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All questions of Reproductive System and Embryogenesis (BIO) for MCAT Exam
Oogenesis and spermatogenesis describe the process of meiosis in females and males respectively. What statement about the two processes is true?- a)Oogenesis forms two polar bodies while spermatogenesis only forms one
- b)Oogenesis produces a larger gamete than spermatogenesis
- c)Both produce the same number of viable gametes
- d)Both are halted in an intermediate step until puberty
Correct answer is option 'B'. Can you explain this answer?
Oogenesis and spermatogenesis describe the process of meiosis in females and males respectively. What statement about the two processes is true?
a)
Oogenesis forms two polar bodies while spermatogenesis only forms one
b)
Oogenesis produces a larger gamete than spermatogenesis
c)
Both produce the same number of viable gametes
d)
Both are halted in an intermediate step until puberty
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Orion Classes answered |
Oogenesis and spermatogenesis are the processes of meiosis that occur in females and males, respectively, leading to the formation of gametes (eggs and sperm). There are several differences between these two processes.
First, oogenesis forms two polar bodies, while spermatogenesis only forms one. During oogenesis, the primary oocyte undergoes two rounds of meiosis, resulting in the formation of one mature egg and two polar bodies. The polar bodies do not have the potential to be fertilized and serve mainly to discard excess genetic material. In spermatogenesis, each primary spermatocyte undergoes meiosis and forms four functional sperm cells.
Second, oogenesis produces a larger gamete than spermatogenesis. The female gamete, or egg, is much larger and contains abundant cytoplasm to support the developing embryo. In contrast, the male gamete, or sperm, is relatively small and streamlined to facilitate motility and reach the egg for fertilization.
Third, both processes do not produce the same number of viable gametes. Oogenesis results in the formation of one viable egg cell per primary oocyte, whereas spermatogenesis produces four viable sperm cells per primary spermatocyte.
Lastly, both oogenesis and spermatogenesis are arrested at an intermediate step until puberty. In females, oogenesis is arrested at prophase I until puberty, while in males, spermatogenesis is arrested at the primary spermatocyte stage. The onset of puberty triggers the continuation of these processes.
Oogenesis, the process that creates female gametes, is halted at prophase I until puberty. Which of the following describes the DNA content of a female’s gametes during her childhood?- a)46 chromosomes, 46 chromatids
- b)46 chromosomes, 92 chromatids
- c)23 chromosomes, 46 chromatids
- d)23 chromosomes, 23 chromatids
Correct answer is option 'B'. Can you explain this answer?
Oogenesis, the process that creates female gametes, is halted at prophase I until puberty. Which of the following describes the DNA content of a female’s gametes during her childhood?
a)
46 chromosomes, 46 chromatids
b)
46 chromosomes, 92 chromatids
c)
23 chromosomes, 46 chromatids
d)
23 chromosomes, 23 chromatids
|
Mason Brooks answered |
Understanding Oogenesis and DNA Content
Oogenesis is the process that produces female gametes, or eggs, in the ovaries. This process begins before birth but is arrested at prophase I of meiosis until puberty.
Key Stages of Oogenesis
- Fetal Development: In females, oogenesis begins during fetal development. Oogonia, the precursor cells, undergo mitosis to increase in number.
- Primary Oocytes: These oogonia develop into primary oocytes and enter prophase I of meiosis. At this stage, they are surrounded by a layer of granulosa cells and are referred to as primordial follicles.
Chromosome and Chromatid Content
- Diploid State: The primary oocytes are diploid, meaning they contain 46 chromosomes. Each chromosome is replicated, resulting in two sister chromatids per chromosome.
- Total Chromatid Count: Since each of the 46 chromosomes has been replicated, the total number of chromatids is 92 (46 chromosomes x 2 chromatids each).
Conclusion: Childhood DNA Content
During childhood, the female's primary oocytes remain in the arrested state with:
- 46 chromosomes (diploid status)
- 92 chromatids (due to replication)
Thus, the correct description of the DNA content of a female's gametes during childhood is:
Option B: 46 chromosomes, 92 chromatids
Oogenesis is the process that produces female gametes, or eggs, in the ovaries. This process begins before birth but is arrested at prophase I of meiosis until puberty.
Key Stages of Oogenesis
- Fetal Development: In females, oogenesis begins during fetal development. Oogonia, the precursor cells, undergo mitosis to increase in number.
- Primary Oocytes: These oogonia develop into primary oocytes and enter prophase I of meiosis. At this stage, they are surrounded by a layer of granulosa cells and are referred to as primordial follicles.
Chromosome and Chromatid Content
- Diploid State: The primary oocytes are diploid, meaning they contain 46 chromosomes. Each chromosome is replicated, resulting in two sister chromatids per chromosome.
