BIO 1408/09 LAB
18th Edition
ISBN: 9781307081701
Author: Mader
Publisher: MCGRAW-HILL HIGHER EDUCATION
expand_more
expand_more
format_list_bulleted
Concept explainers
Videos
Question
Chapter 9.1, Problem 1A
Summary Introduction
Introduction:
Meiosis is reduction division used for the formation of gametes in males and females, sperm and egg respectively. The meiotic division helps in keeping constant chromosome and gene number by producing haploid gamete which upon fertilization forms zygote with normal diploid chromosomal number. Mitosis is the cell division that results in formation of identical daughter cells and occurs throughout in the body for growth and repair of the cells.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
what are the answer from the book
what is lung cancer why plants removes liquid water intead water vapours
*Example 2: Tracing the path of an autosomal dominant trait
Trait: Neurofibromatosis
Forms of the trait:
The dominant form is neurofibromatosis, caused by the production of an abnormal form of the protein
neurofibromin. Affected individuals show spots of abnormal skin pigmentation and non-cancerous
tumors that can interfere with the nervous system and cause blindness. Some tumors can convert to a
cancerous form. i
The recessive form is a normal protein - in other words, no neurofibromatosis.moovi
A typical pedigree for a family that carries neurofibromatosis is shown below. Note that carriers are not
indicated with half-colored shapes in this chart. Use the letter "N" to indicate the dominant neurofibromatosis
allele, and the letter "n" for the normal allele.
Nn
nn
nn
2
nn
Nn
A
3
N-
Chapter 9 Solutions
BIO 1408/09 LAB
Ch. 9.1 - Explain the purpose of meiosisCh. 9.1 - Prob. 2LOCh. 9.1 - Prob. 3LOCh. 9.1 - Prob. 4LOCh. 9.1 - Prob. 1CYPCh. 9.1 - Prob. 2CYPCh. 9.1 - Prob. 3CYPCh. 9.1 - Prob. 4CYPCh. 9.1 - Prob. 1ACh. 9.1 - Prob. 2A
Ch. 9.2 - List the phases of meiosis, and briefly describe...Ch. 9.2 - Prob. 2LOCh. 9.2 - Prob. 1CYPCh. 9.2 - Prob. 2CYPCh. 9.2 - Prob. 3CYPCh. 9.2 - Prob. 3ACh. 9.2 - Prob. 4ACh. 9.2 - Prob. 5ACh. 9.2 - Prob. 6ACh. 9.2 - Prob. 7ACh. 9.2 - Prob. 8ACh. 9.2 - Prob. 9ACh. 9.2 - Prob. 10ACh. 9.2 - Prob. 11ACh. 9.3 - Contrast the events of meiosis I and meiosis II...Ch. 9.3 - Prob. 2LOCh. 9.3 - Prob. 1CYPCh. 9.3 - Prob. 2CYPCh. 9.3 - Summarize the major differences among the stages...Ch. 9.3 - Prob. 12ACh. 9.3 - Prob. 13ACh. 9.3 - Prob. 14ACh. 9.4 - Prob. 1LOCh. 9.4 - Prob. 2LOCh. 9.4 - Prob. 3LOCh. 9.4 - Describe how a zygote can receive an abnormal...Ch. 9.4 - Contrast monosomy with trisomy.Ch. 9.4 - Prob. 3CYPCh. 9.4 - Prob. 15ACh. 9.4 - Prob. 16ACh. 9 - Prob. 1BYBCh. 9 - Prob. 2BYBCh. 9 - Prob. 3BYBCh. 9 - Prob. S4.4BYBCh. 9 - Section 8.2 What events occur during prophase,...Ch. 9 - Prob. S8.2BYBCh. 9 - Although most men with Klinefelter syndrome are...Ch. 9 - Prob. 2TCCh. 9 - Prob. 3TC
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
- I want to be a super nutrition guy what u guys like recommend mearrow_forwardPlease finish the chart at the bottom. Some of the answers have been filled in.arrow_forward9. Aerobic respiration of one lipid molecule. The lipid is composed of one glycerol molecule connected to two fatty acid tails. One fatty acid is 12 carbons long and the other fatty acid is 18 carbons long in the figure below. Use the information below to determine how much ATP will be produced from the glycerol part of the lipid. Then, in part B, determine how much ATP is produced from the 2 fatty acids of the lipid. Finally put the NADH and ATP yields together from the glycerol and fatty acids (part A and B) to determine your total number of ATP produced per lipid. Assume no other carbon source is available. 18 carbons fatty acids 12 carbons 9 glycerol A. Glycerol is broken down to glyceraldehyde 3-phosphate, a glycolysis intermediate via the following pathway shown in the figure below. Notice this process costs one ATP but generates one FADH2. Continue generating ATP with glyceraldehyde-3-phosphate using the standard pathway and aerobic respiration. glycerol glycerol-3- phosphate…arrow_forward
- Normal dive (for diving humans) normal breathing dive normal breathing Oz level CO2 level urgent need to breathe Oz blackout zone high CO2 triggers breathing 6. This diagram shows rates of oxygen depletion and carbon dioxide accumulation in the blood in relation to the levels needed to maintain consciousness and trigger the urgent need to breathe in diving humans. How might the location and slope of the O₂ line differ for diving marine mammals such as whales and dolphins? • How might the location and slope of the CO₂ line differ for diving marine mammals such as whales and dolphins? • • Draw in predicted lines for O2 and CO2, based on your reasoning above. How might the location of the Urgent Need to Breathe line and the O2 Blackout Zone line differ for diving marine mammals? What physiological mechanisms account