Medical Terminology Systems: A Body Systems Approach
8th Edition
ISBN: 9780803676619
Author: Barbara A. Gylys; Mary Ellen Wedding
Publisher: F. A. Davis Company
expand_more
expand_more
format_list_bulleted
Concept explainers
Videos
Question
Chapter 4, Problem 1BSDT
Summary Introduction
The ventral cavity in the human body consists of two sub-division that are abdominopelvic cavity and the thoracic cavity. Abdominopelvic cavity begins from the diaphragm and ends in the pelvic inlet. The abdominal cavity is divided by two methods. Diaphragm divides the abdominopelvic cavity and the thoracic cavity. The thoracic cavity is further divided into pericardial and pleural cavity.
Expert Solution & Answer
Answer to Problem 1BSDT
| Given Term | Meaning |
| Abdominopelvic | h. ventral cavity that contains digestive, reproductive and excretory structures. |
Explanation of Solution
The abdominal cavity and the pelvic cavity constitute the abdominopelvic cavity. It is a body cavity that includes the spleen, liver, gallbladder, pancreas, kidneys, and stomach. This cavity also includes reproductive organs as well as the excretory structure. An example for the excretory structure present in the abdominopelvic cavity is urinary bladder. Examples of reproductive structures present in the abdominopelvic cavity are ovarian ligament and recto-vesical pouch.
Want to see more full solutions like this?
Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
If using animals in medical experiments could save human lives, is it ethical to do so? In your answer, apply at least one ethical theory in support of your position.
You aim to test the hypothesis that the Tbx4 and Tbx5 genes inhibit each other's expression during limb development. With access to chicken embryos and viruses capable of overexpressing Tbx4 and Tbx5, describe an experiment to investigate whether these genes suppress each other's expression in the limb buds. What results would you expect if they do repress each other? What results would you expect if they do not repress each other?
You decide to delete Fgf4 and Fgf8 specifically in the limb bud. Explain why you would not knock out these genes in the entire embryo instead.
Chapter 4 Solutions
Medical Terminology Systems: A Body Systems Approach
Ch. 4 - Prob. 1BSDTCh. 4 - Prob. 2BSDTCh. 4 - Prob. 3BSDTCh. 4 - Prob. 4BSDTCh. 4 - Prob. 5BSDTCh. 4 - Prob. 6BSDTCh. 4 - Prob. 7BSDTCh. 4 - Prob. 8BSDTCh. 4 - Prob. 9BSDTCh. 4 - Prob. 10BSDT
Ch. 4 - Prob. 11BSDTCh. 4 - Prob. 12BSDTCh. 4 - Prob. 13BSDTCh. 4 - Prob. 14BSDTCh. 4 - Prob. 15BSDTCh. 4 - Prob. 1BWECh. 4 - Prob. 2BWECh. 4 - Prob. 3BWECh. 4 - Prob. 4BWECh. 4 - Prob. 5BWECh. 4 - Prob. 6BWECh. 4 - Prob. 7BWECh. 4 - Prob. 8BWECh. 4 - Prob. 9BWECh. 4 - Prob. 10BWECh. 4 - Prob. 11BWECh. 4 - Prob. 12BWECh. 4 - Prob. 13BWECh. 4 - Prob. 14BWECh. 4 - Prob. 15BWECh. 4 - Prob. 16BWECh. 4 - Prob. 17BWECh. 4 - Prob. 18BWECh. 4 - Prob. 19BWECh. 4 - Prob. 20BWECh. 4 - Prob. 1BBTCh. 4 - Prob. 2BBTCh. 4 - Prob. 3BBTCh. 4 - Prob. 4BBTCh. 4 - Prob. 5BBTCh. 4 - Prob. 6BBTCh. 4 - Prob. 7BBTCh. 4 - Prob. 8BBTCh. 4 - Prob. 9BBTCh. 4 - Prob. 10BBTCh. 4 - Prob. 1BMWCh. 4 - Prob. 2BMWCh. 4 - Prob. 3BMWCh. 4 - Prob. 4BMWCh. 4 - Prob. 5BMWCh. 4 - Prob. 6BMWCh. 4 - Prob. 7BMWCh. 4 - Prob. 8BMWCh. 4 - Prob. 9BMWCh. 4 - Prob. 10BMWCh. 4 - Prob. 1DCCh. 4 - Prob. 2DCCh. 4 - Prob. 3DCCh. 4 - Prob. 4DCCh. 4 - Prob. 5DCCh. 4 - Prob. 6DCCh. 4 - Prob. 7DCCh. 4 - Prob. 8DCCh. 4 - Prob. 9DCCh. 4 - Prob. 10DCCh. 4 - Prob. 11DCCh. 4 - Prob. 12DCCh. 4 - Prob. 13DCCh. 4 - Prob. 14DCCh. 4 - Prob. 15DCCh. 4 - Prob. 1PACh. 4 - Prob. 2PACh. 4 - Prob. 3PACh. 4 - Prob. 4PACh. 4 - Prob. 5PACh. 4 - Prob. 6PACh. 4 - Prob. 7PACh. 4 - Prob. 8PACh. 4 - Prob. 9PACh. 4 - Prob. 10PACh. 4 - Prob. 11PACh. 4 - Prob. 12PACh. 4 - Prob. 13PACh. 4 - Prob. 14PACh. 4 - Prob. 15PACh. 4 - Prob. 1.1TCh. 4 - Prob. 1.2TCh. 4 - Prob. 1.3TCh. 4 - Prob. 1.4TCh. 4 - Prob. 1.5TCh. 4 - Prob. 1.6TCh. 4 - Prob. 1.7TCh. 4 - Prob. 1.8TCh. 4 - Prob. 1.1CTCh. 4 - Prob. 1.2CTCh. 4 - Prob. 1.3CTCh. 4 - Prob. 1.4CTCh. 4 - Prob. 1.5CTCh. 4 - Prob. 1.6CTCh. 4 - Prob. 2.1TCh. 4 - Prob. 2.2TCh. 4 - Prob. 2.3TCh. 4 - Prob. 2.4TCh. 4 - Prob. 2.5TCh. 4 - Prob. 2.6TCh. 4 - Prob. 2.7TCh. 4 - Prob. 2.8TCh. 4 - Prob. 2.9TCh. 4 - Prob. 2.10TCh. 4 - Prob. 2.11TCh. 4 - Prob. 2.12TCh. 4 - Prob. 2.13TCh. 4 - Prob. 2.1CTCh. 4 - Prob. 2.2CTCh. 4 - Prob. 2.3CTCh. 4 - Prob. 2.4CTCh. 4 - Prob. 2.5CT