- Total Chromatid Count: Since each of the 46 chromosomes has been replicated, the total number of chromatids is 92 (46 chromosomes x 2 chromatids each).
Conclusion: Childhood DNA Content
During childhood, the female's primary oocytes remain in the arrested state with:
- 46 chromosomes (diploid status)
- 92 chromatids (due to replication)
Thus, the correct description of the DNA content of a female's gametes during childhood is:
Option B: 46 chromosomes, 92 chromatids
What would be a direct consequence of a single point mutation that modifies the proteins inside the acrosome?- a)Inability to dissolve zona pellucida's glycocalyx
- b)Failure to complete spermatogenesis
- c)Impairment of sperm motility
- d)Increased polyspermy
Correct answer is option 'A'. Can you explain this answer?
What would be a direct consequence of a single point mutation that modifies the proteins inside the acrosome?
a)
Inability to dissolve zona pellucida's glycocalyx
b)
Failure to complete spermatogenesis
c)
Impairment of sperm motility
d)
Increased polyspermy
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Orion Classes answered |
The acrosome is an organelle located at the tip of the sperm head and contains enzymes necessary for fertilization. It plays a crucial role in the penetration of the zona pellucida, a glycoprotein layer surrounding the egg. If a single point mutation modifies the proteins inside the acrosome, it can result in an inability to dissolve the zona pellucida's glycocalyx, which is necessary for successful fertilization. This would prevent the sperm from reaching and penetrating the egg, leading to infertility or reduced fertility. Therefore, option A is the correct answer.
Colchicine is a drug used to treat gout. It also happens to arrest cells in anaphase. Which cytoskeletal protein does colchicine effect?- a)Keratin
- b)Actin
- c)Myosin
- d)Tubulin
Correct answer is option 'D'. Can you explain this answer?
Colchicine is a drug used to treat gout. It also happens to arrest cells in anaphase. Which cytoskeletal protein does colchicine effect?
a)
Keratin
b)
Actin
c)
Myosin
d)
Tubulin
|
Zoe Brooks answered |
Effects of Colchicine on Tubulin
Introduction:
Colchicine is a medication primarily used to treat gout, a form of arthritis characterized by severe pain, redness, and swelling in the joints. It works by inhibiting the inflammation caused by the deposition of uric acid crystals in the joints. However, colchicine also has an effect on cell division, specifically on the process of mitosis.
Effect on Tubulin:
Colchicine affects tubulin, which is a key protein involved in the formation of microtubules, a component of the cytoskeleton. Microtubules are dynamic structures involved in various cellular processes, including cell division, intracellular transport, and maintenance of cell shape. They are composed of tubulin protein subunits that polymerize to form long cylindrical structures.
Inhibition of Microtubule Formation:
Colchicine binds to tubulin and inhibits the polymerization of tubulin subunits into microtubules. It does this by binding to the tubulin dimers, preventing their addition to the growing microtubule ends. This leads to the disruption of microtubule formation and destabilization of existing microtubules. Consequently, the cytoskeleton's structure and function are compromised.
Effects on Cell Division:
One of the critical roles of microtubules is to facilitate the separation of chromosomes during cell division. During the process of mitosis, microtubules form the spindle apparatus, which is responsible for aligning and segregating chromosomes into daughter cells. By disrupting microtubule formation, colchicine interferes with the proper functioning of the spindle apparatus, resulting in the arrest of cells in anaphase.
Conclusion:
In summary, colchicine affects tubulin, a cytoskeletal protein involved in microtubule formation. By inhibiting tubulin polymerization, colchicine disrupts the structure and function of the cytoskeleton, including the spindle apparatus involved in cell division. This leads to the arrest of cells in anaphase, a phase of mitosis where chromosomes are supposed to separate into daughter cells. Understanding the effects of colchicine on tubulin provides valuable insights into its mechanism of action and its potential applications in various cellular processes.
Introduction:
Colchicine is a medication primarily used to treat gout, a form of arthritis characterized by severe pain, redness, and swelling in the joints. It works by inhibiting the inflammation caused by the deposition of uric acid crystals in the joints. However, colchicine also has an effect on cell division, specifically on the process of mitosis.
Effect on Tubulin:
Colchicine affects tubulin, which is a key protein involved in the formation of microtubules, a component of the cytoskeleton. Microtubules are dynamic structures involved in various cellular processes, including cell division, intracellular transport, and maintenance of cell shape. They are composed of tubulin protein subunits that polymerize to form long cylindrical structures.