for each of these differences, resulting in the ability of marine mammals to stay submerged for long periods of time?arrow_forwardforaging/diet type teeth tongue stomach intestines cecum Insectivory numerous, spiky, incisors procumbentExample: moleExample: shrew -- simple short mostly lacking Myrmecophagy absent or reduced in numbers, peg-likeExample: tamandua anteater extremely long simple, often roughened short small or lacking Terrestrial carnivory sharp incisors; long, conical canines; often carnassial cheek teeth; may have crushing molarsExample: dog -- simple short small Aquatic carnivory homodont, spiky, numerousExample: common dolphin -- simple or multichambered (cetaceans only) variable small or absent Sanguinivory very sharp upper incisors; reduced cheek teethExample: vampire bat grooved tubular, highly extensible long small or lacking Herbivory (except nectivores) incisors robust or absent; canines reduced or absent; diastema; cheek teeth enlarged with complex occlusal surfacesExample: beaver -- simple (hindgut fermenters) or multichambered (ruminants) long large Filter feeding none…arrow_forward3. Shown below is the dental formula and digestive tract anatomy of three mammalian species (A, B, and C). What kind of diet would you expect each species to have? Support your answers with what you can infer from the dental formula and what you can see in the diagram. Broadly speaking, what accounts for the differences? Species A 3/3, 1/1, 4/4, 3/3 པར『ན་ cm 30 Species B 4/3, 1/1, 2/2, 4/4 cm 10 Species C 0/4, 0/0,3/3, 3/3 020arrow_forward
- 3. Shown below is the dental formula and digestive tract anatomy of three mammalian species (A, B, and C). What kind of diet would you expect each species to have? Support your answers with what you can infer from the dental formula and what you can see in the diagram. Broadly speaking, what accounts for the differences? Species A 3/3, 1/1, 4/4, 3/3 cm 30 Species B 0/4, 0/0, 3/3, 3/3 cm 10 Species C 4/3, 1/1, 2/2, 4/4 E 0 cm 20 AILarrow_forwardNormal dive (for diving humans) normal breathing dive normal breathing Oz level CO₂ level urgent need to breathe Oz blackout zone high CO₂ triggers breathing 6. This diagram shows rates of oxygen depletion and carbon dioxide accumulation in the blood in relation to the levels needed to maintain consciousness and trigger the urgent need to breathe in diving humans. • How might the location and slope of the O2 line differ for diving marine mammals such as whales and dolphins? • How might the location and slope of the CO2 line differ for diving marine mammals such as whales and dolphins? • • Draw in predicted lines for O2 and CO2, based on your reasoning above. How might the location of the Urgent Need to Breathe line and the O2 Blackout Zone line differ for diving marine mammals? What physiological mechanisms account for each of these differences, resulting in the ability of marine mammals to stay submerged for long periods of time?arrow_forwardHow much ATP will be produced during the following metabolic scenario: Aerobic respiration of a 5mM lipid solution that is made up of one glycerol and an 8-carbon fatty acid and 12-carbon fatty acid. Recall that when glycerol breaks down to Glyceraldehyde-3-phosphate it costs one ATP but your get an extra FADH2. Every two carbons of a fatty acid break down to one acetyl-CoA. Units cannot be entered in this style of question but the units of your answer should be in mM of ATP.arrow_forward
- If a bacterium using aerobic respiration was to degrade one small protein molecule into 8 molecules of pyruvic acid, how many ATP would that cell make? Assume there is no other carbon source. Units cannot be entered in this style of question but the units of your answer should be in molecules of ATP.arrow_forwardIf a bacterium using aerobic respiration was to degrade a 30 mM solution of citric acid, how many ATP would that cell make? Assume no other carbon source is available. Units cannot be entered in this style of question but the units of your answer should be in mM of ATP.arrow_forwardHow much ATP will be produced during the following metabolic scenario: Aerobic respiration of a 5mM lipid solution that is made up of one glycerol and an 8-carbon fatty acid and 12-carbon fatty acid. Recall that when glycerol breaks down to Glyceraldehyde-3-phosphate it costs one ATP but your get an extra FADH2. Every two carbons of a fatty acid break down to one acetyl-CoA. (pathways will be provided on the exam) Units cannot be entered in this style of question but the units of your answer should be in mM of ATP.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 Cell Cycle and its Regulation; Author: Professor Dave Explains;https://www.youtube.com/watch?v=eqJqhA8HSJ0;License: Standard YouTube License, CC-BY
Cell Division - Mitosis and Meiosis - GCSE Biology (9-1); Author: Mr Exham Biology;https://www.youtube.com/watch?v=w7vp_uRA8kw;License: Standard YouTube License, CC-BY