To determine:
The type of cell division that is represented in the figure and the relationship between the parts of figure.
Introduction:
Meiosis, also known as the reductional division, is the second type of cell division that takes place in eukaryotes. The cell division is marked by a decrease in the chromosome number to half. This type of cell division allows the recombination of the genes on the chromosome to produce a zygote. The two diploid cells undergo division to produce four haploid daughter cells.
To determine:
The difference between cohesion of centromeres and cohesion of arms.
Introduction:
The chromosomes are condensed structures that are formed during the early phases of cell division from the loose network of chromatin thread and then regain their original structure after being divided into daughter cells.
To determine:
The mechanism that keeps the chromosomes together during metaphase.
Introduction:
The human consists of 23 pairs of chromosomes, condensed form of chromatids which divide during cell division into daughter cells. The human has 22 autosomes and one pair of sex chromosomes. The sex chromosomes determine the sex in an individual based on the type of sex chromosomes that are present in the fusing gametes.
To determine:
The role of shugoshin and the enzyme it inhibits.
Introduction:
The chromosome from chromatids are formed through the process of supercoiling. The chromatin fibres are condensed into thick structures called the chromosomes. This chromosome in bacteria is circular whereas in eukaryotes it is linear.

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Chapter 12 Solutions
GENETICS:FROM GENES TO GENOMES(LL)-PKG
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- 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
- Endospore 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_forwardReactunts C6H12O6 (Glucose) + 2NAD+ + 2ADP 2 Pyruvic acid + 2NADH + 2ATP a. Which of the above are the reactants? b. Which of the above are the products? c. Which reactant is the electron donor? GHz 06 (glucose) d. Which reactant is the electron acceptor? NAD e. Which of the products have been reduced? NADH f. Which of the products have been oxidized? g. Which process was used to produce the ATP? h. Where was the energy initially in this chemical reaction and where is it now that it is finished? i. Where was the carbon initially in this chemical reaction and where is it now that it is finished? j. Where were the electrons initially in this chemical reaction and where is it now that it is finished? 3arrow_forwardThere is ________ the concept of global warming. Very strong evidence to support Some strong evidence to support Evidence both supporting and against Evidence againstarrow_forward
- How many types of reactions can an enzyme perform?arrow_forwardYour goal is to produce black seeds resistant to mold. So you make the same cross again (between a homozygous black seeded, mold susceptible parent and a homozygous white seeded and mold resistant parent), and, again, advance progeny by SSD to create 100 F10 generation plants. Based on the information you obtained from your first crossing experiment (Question #4), how many F10 plants would you expect to have black seeds and be resistant to mold? Assume that a toxin produced by the mold fungus has been isolated. Only mold resistant seeds will germinate in the presence of the toxin. Could you use this toxin screening procedure to have segregation distortion work in your favor in the F2 generation? Explain your answer. Info from Question 4 a. P Locus (Seed Color): Hypothesis: The null hypothesis (H₀) is that seed color is controlled by alleles at a single locus. Observed Data: Total white seeds: 45 (resistant plants) + 6 (susceptible plants) = 51 Total black seeds: 7 (resistant…arrow_forward10. Consider the following enzyme and its substrate where the "+" and "-" indicate cations and anions, respectively. Explain which of the following inhibitors could inhibit this enzyme? Which type of inhibitor would it be and why? (Video 5-2) Substrate Enzyme Potential inhibitorsarrow_forward
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