Inhibition of Microtubule Formation:
Colchicine binds to tubulin and inhibits the polymerization of tubulin subunits into microtubules. It does this by binding to the tubulin dimers, preventing their addition to the growing microtubule ends. This leads to the disruption of microtubule formation and destabilization of existing microtubules. Consequently, the cytoskeleton's structure and function are compromised.
Effects on Cell Division:
One of the critical roles of microtubules is to facilitate the separation of chromosomes during cell division. During the process of mitosis, microtubules form the spindle apparatus, which is responsible for aligning and segregating chromosomes into daughter cells. By disrupting microtubule formation, colchicine interferes with the proper functioning of the spindle apparatus, resulting in the arrest of cells in anaphase.
Conclusion:
In summary, colchicine affects tubulin, a cytoskeletal protein involved in microtubule formation. By inhibiting tubulin polymerization, colchicine disrupts the structure and function of the cytoskeleton, including the spindle apparatus involved in cell division. This leads to the arrest of cells in anaphase, a phase of mitosis where chromosomes are supposed to separate into daughter cells. Understanding the effects of colchicine on tubulin provides valuable insights into its mechanism of action and its potential applications in various cellular processes.
Which embryological stage divides the embryo in two halves establishing bilateral symmetry in mammals?- a)Neurulation
- b)Organogenesis
- c)Cleavage
- d)Blastulation
Correct answer is option 'A'. Can you explain this answer?
Which embryological stage divides the embryo in two halves establishing bilateral symmetry in mammals?
a)
Neurulation
b)
Organogenesis
c)
Cleavage
d)
Blastulation
|
|
Zoe Brooks answered |
Neurulation:
Neurulation is the embryological stage in mammals that divides the embryo in two halves, establishing bilateral symmetry. This process occurs during the third week of gestation in humans.
Importance of Neurulation:
Neurulation is a crucial stage of development as it marks the beginning of the formation of the nervous system. During neurulation, the neural plate folds in on itself to form the neural tube, which gives rise to the brain and spinal cord. This process also establishes the left-right axis of the organism, leading to bilateral symmetry.
Establishing Bilateral Symmetry:
Neurulation is responsible for establishing bilateral symmetry in mammals. Bilateral symmetry means that an organism can be divided into two equal halves along a single plane, resulting in mirror images. This symmetry is essential for the proper development and functioning of complex organisms like mammals.
Conclusion:
In conclusion, neurulation is the embryological stage in mammals that divides the embryo in two halves, establishing bilateral symmetry. This process is critical for the formation of the nervous system and the overall development of the organism.
Neurulation is the embryological stage in mammals that divides the embryo in two halves, establishing bilateral symmetry. This process occurs during the third week of gestation in humans.
Importance of Neurulation:
Neurulation is a crucial stage of development as it marks the beginning of the formation of the nervous system. During neurulation, the neural plate folds in on itself to form the neural tube, which gives rise to the brain and spinal cord. This process also establishes the left-right axis of the organism, leading to bilateral symmetry.
Establishing Bilateral Symmetry:
Neurulation is responsible for establishing bilateral symmetry in mammals. Bilateral symmetry means that an organism can be divided into two equal halves along a single plane, resulting in mirror images. This symmetry is essential for the proper development and functioning of complex organisms like mammals.
Conclusion:
In conclusion, neurulation is the embryological stage in mammals that divides the embryo in two halves, establishing bilateral symmetry. This process is critical for the formation of the nervous system and the overall development of the organism.
A radioactive marker was applied to the mesoderm of a mammalian embryo. Which tissues or organs would have this marker in the adult?- a)Spine and brain
- b)Skin and intestines
- c)Bones and muscles
- d)Muscles and skin
Correct answer is option 'C'. Can you explain this answer?
A radioactive marker was applied to the mesoderm of a mammalian embryo. Which tissues or organs would have this marker in the adult?
a)
Spine and brain
b)
Skin and intestines
c)
Bones and muscles
d)
Muscles and skin
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Orion Classes answered |
The mesoderm is one of the three primary germ layers formed during early embryonic development. It gives rise to various tissues and organs in the adult body, including bones and muscles.
The mesoderm differentiates into mesenchyme, which further develops into different types of connective tissues, including bone tissue. Bones are formed through a process called ossification, where mesenchymal cells differentiate into osteoblasts, which produce the extracellular matrix of bone.
Additionally, the mesoderm also gives rise to muscle tissue. Myoblasts, derived from the mesoderm, differentiate into skeletal muscle, smooth muscle, and cardiac muscle cells.
Therefore, if a radioactive marker is applied to the mesoderm of a mammalian embryo, the tissues or organs that would have this marker in the adult are bones and muscles.
