Concept explainers
Videos
To write:
The steps for cell division to result in an organism with trisomy
Introduction:
Reproductive cells which pass on genetic traits from the parent to child are produced by the process of meiosis. Gametes are sex cells and have half the number of chromosomes.When a male gamete fuses with a female gamete, the process is called fertilization. Fertilization allows
Answer to Problem 14STP
Trisomy occurs due to nondisjunction either in meiosis I or meiosis II.
- Nondisjunction in meiosis I results in two daughter cells, one with ( n -1) number of chromosomes and the other with ( n +1) number of chromosomes. In meiosis II, four daughter cells or gametes are formed, 2 with ( n -1) number of chromosomes and 2 with ( n +1) number of chromosomes. Fertilization of ( n +1) gamete with male gamete will result in trisomy ( 2n +1).
- Nondisjunction in meiosis IIresults in four daughter cells, two normal haploid cells, one with ( n -1) number of chromosomes and the other with ( n +1) number of chromosomes. Fertilization of ( n +1) gamete with male gamete will result in trisomy ( 2n +1).
Explanation of Solution
Nondisjunction during meiosis may result in gametes with abnormal number of chromosomes. Cell division during which sister chromatids fail to separate properly is called nondisjunction. If an offspring has a set of three chromosomes of one kind, it is called trisomy. This can be due to nondisjunction either in meiosis I or meiosis II.
Meiosis I- During anaphase I, if nondisjunction occurs, it will result in failure of tetrads to separate. At the end of meiosis I there will be two daughter cells, one with ( n -1) number of chromosomes and the other with ( n +1) number of chromosomes. Now both these daughter cells will undergo meiosis II resulting in the formation of four daughter cells or gametes, 2 with ( n -1) number of chromosomes and 2 with ( n +1) number of chromosomes.
When ( n +1) gamete fuses with a male gamete during fertilization, it will result in trisomy with ( 2n +1) number of chromosomes. When ( n -1) gamete fuses with a male gamete during fertilization, it will result in monosomy with ( 2n -1) number of chromosomes.
Meiosis II- At the end of meiosis I there will be two daughter cells, each with n number of chromosomes. If nondisjunction occurs during meiosis II, it will result in failure of sister chromatids to separate during anaphase II. This will result in the formation of four daughter cells or gametes, 2 with n number of chromosomes, 1 with ( n +1) number of chromosomes and 1 with n-1) number of chromosomes.
When ( n +1) gamete fuses with a male gamete during fertilization, it will result in trisomy with ( 2n +1) number of chromosomes. When ( n -1) gamete fuses with a male gamete during fertilization, it will result in monosomy with ( 2n -1) number of chromosomes. Two gametes are normal haploid ( n ), so they will fertilize to form diploid zygotes.
Chapter 11 Solutions
EP BIOLOGY 2012-STUDENTWORKS ONLINE
Additional Science Textbook Solutions
Chemistry: The Central Science (14th Edition)
Campbell Biology: Concepts & Connections (9th Edition)
Physics for Scientists and Engineers: A Strategic Approach, Vol. 1 (Chs 1-21) (4th Edition)
Microbiology: An Introduction
Organic Chemistry (8th Edition)
Human Anatomy & Physiology (2nd Edition)
- 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.arrow_forwardYou 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?arrow_forwardYou 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.arrow_forward
- 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
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