Human Biology: Concepts and Current Issues - With Access (Custom)
15th Edition
ISBN: 9781323045237
Author: Johnson
Publisher: PEARSON
expand_more
expand_more
format_list_bulleted
Videos
Question
Chapter 18, Problem 2TY
Summary Introduction
Introduction:
Metastasis is the process of spreading of the pathogenic agent from the primary site to the secondary site in the human body. The secondary malignant growths in cases of cancer may perform metastasis. The most common metastasis locations are lungs, liver, bones, brain, the lining of the abdomen or pelvis and other abdominal organs such as the uterus, bladder, and large intestine. Metastasizing cancer keeps on spreading until it completely overruns all the tissues, organs and organ system.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
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.
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.
Chapter 18 Solutions
Human Biology: Concepts and Current Issues - With Access (Custom)
Ch. 18 - Prob. 1QCCh. 18 - Prob. 2QCCh. 18 - Prob. 3QCCh. 18 - Compare and contrast a benign tumor and a...Ch. 18 - Prob. 2CRCh. 18 - Prob. 3CRCh. 18 - Explain why we have not yet made much progress...Ch. 18 - Prob. 5CRCh. 18 - Prob. 6CRCh. 18 - Prob. 7CR
Ch. 18 -
8. Describe how tumors are diagnosed.
Ch. 18 - Prob. 9CRCh. 18 - Prob. 10CRCh. 18 - Prob. 1TYCh. 18 - Prob. 2TYCh. 18 - Prob. 3TYCh. 18 -
4. Which of the following statements regarding...Ch. 18 - Prob. 5TYCh. 18 - Prob. 6TYCh. 18 - Which of the following cancer treatments would be...Ch. 18 - Prob. 8TYCh. 18 - Prob. 9TYCh. 18 -
10. The ABCD rule refers to the evaluation of:
a....Ch. 18 - The most common cause(s) of cancer deaths in the...Ch. 18 - Which of the following statements about breast...Ch. 18 - Prob. 13TYCh. 18 - Prob. 14TYCh. 18 - Prob. 15TYCh. 18 -
1. Why do you suppose that the death rate from...Ch. 18 - Prob. 2AWKCh. 18 - Prob. 3AWKCh. 18 - Prob. 4AWKCh. 18 - Common therapies for cancer include chemotherapy...Ch. 18 - The first cancer that can be nearly completely...
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
- Design 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_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_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_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_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
- Which 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_forwardDuring 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_forward
- Explain 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_forwardEndospore stain: Consider tube 2 of the 7-day bacillus culture. After is was heated, it was incubated for 24 hours then refrigerated. Do you think the cloudiness in this tube is due mostly to vegetative cells or to endospores? Explain your reasoningarrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Comprehensive Medical Assisting: Administrative a...NursingISBN:9781305964792Author:Wilburta Q. Lindh, Carol D. Tamparo, Barbara M. Dahl, Julie Morris, Cindy CorreaPublisher:Cengage Learning
Medical Terminology for Health Professions, Spira...Health & NutritionISBN:9781305634350Author:Ann Ehrlich, Carol L. Schroeder, Laura Ehrlich, Katrina A. SchroederPublisher:Cengage Learning
Human Biology (MindTap Course List)BiologyISBN:9781305112100Author:Cecie Starr, Beverly McMillanPublisher:Cengage Learning
Concepts of BiologyBiologyISBN:9781938168116Author:Samantha Fowler, Rebecca Roush, James WisePublisher:OpenStax College
Comprehensive Medical Assisting: Administrative a...
Nursing
ISBN:9781305964792
Author:Wilburta Q. Lindh, Carol D. Tamparo, Barbara M. Dahl, Julie Morris, Cindy Correa
Publisher:Cengage Learning
Medical Terminology for Health Professions, Spira...
Health & Nutrition
ISBN:9781305634350
Author:Ann Ehrlich, Carol L. Schroeder, Laura Ehrlich, Katrina A. Schroeder
Publisher:Cengage Learning
Human Biology (MindTap Course List)
Biology
ISBN:9781305112100
Author:Cecie Starr, Beverly McMillan
Publisher:Cengage Learning
Concepts of Biology
Biology
ISBN:9781938168116
Author:Samantha Fowler, Rebecca Roush, James Wise
Publisher:OpenStax College
cell division of meiosis and mitosis; Author: Stated Clearly;https://www.youtube.com/watch?v=A-mFPZLLbHI;License: Standard youtube license