Which of the following would you expect to see in a cancerous mass?- a)Loss of function mutations in apoptosis-controlling proteins
- b)Increased time in G0
- c)Gain of function mutations in checkpoint proteins
- d)Decreased ATP use
Correct answer is option 'A'. Can you explain this answer?
Which of the following would you expect to see in a cancerous mass?
a)
Loss of function mutations in apoptosis-controlling proteins
b)
Increased time in G0
c)
Gain of function mutations in checkpoint proteins
d)
Decreased ATP use
|
Jackson Wright answered |
Loss of function mutations in apoptosis-controlling proteins
Loss of function mutations in apoptosis-controlling proteins would be expected to see in a cancerous mass for the following reasons:
- Apoptosis resistance: Cancer cells often acquire mutations that inhibit the programmed cell death process, allowing them to survive and proliferate uncontrollably.
- Unregulated cell growth: Apoptosis-controlling proteins play a crucial role in regulating cell death. Mutations that disrupt this regulation can lead to uncontrolled cell growth, a hallmark of cancer.
- Survival advantages: By avoiding apoptosis, cancer cells can evade the body's natural defense mechanisms that would normally eliminate abnormal cells.
Overall, the presence of loss of function mutations in apoptosis-controlling proteins contributes to the development and progression of cancer by promoting cell survival and uncontrolled growth.
Which of the following is not a reason cancer is hard to treat?- a)Increased contact inhibition
- b)Increased growth rate
- c)Increased mutation rate
- d)Increased metastasis
Correct answer is option 'A'. Can you explain this answer?
Which of the following is not a reason cancer is hard to treat?
a)
Increased contact inhibition
b)
Increased growth rate
c)
Increased mutation rate
d)
Increased metastasis
|
|
James Rodriguez answered |
Understanding Cancer Treatment Challenges
Cancer presents significant challenges in treatment due to various characteristics of tumor cells. Here’s a breakdown of why "increased contact inhibition" is not a reason cancer is hard to treat.
What is Contact Inhibition?
- Contact inhibition is a normal cellular mechanism that prevents cells from growing uncontrollably when they come into contact with other cells.
- Healthy cells will stop dividing when they touch other cells, maintaining tissue architecture and function.
How Cancer Cells Differ
- Cancer cells often lose the ability to undergo contact inhibition.
- This loss allows them to grow and proliferate uncontrollably, leading to tumor formation.
Reasons Why Cancer is Hard to Treat
- Increased Growth Rate: Cancer cells often divide more rapidly than normal cells, making them harder to target effectively with treatments that aim to kill or slow down cell division.
- Increased Mutation Rate: Cancer cells frequently mutate, leading to genetic diversity within tumors. This variability can result in resistance to treatments, making it difficult to find a one-size-fits-all approach.
- Increased Metastasis: Many cancers can spread from their original site to other parts of the body. This process complicates treatment as it requires targeting multiple tumors rather than a single location.
Conclusion
In summary, "increased contact inhibition" is not a reason cancer is hard to treat because cancer cells typically lose this ability, leading to their aggressive growth. The actual challenges stem from their rapid growth, high mutation rates, and tendency to metastasize. Understanding these factors is crucial for developing effective treatment strategies.
Cancer presents significant challenges in treatment due to various characteristics of tumor cells. Here’s a breakdown of why "increased contact inhibition" is not a reason cancer is hard to treat.
What is Contact Inhibition?
- Contact inhibition is a normal cellular mechanism that prevents cells from growing uncontrollably when they come into contact with other cells.
- Healthy cells will stop dividing when they touch other cells, maintaining tissue architecture and function.
How Cancer Cells Differ
- Cancer cells often lose the ability to undergo contact inhibition.
- This loss allows them to grow and proliferate uncontrollably, leading to tumor formation.
Reasons Why Cancer is Hard to Treat
- Increased Growth Rate: Cancer cells often divide more rapidly than normal cells, making them harder to target effectively with treatments that aim to kill or slow down cell division.
- Increased Mutation Rate: Cancer cells frequently mutate, leading to genetic diversity within tumors. This variability can result in resistance to treatments, making it difficult to find a one-size-fits-all approach.
- Increased Metastasis: Many cancers can spread from their original site to other parts of the body. This process complicates treatment as it requires targeting multiple tumors rather than a single location.
Conclusion
In summary, "increased contact inhibition" is not a reason cancer is hard to treat because cancer cells typically lose this ability, leading to their aggressive growth. The actual challenges stem from their rapid growth, high mutation rates, and tendency to metastasize. Understanding these factors is crucial for developing effective treatment strategies.
Centrioles can be found in:- a)Only eukaryotic cells
- b)Both eukaryotic and prokaryotic cells
- c)Only prokaryotic cells
- d)In eukaryotic cells and in prokaryotic cells undergoing genetic recombination
Correct answer is option 'A'. Can you explain this answer?
