase the genetic diversity. This is an example of _______
a. Genetic drift
b. Mutations
c. Non-random mating
d. Gene flow
44) Natural selection states that organisms are best suited for their environment if they __________________ and __________________ .
a. Survive and Thrive
b. Survive and Reproduce
c. Survive and Adapt
d. Survive and Predate
45) Which of the following is not an example of prezygotic reproductive isolation?
a. Species breeding at different times of the year.
b. Morphological differences preventing successful mating attempt.
c. Physical barriers prevent to species from encountering one
another to mate.
d. A horse and a donkey produce a sterile mule.
e. The sperm of a sea star is unable to fertilize the egg of a sea
urchin.
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43) Introducing a new population of reindeer to increase genetic diversity is an example of gene flow.44) Natural selection states that organisms best suited for their environment are those that can survive and reproduce. 45) The production of a sterile mule by a horse and a donkey is not an example of prezygotic reproductive isolation.
44. Reindeer herders in Norway introduce a new population of reindeer from Russia to breed with their existing herd in order to increase the genetic diversity. This is an example of Gene flow. The correct answer is d. Introducing a new population of reindeer from Russia to breed with the existing herd is an example of gene flow. Gene flow occurs when genetic material is transferred between different populations, increasing the genetic diversity within a population.
44. Natural selection states that organisms are best suited for their environment if they Survive and Reproduce. The correct answer is b. Natural selection states that organisms that can survive in their environment and successfully reproduce are more likely to pass on their advantageous traits to the next generation. This leads to the gradual adaptation of species to their specific environment over time.
45. A horse and a donkey produce a sterile muleis not an example of prezygotic reproductive isolation . The correct answer is d. Prezygotic reproductive isolation refers to mechanisms that prevent the formation of a viable zygote (fertilized egg) between different species. In this case, a horse and a donkey can mate and produce a hybrid offspring (mule), but the mule is sterile and cannot produce viable offspring. The other options (a, b, c, e) describe different types of prezygotic reproductive isolation mechanisms.
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Animals and plants both use signaling chemicals. Give two examples for each animal and plant and the effect of each mechanism described. Be sure to include what the signal trigger is, what the signaling chemical is, where it travels to and from, and the effect at its final destination.
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Animals and plants both utilize signaling chemicals to communicate and coordinate various physiological processes. Here are two examples of signaling mechanisms and their effects in both animals and plants:
Animals:
a. Nervous System Signaling:
Signal Trigger: Stimulus such as touch, light, or temperature.
Signaling Chemical: Neurotransmitters (e.g., acetylcholine, serotonin, dopamine).
Travel: Signals travel along neurons.
Effect: Neurotransmitters transmit signals across synapses, allowing for rapid communication between neurons. This signaling mechanism regulates processes such as muscle contraction, sensory perception, and cognition.
b. Hormonal Signaling:
Signal Trigger: Internal or external stimuli, such as stress, environmental cues, or developmental changes.
Signaling Chemical: Hormones (e.g., adrenaline, insulin, estrogen).
Travel: Hormones are released into the bloodstream and travel throughout the body.
Effect: Hormones act as chemical messengers, binding to specific receptors on target cells. This signaling mechanism regulates processes like growth, reproduction, metabolism, and behavior.
Plants:
a. Phototropism:
Signal Trigger: Light, particularly blue light.
Signaling Chemical: Auxin.
Travel: Auxin is synthesized in the plant's shoot tip and travels downward.
Effect: Auxin promotes elongation of cells on the darker side of the plant, causing it to bend towards the light source. This mechanism allows plants to optimize photosynthesis by orienting their leaves toward sunlight.
b. Defense Signaling:
Signal Trigger: Presence of pathogens or herbivores.
Signaling Chemical: Jasmonic acid (JA) and salicylic acid (SA).
Travel: JA and SA are produced in response to attack and travel within the plant.
Effect: JA and SA activate defense responses, such as the production of toxic compounds, strengthening of cell walls, and activation of defense-related genes. These signaling mechanisms help plants defend against pathogens and herbivores.
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What kind of inheritance does the following trait
show?
Genotype Male Female
AA Mahogany Mahogany
Aa Mahogany Red aa Red Red
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The trait described shows an example of autosomal inheritance, specifically a co-dominant inheritance pattern.
The given trait has two possible phenotypes: "Mahogany" and "Red." In the genotype, "A" represents the allele for Mahogany, and "a" represents the allele for Red. Both males and females can have the Mahogany phenotype if they carry at least one Mahogany allele (AA or Aa), while individuals with two Red alleles (aa) will have the Red phenotype.
