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Performance Benchmark L.12.A.1
Students know genetic information passed from parents to offspring is coded in the DNA
molecule. E/S
During the first half of the 20th century a debate among biologists centered around whether
proteins or deoxyribonucleic acid (DNA) was the molecule of inheritance. In 1928, Fredrick
Griffith first proposed that the transfer of DNA between bacteria caused transformation, but
could not provide convincing proof. In the 1940s Oswald Avery and others were able to devise
experiments that provided the evidence that Griffith lacked. However it took an experiment by
Alfred Hershey and Martha Chase in 1952 to finally demonstrate to the scientific community that
DNA and not proteins was the molecule of inheritance.
Today we know that the coded message for our traits is based in the four nucleotides of DNA.
These four bases adenine (A), thymine (T), guanine (G), and cytosine (C) are divided into two
groups called purines (A & G) and pyrimidines (T & C). The Human Genome Project has
confirmed that the DNA in a typical human cell contains over 3 billion base pairs (bp). In these 3
billion bp are 20,000 to 25,000 genes that code for proteins, which in turn code for our traits.
Each gene is a specific sequence of nucleotides located on one of the DNA strands.
The DNA in the human body is spread over 24 distinct chromosomes which range in size from
50 million to 250 million bp.
To learn more about the findings of the Human Genome Project go to
http://www.ornl.gov/sci/techresources/Human_Genome/home.shtml
DNA is a nucleic acid that contains nucleotides running in two strands and twisted into a double
helix. Each nucleotide contains three molecules (a base, a deoxyribose sugar and a phosphate
group). The nucleotides pairings are shown in the illustration below. These pairings also referred
as complementary base pairs.
Figure 1: The DNA molecule showing base pairing.
http://student.ccbcmd.edu/~gkaiser/biotutorials/dna/images/u4fg8f.jpg
In the illustration above you will see that the base adenine is paired with the base thymine and
the base guanine is paired with the base cytosine. The idea of these pairing was first discovered
by Edwin Chargaff in the late 1940’s. In working with cells from different organisms Chargaff
discovered that the percentages of adenine in a cell were equal to the percentages of thymine,
and the percentages of guanine were equal to the percentages of cytosine in the nucleus. These
pairing have become known as “Chargaff’s Rule”. However the meaning of this discovery was
not clear until Watson and Crick developed their mode of DNA in 1953.
The genetic code in DNA is passed along via mitosis, meiosis or binary fission. Prior to these
processes the molecules of DNA in the parent cell must be copied via DNA replication. In their
1953 paper Watson and Crick proposed that each strand of the DNA molecule makes a
complementary copy of itself through DNA replication prior to cell division. In 1957, Meselson
and Stahl devised an experiment that demonstrated this semi-conservation nature of DNA
replication as first proposed by Watson and Crick. In this process each strand from the original
DNA molecule gets a new complementary strand. Thus each new DNA molecule has 1 (one)
strand from the “old” molecule and one “new” strand that is an exact copy of the original.
Figure 2: The semi-conservative nature of DNA replication. http://fig.cox.miami.edu/~cmallery/150/gene/sf12x1.jpg
To learn more about the history of DNA’s discovery go to
http://www.dnai.org/
Before the genetic code is passed from parents to offspring via meiosis or from one cell to
another new cell via mitosis the DNA must be replicated. As illustrated in Figure 2 each strand
of the DNA makes a complementary copy of itself.
In a simplified view the copying can be seen in Figure 3. The DNA “unzips” and complementary
bases are brought to each strand and the new stands “rezip” forming two identical copies of the
original DNA molecule.
Figure 3: A simplified view of DNA replication. http://library.thinkquest.org/18617/media/replication-simple.gif
Biologists have discovered that the actual process is far more complicated. First a molecule
called helicase unwinds the DNA double helix. Before DNA polymerase travels along each
strand matching complementary bases, a short sequence of RNA nucleotides is matched with the
separated strands by RNA primase. The copying of each strand, however, is different. One strand
called the leading strand is copied in a continuous fashion, while the other called the lagging
strand is copied in fragments called Okazaski fragments as seen in the illustration below. Later
DNA ligase will join these fragments.
Figure 4: The replication of DNA along the leading and lagging strands.
http://fig.cox.miami.edu/~cmallery/150/gene/c7.16.14.fork.jpg
The rate at which new nucleotides are added is about 50 per second and would take 53 days to
replicate the largest human chromosome if replication began at one end and proceeded to the
other end. As such the replication of any chromosome begins at many origins along the
chromosome. In fruit fly chromosomes there are some 3500 origin sites where DNA replication
begins simultaneously.