Centrioles can be found in:
a)
Only eukaryotic cells
b)
Both eukaryotic and prokaryotic cells
c)
Only prokaryotic cells
d)
In eukaryotic cells and in prokaryotic cells undergoing genetic recombination
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Orion Classes answered |
Centrioles are small, cylindrical structures found in eukaryotic cells. They are typically found in pairs and are located within a specialized region of the cell called the centrosome. Centrioles play a crucial role in cell division, particularly in the formation of the mitotic spindle during mitosis and the organization of microtubules.
Prokaryotic cells, on the other hand, do not possess centrioles. Prokaryotes, such as bacteria, lack membrane-bound organelles typically found in eukaryotic cells. They undergo cell division through a process called binary fission, which does not involve the formation of a mitotic spindle or the presence of centrioles.
The lungs, spine and muscles are derived from 3 different germ layers. What would be the correct order of formation of the embryological cells from which these tissues are derived from?- a)Embryological cells that will form spine will form first, followed by the muscles and then lungs
- b)Embryological cells that will form muscles will form first, followed by the lungs and then spine
- c)Embryological cells that will form muscles will form first, followed by the spine and then lungs
- d)mbryological cells that will form spine will form first, followed by the lungs and then muscles
Correct answer is option 'D'. Can you explain this answer?
The lungs, spine and muscles are derived from 3 different germ layers. What would be the correct order of formation of the embryological cells from which these tissues are derived from?
a)
Embryological cells that will form spine will form first, followed by the muscles and then lungs
b)
Embryological cells that will form muscles will form first, followed by the lungs and then spine
c)
Embryological cells that will form muscles will form first, followed by the spine and then lungs
d)
mbryological cells that will form spine will form first, followed by the lungs and then muscles
|
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Orion Classes answered |
Embryological cells that will form the spine will form first, followed by the lungs, and then the muscles.
In this case, the order of formation is:
- Embryological cells that will form the spine (derived from mesoderm)
- Embryological cells that will form the lungs (derived from endoderm)
- Embryological cells that will form the muscles (also derived from mesoderm)
During which phase does the nuclear envelope begin to disappear?- a)Prometaphase
- b)Anaphase
- c)S phase
- d)Cytokinesis
Correct answer is option 'A'. Can you explain this answer?
During which phase does the nuclear envelope begin to disappear?
a)
Prometaphase
b)
Anaphase
c)
S phase
d)
Cytokinesis
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|
Orion Classes answered |
During cell division, the nuclear envelope, which surrounds the nucleus, needs to break down to allow for proper segregation of chromosomes. This breakdown occurs during prometaphase, which is a stage of mitosis or meiosis. Prometaphase follows prophase and precedes metaphase.
In prometaphase, the following events take place:
- The nuclear envelope starts to disintegrate, and the membrane breaks down into smaller vesicles.
- The microtubules of the mitotic spindle attach to the kinetochores of the chromosomes.
- The chromosomes become condensed and are prepared for proper alignment during metaphase.
- The disappearance of the nuclear envelope during prometaphase allows the microtubules to interact with the chromosomes directly, facilitating their movement and alignment. This breakdown is crucial for the subsequent stages of cell division.
Anaphase (option B) is the stage during which sister chromatids separate and move towards opposite poles of the cell. The nuclear envelope is already absent during this phase.
S phase (option C) is the synthesis phase of the cell cycle, where DNA replication occurs. The nuclear envelope is still intact during this phase.
Cytokinesis (option D) is the final stage of cell division, where the cytoplasm divides, and two daughter cells are formed. The nuclear envelope reforms around the separated chromosomes in each daughter cell during cytokinesis.
Malformation of the peripheral nervous system is linked to specific groups of cells formed during neurulation, and these cells are derived from one of 3 embryonic germ layers. What is this specific group of cells, and which embryonic layers are they derived from?- a)Neural layer cells, mesoderm
- b)Neural tube cells, ectoderm
- c)Neural crest cells, ectoderm
- d)Neural fold cells, endoderm
Correct answer is option 'C'. Can you explain this answer?
Malformation of the peripheral nervous system is linked to specific groups of cells formed during neurulation, and these cells are derived from one of 3 embryonic germ layers. What is this specific group of cells, and which embryonic layers are they derived from?
a)
Neural layer cells, mesoderm
b)
Neural tube cells, ectoderm
c)
Neural crest cells, ectoderm
d)
Neural fold cells, endoderm
|
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Orion Classes answered |
Malformation of the peripheral nervous system is linked to specific groups of cells formed during neurulation. These cells are known as neural crest cells, and they are derived from the ectoderm, which is one of the three embryonic germ layers.