This pattern of inheritance is known as co-dominance, where both alleles are expressed equally and simultaneously in the heterozygous state (Aa). Neither allele is dominant over the other, resulting in a distinct phenotype for each allele.
Furthermore, this trait follows autosomal inheritance, meaning it is located on an autosome (a non-sex chromosome) and can be inherited by both males and females. The inheritance of this trait is not influenced by the sex of the individual, and both males and females can carry and pass on the Mahogany or Red alleles to their offspring.
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Review Questions 1. ______ is the net movement of water through a selectively permeable membrane from an area of low solute concentration to an area of high solute concentration. 2. a. _______ Did the color change in the beaker, the dialysis bag, or both in Procedure 6.17 b. Explain why 3. a. ______ For which dialysis bags in Procedure 6.2 did water move across the membrane? b. Explain how you determined this based on your results.
4. a. ______ What salt solution (0%, 9%, or 5%) is closest to an isotonic solution to the potato cells in Procedure 6.5? b. Explain how you determined this based on your results. 5. _______ Would you expect a red blood cell to swell, shrink, or remain the same if placed into distilled water? 6. Explain why hypotonic solutions affect plant and animal cells differently. 7. Explain how active transport is different than passive transport. 8. Phenolphthalein is a pH indicator that turns red in basic solutions. You set up an experiment where you place water and phenolphthalein into a dialysis bag. After closing the bag and rinsing it in distilled water, you place the dialysis bag into a beaker filled with sodium hydroxide (a basic/alkaline solution). You observe at the beginning of the experiment both the dialysis bag and the solution in the beaker are clear. After 30 minutes you observe that the contents of the dialysis bag have turned pink but the solution in the beaker has remained clear. What can you conclude in regards to the movements of phenolphthalein and sodium hydroxide?
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Osmosis is the net movement of water through a selectively permeable membrane from an area of low solute concentration to an area of high solute concentration.
In Procedure 6.17, where did the color change occur and why?
In Procedure 6.17, the color change can occur in the beaker, the dialysis bag, or both. The color change indicates the movement of solute particles across the membrane.
If the color changes in the beaker, it suggests that the solute molecules have diffused out of the dialysis bag into the surrounding solution.
If the color changes in the dialysis bag, it indicates that the solute molecules have passed through the membrane and entered the bag.
The occurrence of color change in both the beaker and the dialysis bag suggests that there is movement of solute in both directions.
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Pre-mRNA from eukaryotes (prior to processing) contains the following elements except: A. a 5' UTR. B. a ribosome binding site. C. a transcription factor binding site. D. introns. E. a polyadenylation signal.
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Pre-mRNA from eukaryotes (prior to processing) contains the following elements except a ribosome binding site. So, option B is accurate.
Pre-mRNA from eukaryotes, prior to processing, contains several elements involved in gene expression and post-transcriptional modification. These elements include a 5' UTR (untranslated region), which is a non-coding region upstream of the coding sequence, providing regulatory and structural functions. It also contains a transcription factor binding site, where transcription factors bind to regulate gene expression. Pre-mRNA contains introns, non-coding sequences that are removed during RNA splicing to generate mature mRNA. Additionally, it includes a polyadenylation signal, which is a specific sequence that marks the end of the transcript and is essential for the addition of a poly(A) tail during mRNA processing. However, a ribosome binding site, also known as a Shine-Dalgarno sequence, is a feature found in prokaryotic mRNA but not in eukaryotic pre-mRNA.
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where is the b cell located when it is activated by cd4 t cell?
a. primary follicle
b. germinal center (dark)
c. secondary follicle
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When a B-cell is activated by CD4 T-cells, it is located in the germinal center (dark). A germinal center is a site in which activated B cells undergo rapid proliferation and differentiation into memory B.
Cells or plasma cells that produce high-affinity antibodies to the targeted antigen.More than 100 B-cell clones form as a result of the activation of B cells by CD4 T cells. A germinal center is the location where this procedure occurs. It is a site within a secondary follicle in the lymph node cortex, distinguished by the existence of mitotically active B cells that have a large quantity of cell divisions and mutations, as well as a dark appearance under the microscope.
The germinal center of a secondary follicle is where B cells undergo selection, differentiation, and proliferation to generate high-affinity antibodies to a particular antigen. T cells work with B cells at this location to improve antibody production.