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, biology and related others by exploring similar questions and additional content below.Similar questions
- You implant an FGF10-coated bead into the anterior flank of a chicken embryo, directly below the level of the wing bud. What is the phenotype of the resulting ectopic limb? Briefly describe the expected expression domains of 1) Shh, 2) Tbx4, and 3) Tbx5 in the resulting ectopic limb bud.arrow_forwardDesign a grafting experiment to determine if limb mesoderm determines forelimb / hindlimb identity. Include the experiment, a control, and an interpretation in your answer.arrow_forwardThe Snapdragon is a popular garden flower that comes in a variety of colours, including red, yellow, and orange. The genotypes and associated phenotypes for some of these flowers are as follows: aabb: yellow AABB, AABb, AaBb, and AaBB: red AAbb and Aabb: orange aaBB: yellow aaBb: ? Based on this information, what would the phenotype of a Snapdragon with the genotype aaBb be and why? Question 21 options: orange because A is epistatic to B yellow because A is epistatic to B red because B is epistatic to A orange because B is epistatic to A red because A is epistatic to B yellow because B is epistatic to Aarrow_forward
- A sample of blood was taken from the above individual and prepared for haemoglobin analysis. However, when water was added the cells did not lyse and looked normal in size and shape. The technician suspected that they had may have made an error in the protocol – what is the most likely explanation? The cell membranes are more resistant than normal. An isotonic solution had been added instead of water. A solution of 0.1 M NaCl had been added instead of water. Not enough water had been added to the red blood cell pellet. The man had sickle-cell anaemia.arrow_forwardA sample of blood was taken from the above individual and prepared for haemoglobin analysis. However, when water was added the cells did not lyse and looked normal in size and shape. The technician suspected that they had may have made an error in the protocol – what is the most likely explanation? The cell membranes are more resistant than normal. An isotonic solution had been added instead of water. A solution of 0.1 M NaCl had been added instead of water. Not enough water had been added to the red blood cell pellet. The man had sickle-cell anaemia.arrow_forwardWith reference to their absorption spectra of the oxy haemoglobin intact line) and deoxyhemoglobin (broken line) shown in Figure 2 below, how would you best explain the reason why there are differences in the major peaks of the spectra? Figure 2. SPECTRA OF OXYGENATED AND DEOXYGENATED HAEMOGLOBIN OBTAINED WITH THE RECORDING SPECTROPHOTOMETER 1.4 Abs < 0.8 06 0.4 400 420 440 460 480 500 520 540 560 580 600 nm 1. The difference in the spectra is due to a pH change in the deoxy-haemoglobin due to uptake of CO2- 2. There is more oxygen-carrying plasma in the oxy-haemoglobin sample. 3. The change in Mr due to oxygen binding causes the oxy haemoglobin to have a higher absorbance peak. 4. Oxy-haemoglobin is contaminated by carbaminohemoglobin, and therefore has a higher absorbance peak 5. Oxy-haemoglobin absorbs more light of blue wavelengths and less of red wavelengths than deoxy-haemoglobinarrow_forward