During neurulation, which is the process of neural tube formation, the neural plate folds and forms the neural tube, which gives rise to the central nervous system. However, at the edges of the neural plate, some cells undergo a transformation and detach from the neural tube. These cells are the neural crest cells.
The neural crest cells migrate extensively throughout the developing embryo and give rise to a wide range of cell types, including neurons and glial cells of the peripheral nervous system, as well as other cell types in various tissues such as the craniofacial structures, melanocytes, and adrenal medulla.
Therefore, the specific group of cells linked to the malformation of the peripheral nervous system during neurulation is the neural crest cells, and they are derived from the ectoderm.
The correct answer is option C: Neural crest cells, ectoderm.
What would be the correct order of the following events during fertilization? - Digestion of zona pellucida by acrosomal enzymes
- Sperm bypasses corona radiata
- Cortical granules release enzymes
- Sperm binding with vitelline membrane
- a)3, 2, 4, 1
- b)2, 1, 3, 4
- c)1, 3, 2, 4
- d)2, 3, 4, 1
Correct answer is option 'B'. Can you explain this answer?
What would be the correct order of the following events during fertilization?
- Digestion of zona pellucida by acrosomal enzymes
- Sperm bypasses corona radiata
- Cortical granules release enzymes
- Sperm binding with vitelline membrane
a)
3, 2, 4, 1
b)
2, 1, 3, 4
c)
1, 3, 2, 4
d)
2, 3, 4, 1
|
|
Orion Classes answered |
- Sperm bypasses corona radiata: The corona radiata is the layer of follicular cells surrounding the oocyte. The sperm needs to penetrate through this layer to reach the oocyte.
- Sperm binding with vitelline membrane: Once the sperm reaches the oocyte, it binds to the receptor proteins on the vitelline membrane, which is the outer layer of the oocyte.
- Cortical granules release enzymes: After the sperm binding, cortical granules in the oocyte release their contents. These enzymes modify the zona pellucida, the glycoprotein layer surrounding the oocyte, preventing polyspermy.
- Digestion of zona pellucida by acrosomal enzymes: The acrosome of the sperm contains enzymes that help in the penetration of the zona pellucida. These enzymes digest the zona pellucida, allowing the sperm to reach the oocyte's plasma membrane.
Therefore, the correct order is 2, 1, 3, 4.
Placenta accreta is a medical condition in which the placenta attaches abnormally to the myometrium. What is the earliest embryonic stage that this could occur, which cells are responsible for this attachment?- a)Blastulation, trophoblast
- b)Early cleavage, trophoblast
- c)Blastulation, epiblast
- d)Early cleavage, embryoblast
Correct answer is option 'A'. Can you explain this answer?
Placenta accreta is a medical condition in which the placenta attaches abnormally to the myometrium. What is the earliest embryonic stage that this could occur, which cells are responsible for this attachment?
a)
Blastulation, trophoblast
b)
Early cleavage, trophoblast
c)
Blastulation, epiblast
d)
Early cleavage, embryoblast
|
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Orion Classes answered |
Placenta accreta is a condition where the placenta attaches abnormally to the myometrium, the muscular layer of the uterus. The attachment of the placenta occurs during the early stages of embryonic development.
In early embryonic development, after fertilization, the zygote undergoes cleavage, resulting in a solid ball of cells called the morula. Subsequently, the morula undergoes further cell division and forms a fluid-filled cavity called the blastocyst. The blastocyst consists of two main cell populations: the trophoblast and the inner cell mass (ICM).
The trophoblast is responsible for implantation and the subsequent formation of the placenta. It is the outer layer of cells in the blastocyst that interacts with the maternal tissues. The trophoblast differentiates into two distinct layers: the outer syncytiotrophoblast and the inner cytotrophoblast.
During blastulation, which is the process of blastocyst formation, the trophoblast cells play a crucial role in attaching the blastocyst to the uterine wall. The trophoblast cells invade and attach to the endometrium of the uterus, facilitating implantation.
Therefore, the earliest embryonic stage at which placenta accreta could occur is during blastulation, specifically during the attachment of trophoblast cells to the uterine wall.
The correct answer is option A: Blastulation, trophoblast.
Cortical reaction and acrosomal reaction are analogous, but disruption of these two have different effects. What could occur if these reactions fail?- a)Both would lead to sterility
- b)Sterility and higher risk of conjoint twins
- c)Both would lead to polyspermy
- d)Higher trisomy risk and sterility, respectively
Correct answer is option 'D'. Can you explain this answer?