To learn more and to view an animation of DNA replication go to
http://www.johnkyrk.com/DNAreplication.html
and, http://www.stolaf.edu/people/giannini/flashanimat/molgenetics/dna-rna2.swf
and, http://www.emc.maricopa.edu/faculty/farabee/biobk/BioBookDNAMOLGEN.html
The later is a part of the online textbook “Online Biology Book”. Numerous diagrams are
included along with a description of DNA structure and replication. Links are also provided to
other websites.
Performance Benchmark L.12.A.1
Students know genetic information passed from parents to offspring is coded in the DNA
molecule. E/S
Common misconceptions associate with this benchmark:
1. Students incorrectly assume that DNA is vastly different amongst members of the same
species.
Despite the fact that humans contain over 3 billion bp in their DNA, researchers have found that
most of DNA is quite similar. Based on sequencing to date it appears that on average two
unrelated people have one different nucleotide per 1000 bases. Thus with 3 billion bp total bases
this means there are 3 million differences between individuals or less than 0.01% difference
between individuals. These differences are called single nucleotide polymorphisms or SNPs
(pronounced “snips”)
Aside from the fact each human has a unique combination of genes; these genes are shared by all
members of the human species. The goal of the Human Genome Project is to identify these genes
and than determine what each gene codes for in humans.
To learn more about SNPs go to
http://www.ornl.gov/sci/techresources/Human_Genome/faq/snps.shtml
2. Students mistakenly assume that DNA coding for molecules is different between
different species.
While a species has a unique gene pool that defines that species, many genes are shared by
humans and other organisms. For example 45% of the genes found in fruit flies are also found in
humans and we share approximately 96% of our genes with chimpanzees. This should not be
surprising considering the number of biochemical pathways that are commonly found in
organisms. For example most organisms obtain energy or ATP by cellular respiration and the
enzymes (coded for by DNA) involved in this biochemical pathway are found in most
organisms.
To learn more about the similarity of human and chimpanzee DNA go to
http://news.nationalgeographic.com/news/2005/08/0831_050831_chimp_genes.html
3. Students incorrectly assume that mutations in DNA are always harmful.
Single base errors in DNA copying are called point mutations, however, these are rare. During
DNA replication the error rate is 1 in 10,000 bases being copied. Most of these errors are
corrected by DNA proof readers. Secondly if an error is not corrected, because of the redundancy
of the genetic code the same amino acid may be coded for by the codon. This is called a silent
mutation. Or an amino acid with similar properties can be coded for by the “mutant” codon
which is sometimes called a neutral mutation. Note the table below.
At the same time some point mutations can be harmful. In the table below the DNA triplet and
mRNA codon are shown for the 6th amino acid for normal hemoglobin. In sickle-cell anemia a
mutation from T to A leads to the replacing of glutamic acid with valine. The result is an
abnormally shaped hemoglobin molecule during low oxygen concentration in the blood. And if
the first C is replace by A, the result is the termination of protein synthesis and no hemoglobin
molecule is produced.
DNA
Triplet
CTC
CTT
CTA
CAC
ATC
mRNA
codon
GAG
GAA
GAU
GUG
UAG
Amino Acid
Properties
Mutation Type
Glutamic acid
Glutamic acid
Aspartic acid
Valine
Stop
Hydrophobic
Hydrophobic
Hydrophobic
Hydrophilic
Termination
Normal codon
Neutral
Silent
Missense
Nonsense
All humans with blood type O are also carrying a mutation. The genes for blood type code for
proteins found on the red blood cell. Many inherit the genes for the A and B proteins. However
due to a point mutation in our ancestral past the coding for these proteins was lost, thus those
who inherit the alleles for O lack coding for either protein.
To learn more about DNA mutations go to
http://www.genetichealth.com/G101_Changes_in_DNA.shtml#Anchor2
http://evolution.berkeley.edu/evolibrary/article/0_0_0/mutations_01
4. Students incorrectly assume that DNA and chromosomes are not the same.
Both of these are names for the same molecule. Humans have 46 chromosomes and each
chromosome contains a specific sequence of DNA nucleotides which are codes for our genes.
However, each chromosome contains a unique combinations of genes. Genes found in
Chromosome #1 are not the same as those found in chromosome #2..
To learn more about what genes have been discovered on each human chromosome and to order
a free poster showing human genome landmarks go to
http://www.ornl.gov/sci/techresources/Human_Genome/posters/chromosome/
5. Students inaccurately assume that the amount of genetic material is equal in males and
females.