- With reference to their absorption spectra of the oxy haemoglobin intact line) and deoxyhemoglobin (broken line) shown in Figure 2 below, how would you best explain the reason why there are differences in the major peaks of the spectra? Figure 2. SPECTRA OF OXYGENATED AND DEOXYGENATED HAEMOGLOBIN OBTAINED WITH THE RECORDING SPECTROPHOTOMETER 1.4 Abs < 0.8 06 0.4 400 420 440 460 480 500 520 540 560 580 600 nm 1. The difference in the spectra is due to a pH change in the deoxy-haemoglobin due to uptake of CO2- 2. There is more oxygen-carrying plasma in the oxy-haemoglobin sample. 3. The change in Mr due to oxygen binding causes the oxy haemoglobin to have a higher absorbance peak. 4. Oxy-haemoglobin is contaminated by carbaminohemoglobin, and therefore has a higher absorbance peak 5. Oxy-haemoglobin absorbs more light of blue wavelengths and less of red wavelengths than deoxy-haemoglobinarrow_forwardWhich ONE of the following is FALSE regarding haemoglobin? It has two alpha subunits and two beta subunits. The subunits are joined by disulphide bonds. Each subunit covalently binds a haem group. Conformational change in one subunit can be transmitted to another. There are many variant ("mutant") forms of haemoglobin that are not harmful.arrow_forwardWhich ONE of the following is FALSE regarding haemoglobin? It has two alpha subunits and two beta subunits. The subunits are joined by disulphide bonds. Each subunit covalently binds a haem group. Conformational change in one subunit can be transmitted to another. There are many variant ("mutant") forms of haemoglobin that are not harmful.arrow_forward
- During a routine medical check up of a healthy man it was found that his haematocrit value was highly unusual – value of 60%. What one of the options below is the most likely reason? He will have a diet high in iron. He is likely to be suffering from anaemia. He lives at high altitude. He has recently recovered from an accident where he lost a lot of blood. He has a very large body size.arrow_forwardExplain what age of culture is most likely to produce an endospore?arrow_forwardExplain why hot temperatures greater than 45 degrees celsius would not initiate the sporulation process in endospores?arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Human Anatomy & Physiology (11th Edition)BiologyISBN:9780134580999Author:Elaine N. Marieb, Katja N. HoehnPublisher:PEARSON
Biology 2eBiologyISBN:9781947172517Author:Matthew Douglas, Jung Choi, Mary Ann ClarkPublisher:OpenStax
Anatomy & PhysiologyBiologyISBN:9781259398629Author:McKinley, Michael P., O'loughlin, Valerie Dean, Bidle, Theresa StouterPublisher:Mcgraw Hill Education,
Molecular Biology of the Cell (Sixth Edition)BiologyISBN:9780815344322Author:Bruce Alberts, Alexander D. Johnson, Julian Lewis, David Morgan, Martin Raff, Keith Roberts, Peter WalterPublisher:W. W. Norton & Company
Laboratory Manual For Human Anatomy & PhysiologyBiologyISBN:9781260159363Author:Martin, Terry R., Prentice-craver, CynthiaPublisher:McGraw-Hill Publishing Co.
Inquiry Into Life (16th Edition)BiologyISBN:9781260231700Author:Sylvia S. Mader, Michael WindelspechtPublisher:McGraw Hill Education
Human Anatomy & Physiology (11th Edition)
Biology
ISBN:9780134580999
Author:Elaine N. Marieb, Katja N. Hoehn
Publisher:PEARSON
Biology 2e
Biology
ISBN:9781947172517
Author:Matthew Douglas, Jung Choi, Mary Ann Clark
Publisher:OpenStax
Anatomy & Physiology
Biology
ISBN:9781259398629
Author:McKinley, Michael P., O'loughlin, Valerie Dean, Bidle, Theresa Stouter
Publisher:Mcgraw Hill Education,
Molecular Biology of the Cell (Sixth Edition)
Biology
ISBN:9780815344322
Author:Bruce Alberts, Alexander D. Johnson, Julian Lewis, David Morgan, Martin Raff, Keith Roberts, Peter Walter
Publisher:W. W. Norton & Company
Laboratory Manual For Human Anatomy & Physiology
Biology
ISBN:9781260159363
Author:Martin, Terry R., Prentice-craver, Cynthia
Publisher:McGraw-Hill Publishing Co.
Inquiry Into Life (16th Edition)
Biology
ISBN:9781260231700
Author:Sylvia S. Mader, Michael Windelspecht
Publisher:McGraw Hill Education
The Psychology of Violent Behaviour; Author: Simon Fraser University;https://www.youtube.com/watch?v=iTdqo_7_qLE;License: Standard Youtube License