Cortical reaction and acrosomal reaction are analogous, but disruption of these two have different effects. What could occur if these reactions fail?
a)
Both would lead to sterility
b)
Sterility and higher risk of conjoint twins
c)
Both would lead to polyspermy
d)
Higher trisomy risk and sterility, respectively
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Orion Classes answered |
If the acrosomal reaction fails, it can indeed lead to higher trisomy risk and sterility, respectively. The acrosomal reaction is essential for the sperm to penetrate the zona pellucida and reach the egg for fertilization. If this process is impaired or disrupted, it can result in difficulties in sperm-egg interaction and fertilization.
Without a successful acrosomal reaction, the sperm may not be able to penetrate the zona pellucida effectively, leading to a higher risk of polyspermy and subsequent abnormalities in chromosome number (trisomy). Additionally, if the acrosomal reaction is consistently unsuccessful, it can contribute to infertility as fertilization cannot occur.
Therefore, option D, higher trisomy risk and sterility, respectively, is the correct answer.
A karyotype can be used to check for certain chromosomal mutations. In what cell cycle phase can cells be karyotyped?- a)G0
- b)S
- c)Metaphase
- d)Prophase
Correct answer is option 'C'. Can you explain this answer?
A karyotype can be used to check for certain chromosomal mutations. In what cell cycle phase can cells be karyotyped?
a)
G0
b)
S
c)
Metaphase
d)
Prophase
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Orion Classes answered |
Karyotyping involves the analysis of chromosomes to detect any abnormalities or mutations in their structure or number. This technique is typically performed during metaphase of the cell cycle.
During metaphase, chromosomes are fully condensed and aligned along the equatorial plane of the cell. At this stage, the chromosomes are most visible and can be easily observed and analyzed under a microscope. Cells undergoing mitosis or meiosis can be arrested at metaphase by various methods, allowing for the collection and examination of chromosomes for abnormalities.
Which statement best describes chromosomal movement during mitosis?- a)Centrioles send out microtubules, which bind to kinetochores to pull chromatids apart.
- b)Microtubules bind tubulin polymers on centromeres to initiate homologous chromosome separation.
- c)Centrioles on chromatids bind kinetochores to rays sent by the asters.
- d)ach centrosome binds an individual chromatid to ensure equal division.
Correct answer is option 'A'. Can you explain this answer?
Which statement best describes chromosomal movement during mitosis?
a)
Centrioles send out microtubules, which bind to kinetochores to pull chromatids apart.
b)
Microtubules bind tubulin polymers on centromeres to initiate homologous chromosome separation.
c)
Centrioles on chromatids bind kinetochores to rays sent by the asters.
d)
ach centrosome binds an individual chromatid to ensure equal division.
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Orion Classes answered |
During mitosis, the movement of chromosomes is facilitated by the interaction between microtubules and kinetochores. Centrioles, located in the centrosomes, organize and send out microtubules that form the mitotic spindle. The microtubules attach to kinetochores, which are protein structures located at the centromeres of sister chromatids.
As the mitotic spindle forms, the microtubules extend and connect to the kinetochores of each sister chromatid. The microtubules exert forces on the kinetochores, pulling the sister chromatids apart towards opposite poles of the cell. This separation ensures that each daughter cell receives an equal and complete set of chromosomes.
Option B is incorrect because homologous chromosome separation occurs during meiosis, not mitosis. Option C is incorrect as centrioles on chromatids do not exist. Option D is also incorrect because each centrosome does not bind an individual chromatid, but rather the microtubules from both centrosomes collectively interact with all the chromatids.
Retinoblastoma protein, RB, is an important cell cycle checkpoint protein that prevents the transition from G1 to S phase. The presence of the cyclin/CDK complex releases it from the genes that it binds. Which of the following mutation scenarios is least likely to lead to the formation of a neoplasm?- a)Loss of functional RB and CDK that does not respond to negative feedback
- b)No RB mutation and CDK that does not respond to negative feedback
- c)Loss of functional RB and no CDK mutation
- d)No RB mutation and CDK that cannot bind to cyclin
Correct answer is option 'D'. Can you explain this answer?
Retinoblastoma protein, RB, is an important cell cycle checkpoint protein that prevents the transition from G1 to S phase. The presence of the cyclin/CDK complex releases it from the genes that it binds. Which of the following mutation scenarios is least likely to lead to the formation of a neoplasm?
a)
Loss of functional RB and CDK that does not respond to negative feedback
b)
No RB mutation and CDK that does not respond to negative feedback
c)
Loss of functional RB and no CDK mutation
d)
No RB mutation and CDK that cannot bind to cyclin
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Orion Classes answered |
In option D, there is no RB mutation, meaning that the RB protein is intact and able to properly inhibit the cell cycle progression. Additionally, the CDK in this scenario cannot bind to cyclin. Without the formation of the cyclin/CDK complex, CDK activity is inhibited, preventing uncontrolled cell cycle progression. This scenario, where RB is functional and CDK cannot bind to cyclin, is less likely to lead to the formation of a neoplasm compared to the other options.
