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Chapter 2: Problem 16

The broad bean (Vicia faba) is diploid and \(2 n=18\). Each haploid chromosome set contains approximately \(4 \mathrm{m}\) of DNA. The average size of each chromosome during metaphase of mitosis is \(13 \mu \mathrm{m}\). What is the average packing ratio of DNA at metaphase? (Packing ratio = length of chromosome/length of DNA molecule therein.) How is this packing achieved?

Short Answer

Expert verified
The average packing ratio of DNA at metaphase is approximately \(2.925 \times 10^{-5}\). This compact structure is achieved through coiling of DNA around histones, forming nucleosomes, and higher-order coiling.

Step by step solution

01

Identify the DNA Length in Chromosomes

The total length of the DNA in one haploid set of chromosomes for Vicia faba is given as approximately 4 meters. Since the organism is diploid, there are two such sets in each somatic cell, so the total DNA length in a diploid cell is approximately \(2 \times 4 = 8\text{ meters}.\)
02

Calculate the Total Chromosome Length at Metaphase

Given that each chromosome during metaphase is on average 13 micrometers in length, and there are 18 chromosomes in total in a diploid cell (\(2n = 18\)), the total length of all chromosomes at metaphase is \(18 \times 13 = 234 \text{ micrometers}.\)
03

Convert Units for Consistency

To compare the lengths, convert the total metaphase chromosome length from micrometers to meters. Since \(1 \mu\text{m} = 1 \times 10^{-6} \text{m}\), the total metaphase chromosome length in meters is \(234 \times 10^{-6} \text{ meters} = 0.000234 \text{ meters}.\)
04

Compute the Packing Ratio

Calculate the packing ratio using the formula: \( \text{Packing ratio} = \frac{\text{Total metaphase chromosome length (m)}}{\text{Total DNA length (m)}} = \frac{0.000234}{8} \approx 2.925 \times 10^{-5}.\)
05

Explain How The Packing is Achieved

The packing of DNA into chromosomes involves several levels of coiling and compaction: 1. DNA wraps around histone proteins to form nucleosomes ("beads on a string" structure). 2. Nucleosomes coil to form a 30 nm chromatin fiber. 3. The 30 nm fiber loops to form loops attached to a scaffold, further coiling during mitosis to form visible metaphase chromosomes.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chromosome Metaphase
When cells undergo mitosis, they enter a stage called metaphase. During this stage, chromosomes align in the center of the cell, preparing for separation.
This orderly alignment is critical for equal distribution of genetic material to the daughter cells. In this phase, chromosomes are condensed and highly organized, making them visible under a microscope.
In the context of our problem, the key point is the remarkable degree of DNA compaction visible during metaphase. Understanding how this compaction occurs helps unravel the intricacies of genetic regulation and stability in cell division.
Nucleosome Structure
DNA's immense length poses a storage challenge within the cell nucleus. Nature resolves this with the nucleosome structure. Think of nucleosomes as nature鈥檚 microscopic filing system.
Within the nucleus, DNA wraps around proteins called histones. This wrapping creates bead-like structures along the DNA strand, known as nucleosomes or the "beads on a string" formation.
This fundamental unit plays an instrumental role in organizing DNA into a more compact form. This wrapping also aids in the regulation of gene expression and access to genetic information when needed by the cell.
Chromatin Fiber
After forming nucleosomes, DNA undergoes further organization into a thicker structure called chromatin fiber. It helps in managing DNA within the confined space of the nucleus.
These nucleosomes coil further to form a 30-nanometer fiber, adding another layer of DNA compaction. Imagine this as coiling a string of beads into a more dense thread.
The coiling of chromatin fibers is dynamic, allowing DNA to be packed during cell division while enabling access for necessary genetic processes during other times. This delicately balanced structure supports the cell's genetic needs at various stages of its life cycle.
DNA Molecule Compaction
The packaging of DNA from its extended length into a compact chromosome is an extraordinary process. Initially, the DNA forms nucleosomes by wrapping around histone proteins.
Further compaction is achieved as nucleosomes coil into chromatin fibers. These fibers are then looped and folded into even more compact structures during cell division.
Finally, during metaphase, chromosomes reach their most compact and structured form. This high-level organization ensures that the long DNA strands efficiently fit within the cell nucleus, ready for orderly distribution to daughter cells during cell division.
Thus, through a hierarchical structure of folding and coiling, life resolves its cellular storage challenge.

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Most popular questions from this chapter

A geneticist working on peas has a single plant monohybrid \(Y / y\) (yellow) plant and, from a self of this plant, wants to produce a plant of genotype \(y / y\) to use as a tester. How many progeny plants need to be grown to be 95 percent sure of obtaining at least one in the sample?

Human cells normally have 46 chromosomes. For each of the following stages, state the number of nuclear DNA molecules present in a human cell: a. Metaphase of mitosis b. Metaphase I of meiosis c. Telophase of mitosis d. Telophase I of meiosis e. Telophase II of meiosis

The recessive allele \(s\) causes Drosophila to have small wings, and the \(s^{+}\) allele causes normal wings. This gene is known to be X linked. If a small-winged male is crossed with a homozygous wild-type female, what ratio of normal to small-winged flies can be expected in each sex in the \(\mathrm{F}_{1}\) ? If \(\mathrm{F}_{1}\) flies are intercrossed, what \(\mathrm{F}_{2}\) progeny ratios are expected? What progeny ratios are predicted if \(\mathrm{F}_{1}\) females are backcrossed with their father?

A Neurospora colony at the edge of a plate seemed to be sparse (low density) in comparison with the other colonies on the plate. This colony was thought to be a possible mutant, and so it was removed and crossed with a wild type of the opposite mating type. From this cross, 100 ascospore progeny were obtained. None of the colonies from these ascospores was sparse, all appearing to be normal. What is the simplest explanation of this result? How would you test your explanation? (Note: Neurospora is haploid.)

A plant geneticist has two pure lines, one with purple petals and one with blue. She hypothesizes that the phenotypic difference is due to two alleles of one gene. To test this idea, she aims to look for a 3: 1 ratio in the \(\mathrm{F}_{2} .\) She crosses the lines and finds that all the \(\mathrm{F}_{1}\) proge ny are purple. The \(\mathrm{F}_{1}\) plants are selfed, and \(400 \mathrm{F}_{2}\) plants are obtained. Of these \(\mathrm{F}_{2}\) plants, 320 are purple and 80 are blue. Do these results fit her hypothesis well? If not, suggest why.
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