BIOLOGY
5th Edition
ISBN: 9781264104680
Author: BROOKER
Publisher: MCG
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Chapter 17.2, Problem 1CS
Summary Introduction
To explain: The relationship between sexual reproduction and homologous chromosomes.
Introduction: Sexual reproduction is the process of producing a living organism by combining genetic information from two separate organisms such as one male and one female. In this type of reproduction, both the parents synthesize a haploid cell that contains a single set of the chromosome and these haploid cells fuse with each other to form a diploid cell. This type of reproduction is importance fo promoting variations in the genetic material of the population.
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Lab Prelab 1.4
Name:
Define the following terms:
Note: the definitions don't necessarily need to be long or even complete sentences as long as they are correct. Also
some of these terms have different definitions depending on the source you choose. If you see multiple definitions,
choose the one that makes the most sense to you.
Chromosome:
Sister chromatid:
Gene:
Genome:
Genotype:
Phenotype:
Mitosis:
Meiosis:
Centromere:
Centrosome:
What is the difference between the diploid and haploid numbers of chromosomes?
INSTRUCTION
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QUESTION
1. Rarely, both sister chromatids of a replicated chromosome end up in one daughter cell. How might this happen? What could be the consequences of such a mitotic error?
MITOSIS & MEIOSIS COMPARISON WORKSHEET
MITOSIS
MEIOSIS
COMPARISON
number of parent cells
number of divisions
# of daughter cells produced
(in animals)
size of daughter cells relative
to parent cell
# of chromosomes in parent
# of chromosomes per
daughter cell at end of
process
genetic comparison:
chromosomes in parent cell
versus daughter cell
location of process in
organism
(provide examples of specific
cells)
function of process
sexual or asexual
reproduction?
advantages
disadvantages
Chapter 17 Solutions
BIOLOGY
Ch. 17.1 - Prob. 1CCCh. 17.1 - Prob. 2CCCh. 17.1 - Prob. 3CCCh. 17.1 - Prob. 4CCCh. 17.1 - Mendels Laws of Inheritance Concept Check: What...Ch. 17.2 - Prob. 1CSCh. 17.2 - Prob. 2CSCh. 17.2 - Prob. 1CCCh. 17.3 - Prob. 1CCCh. 17.3 - Prob. 2CC
Ch. 17.4 - Prob. 1CCCh. 17.4 - Prob. 1EQCh. 17.4 - Prob. 2EQCh. 17.4 - Prob. 3EQCh. 17.5 - Prob. 1CSCh. 17.5 - Prob. 1CCCh. 17.6 - Prob. 1CCCh. 17 - Prob. 1TYCh. 17 - During which phase of nuclear division does the...Ch. 17 - Prob. 3TYCh. 17 - Which of Mendels laws cannot be observed in a...Ch. 17 - During a __________blank, an individual with the...Ch. 17 - Prob. 6TYCh. 17 - Prob. 7TYCh. 17 - A hypothetical flowering plant species produces...Ch. 17 - Genes located on a sex chromosome are said to be...Ch. 17 - Prob. 10TYCh. 17 - Prob. 1CQCh. 17 - A cross is made between individuals having the...Ch. 17 - Core Concept: Systems We can view life as a...Ch. 17 - Discuss the principles of the chromosome theory of...Ch. 17 - When examining a human pedigree, what patterns do...
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- Discussion 1 pts Meets Standards A complete and accurate description of the difference between sister chromatids and homologous chromosomes is provided. Accurate definitions for sister chromatids and homologous chromosomes are provided.arrow_forwardObjective: Get a sense of how genomics, the study of the genome in its entirety,needs to think about how to go about its research. Geonomic DNA is broken up into fragments. The 5’ and 3’ ends of each fragment(a “read”) are sequenced. The sequenced reads are assembled together intocontiguous sequences (“contigs”) based on sequence similarity. The idea is to sequence enough random fragments so that every nucleotide in thegenome is represented on some read. The number of such fragments needed iscalled the coverage, c. The coverage c can be calculated by the formula RL/G, where R is the number ofreads sequenced, L is the average length of a read and G is the total length of thegenome. The units of length are bases (b) or base pairs (bp). Consider a genome whose length is 1000 bp. “Shotgun” sequencing techniquesare applied to the genome, resulting in 20 reads, with an average length of 50 bp.A very important point is that, even though 20 x 50 = 1000, there is no guaranteethat ALL…arrow_forwardObjective: Get a sense of how genomics, the study of the genome in its entirety,needs to think about how to go about its research. Geonomic DNA is broken up into fragments. The 5’ and 3’ ends of each fragment(a “read”) are sequenced. The sequenced reads are assembled together intocontiguous sequences (“contigs”) based on sequence similarity. The idea is to sequence enough random fragments so that every nucleotide in thegenome is represented on some read. The number of such fragments needed iscalled the coverage, c. The coverage c can be calculated by the formula RL/G, where R is the number ofreads sequenced, L is the average length of a read and G is the total length of thegenome. The units of length are bases (b) or base pairs (bp). Consider a genome whose length is 1000 bp. “Shotgun” sequencing techniquesare applied to the genome, resulting in 20 reads, with an average length of 50 bp.A very important point is that, even though 20 x 50 = 1000, there is no guaranteethat ALL…arrow_forward
- Why is meiosis useful? Select all that apply 1. I produces genetically variable offspring 2. it maintains the chromosome number between generations 3. it reduces the frequency of mutations 4. it provides an alternative route to cell division for cells that fail to complete mitosis 5. It does not require proteinsarrow_forward• Deletion chromosomes are useful for what purpose?arrow_forwardmat Arrange Tools Slide Show Window Help A 12% D Sat 2:02 PM 2 Chapter 13 clicker questions imations Slide Show Review View Shape Format O Tell me e Share V Play Narrations Always Use Subtitles V Use Timings Rehearse Record Timings Slide Show V Show Media Controls A Subtitle Settings v What allows sister chromatids to finally separate, and in which phase of meiosis does this occur? a) release of cohesin along sister chromatid arms in anaphase I b) crossing over of chromatids in prophase I c) release of ohesin at centromeres in anaphase I d) release of cohesin at centromeres in anaphase II e) crossing over of homologs in prophase I O 2017 Pearson Education, Inc 103% :: E Notes Comments étv MacBook Pro 80 888 * ( & $ 8 9 6 7arrow_forward
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