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1. Which ONE of the following statements is INCORRECT?
a. The oxidative decarboxylation of pyruvate and formation of acetyl-CoA is catalysed by the pyruvate dehydrogenase (PDH) complex
b. Lipoic acid is non-covalently bound to E2, while NAD+ and FAD are both covalently bound to E3 in the PDH complex
c. Thiamine pyrophosphate (TPP) is a derivative of vitamin B1
d, ydroxyethyl-TPP is one intermediate formed in the PDH complex reaction
With respect to the above question, explain why the statement you selected is incorrect (no more than two sentences).
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The statement that is incorrect is: Lipoic acid is non-covalently bound to E2, while NAD+ and FAD are both covalently bound to E3 in the PDH complex. So, Option B is the correct alternative.
The incorrect statement is incorrect because lipoic acid is actually covalently bound to E2, while NAD+ and FAD are non-covalently bound to E3 in the pyruvate dehydrogenase (PDH) complex. Lipoic acid plays a crucial role in the PDH complex as it acts as a coenzyme, forming a covalent linkage with E2 to transfer acetyl groups during the reaction. On the other hand, NAD+ and FAD function as electron carriers in the complex but are not covalently bound to the enzymes.
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Explain in detail how circulating antibodies are produced in the body.
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Circulating antibodies, also known as immunoglobulins, are produced by specialized cells of the immune system called B lymphocytes or B cells.
The process of antibody production, known as antibody synthesis or humoral immune response, involves several stages: Antigen Recognition: B cells are capable of recognizing specific antigens, which are molecules or components found on the surface of pathogens such as bacteria, viruses, or other foreign substances. Each B cell has a unique receptor on its surface that can bind to a specific antigen. Antigen Presentation and Activation: When a B cell encounters its specific antigen, the antigen binds to the B cell receptor, triggering internal signaling processes. The B cell engulfs the antigen, processes it, and displays fragments of the antigen on its surface using a protein called major histocompatibility complex class II (MHC II). T Cell Interaction: The antigen-presenting B cell interacts with helper T cells, which recognize the displayed antigen fragments. This interaction stimulates the helper T cells to release signaling molecules called cytokines, which provide additional activation signals to the B cell. B Cell Activation and Clonal Expansion: The interaction with helper T cells, along with the cytokine signals, activates the B cell.
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168 Anatomy and Physiology I MJB01 302 (Summer 2022) Microscopically, muscle fibers contain parallel myofibrils, banded by repeating units. Each unit is called a/an Select one: a. sarcomere b. sarcopl
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Muscle fibers are microscopically characterized by parallel myofibrils, which are banded by repeating units. Each unit is referred to as a sarcomere.
A sarcomere is a structural and functional unit of a myofibril, and it is responsible for the contraction of the muscle fiber when it receives a signal from the nervous system. The sarcomere comprises thick filaments of myosin and thin filaments of actin, which are arranged in a very specific pattern. It is the arrangement of these filaments that provides the striated appearance of skeletal muscle.
The sarcomere contains two Z-discs, which define its boundaries, and a M-line that runs through the center of the sarcomere. When the muscle fiber is stimulated, the actin and myosin filaments slide over each other, causing the sarcomere to shorten and generating the force of contraction. In summary, a sarcomere is a repeating unit of a myofibril, and it is the basic functional unit of skeletal muscle.
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List the three stages of telomerase activity and briefly describe each one, along with the two other enzymes involved in the process of telomerisation.
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The three stages of telomerase activity are recruitment, extension, and translocation. The two enzymes involved in the process of telomerization are telomerase reverse transcriptase (TERT) and telomerase .
Recruitment: Telomerase is recruited to the telomeres, which are the protective caps at the ends of chromosomes. This step involves the binding of telomerase to the telomeric DNA sequence.
Once recruited, telomerase adds additional telomeric repeats to the chromosome ends using its catalytic component called telomerase reverse transcriptase (TERT). The TERT enzyme extends the telomeric DNA strand by adding new nucleotides in a reverse transcriptase-like manner.
Translocation: After extension, telomerase translocates to a new position along the telomere to repeat the process of adding telomeric repeats. This translocation allows telomerase to continue lengthening the telomeres.
Apart from telomerase, two other enzymes are involved in the process of telomerization:
Telomerase RNA component (TERC): This non-coding RNA molecule provides the template for the synthesis of the telomeric DNA repeats during the extension stage.
DNA polymerase: After telomerase adds telomeric repeats, DNA polymerase synthesizes the complementary strand to complete the replication of the telomere.
In summary, telomerase activity involves recruitment to the telomeres, extension of telomeric repeats using TERT and TERC, and translocation for further lengthening. The process also requires the involvement of DNA polymerase to complete telomere replication.
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