At which point in development do we start to observe a significant increase of the size of embryo?- a)During blastulation
- b)During organogenesis
- c)During fertilization
- d)During early cleavage
Correct answer is option 'A'. Can you explain this answer?
At which point in development do we start to observe a significant increase of the size of embryo?
a)
During blastulation
b)
During organogenesis
c)
During fertilization
d)
During early cleavage
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Naomi Reed answered |
During blastulation
- Blastulation is the process that occurs during the early stages of embryonic development, specifically after fertilization.
- During blastulation, the zygote undergoes several rapid cell divisions, resulting in the formation of a hollow ball of cells called the blastula.
- The blastula consists of an outer layer of cells called the trophoblast and an inner cell mass called the embryoblast.
- The trophoblast will eventually give rise to the placenta, while the embryoblast will develop into the embryo itself.
- As blastulation progresses, the size of the embryo increases significantly due to the rapid cell division and the accumulation of cells within the blastula.
- This increase in size is primarily due to the proliferation of cells within the embryoblast.
Why not the other options?
- During organogenesis (option B), the embryo undergoes the process of organ formation, but this occurs after blastulation. Therefore, it is not the point at which a significant increase in embryo size is observed.
- Fertilization (option C) is the fusion of sperm and egg to form a zygote, which is the first step in embryonic development. However, the embryo is not formed at this point, so there is no significant increase in size.
- Early cleavage (option D) is the division of the zygote into smaller cells through mitosis. While this process is essential for the development of the embryo, it does not result in a significant increase in size. The cells produced during early cleavage are still relatively small and compact.
In summary, the significant increase in the size of the embryo is observed during blastulation, specifically as the blastula forms and the cells within the embryoblast proliferate.
- Blastulation is the process that occurs during the early stages of embryonic development, specifically after fertilization.
- During blastulation, the zygote undergoes several rapid cell divisions, resulting in the formation of a hollow ball of cells called the blastula.
- The blastula consists of an outer layer of cells called the trophoblast and an inner cell mass called the embryoblast.
- The trophoblast will eventually give rise to the placenta, while the embryoblast will develop into the embryo itself.
- As blastulation progresses, the size of the embryo increases significantly due to the rapid cell division and the accumulation of cells within the blastula.
- This increase in size is primarily due to the proliferation of cells within the embryoblast.
Why not the other options?
- During organogenesis (option B), the embryo undergoes the process of organ formation, but this occurs after blastulation. Therefore, it is not the point at which a significant increase in embryo size is observed.
- Fertilization (option C) is the fusion of sperm and egg to form a zygote, which is the first step in embryonic development. However, the embryo is not formed at this point, so there is no significant increase in size.
- Early cleavage (option D) is the division of the zygote into smaller cells through mitosis. While this process is essential for the development of the embryo, it does not result in a significant increase in size. The cells produced during early cleavage are still relatively small and compact.
In summary, the significant increase in the size of the embryo is observed during blastulation, specifically as the blastula forms and the cells within the embryoblast proliferate.
Consider the whole gestational period. When does fertilization normally occur?- a)Prior to gestation occur
- b)Week 2 of gestation
- c)At the very beginning of gestation
- d)Week 1 of gestation
Correct answer is option 'B'. Can you explain this answer?
Consider the whole gestational period. When does fertilization normally occur?
a)
Prior to gestation occur
b)
Week 2 of gestation
c)
At the very beginning of gestation
d)
Week 1 of gestation
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Orion Classes answered |
During the first week of gestation, the fertilized egg undergoes several divisions and forms a solid ball of cells called a morula. By the end of the first week, the morula develops into a blastocyst, which is a hollow ball of cells with an inner cell mass and an outer layer called the trophoblast.
Fertilization typically occurs in the fallopian tubes within 24 hours after ovulation. After fertilization, the zygote undergoes cleavage divisions during the first week, eventually forming the blastocyst. The blastocyst then travels through the fallopian tube and reaches the uterus by the end of Week 1 or the beginning of Week 2, where it implants into the uterine lining.
Therefore, the correct answer is B. Week 2 of gestation.
Chapter doubts & questions for Reproductive System and Embryogenesis (BIO) - MCAT Biological and Biochemical Foundations 2025 is part of MCAT exam preparation. The chapters have been prepared according to the MCAT exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for MCAT 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
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