While each offspring, regardless of sex, receives 23 chromosomes from each parent, the amount
of DNA and the genes they receive is not equal. Both males and female receive the same 22 pairs
of autosomes (non-sex chromosomes) and therefore each receives equal number of genes for
non-sexual characteristics found on these chromosomes. However the sex chromosomes (X and
Y) do not contain equal number of genes or types of genes.
The X chromosome represents approximately 5% of the total DNA in cells and contains
approximately 1300 genes. These 1300 genes not only include genes for femaleness, but genes
for such traits as blood clotting and color vision. On the other hand the Y chromosome contains
approximately 2% of the DNA in a cell or approximately 300 genes which will be inherited by
males only, since females do not inherit a Y chromosome. Thus males inherit both the X and Y
chromosomes, they will inherit all genes for the same traits as females, but females will lack any
genes found on the Y chromosome.
To learn more about the X and Y chromosomes go to
http://ghr.nlm.nih.gov/ghr/chromosomes
6. Students incorrectly believe that the amount of DNA varies in organisms based upon
their “complexity”
While is true that humans have approximately 1000 times more DNA than a typical bacterium,
the Human Genome Project has reveled some interesting surprises. Humans have approximately
3.0 X 109 base pairs (bp) and 20,000 to 25,000 genes. But notice the figures for other organisms.
Organism
Rice
Maize
Mouse
Whisk fern
Sea urchin
C. elegans
DNA base pairs
3.9 X 109
2.5 X 109
2.5 X 109
2.5 X 1011
8.14 X108
1 X 108
Approximate # of Genes
37,000
Over 50,000
22,500
?
23,000
19,000
Chromosome #
12
20
40
?
44
12
To view more examples, go to
http://users.rcn.com/jkimball.ma.ultranet/BiologyPages/G/GenomeSizes.html#Anopheles
Performance Benchmark L.12.A.1
Students know genetic information passed from parents to offspring is coded in the DNA
molecule. E/S
Sample Test Questions
1. In Chargaff’s rule, the base-pairing rule, state that in DNA
a. the amount of adenine equals the amount of thymine.
b. the amount of guanine equals the amount of thymine.
c. the amount of cytosine equals the amount of thymine.
d. the amount of adenine equals the amount of guanine
2. Which one of the following nucleotide pair bonds would be found in a DNA molecule?
a. cytosine-guanine
b. adenine-cytosine
c. cytosine-uracil
d. adenine-guanine
3. The base-pairing rules state that the following are base pairs in DNA:
a. adenine—thymine; uracil—cytosine.
b. adenine—thymine; guanine—cytosine.
c. adenine—guanine; thymine—cytosine.
d. uracil—thymine; guanine—cytosine.
4. During DNA replication, a complementary strand of DNA is made for each original DNA
strand. Thus, if a portion of the original strand is CCTAGCT, then the new strand will be
a. TTGCATG
b. CCTAGCT
c. AAGTATC
d. GGATCGA
5. Which statement below correctly describe the relationship between genome size and organism
“complexity”
a. As organisms get more complex their genomes get larger.
b. Organism of similar complexity have genomes of similar size.
c. There appears to be little relationship between complexity and genome size.
d. Individuals of the same species have vastly different genomes.
Performance Benchmark L.12.A.1
Students know genetic information passed from parents to offspring is coded in the DNA
molecule. E/S
Answers to Sample Test Questions
1.
2.
3.
4.
5.
(a)
(a)
(b)
(d)
(b)
Performance Benchmark L.12.A.1
Students know genetic information passed from parents to offspring is coded in the DNA
molecule. E/S
Intervention strategies and resources
The following is a list of intervention strategies and resources that will facilitate student
understanding of this benchmark.
1. Movie on the discovery of DNA and its molecular structure
To understand and appreciate the discovery of DNA as the molecule of inheritance teachers can
begin by showing the video Biologix: Development of Molecular Genetics which last 29 minutes.
It is divide into two parts. The first part talks about the works of Griffith, Avery, and
Hershey/Chase that lead to the establishment of DNA as the molecule of inheritance. In the
second part the work Levene, Chargaff, Franklin and Watson/Crick is and their efforts to
establish the structure of DNA. This video is useful for all levels of biology.
To find this video got the KLVX Video Streaming website at
http://www.klvx.org/
2. The building of the DNA model
One of the important aspects of science is to build model molecular structures that scientists are
trying to understand. In their efforts to discover the structure of DNA Watson and Crick resorted
to a variety of models. To help students appreciate this aspect of science students can also build
their own models of DNA.
To learn more about building DNA models and for lesson plans go to
http://www.ncbe.reading.ac.uk/DNA50/cutout.html
http://www.ncbe.reading.ac.uk/NCBE/PROTOCOLS/DNA/modelling.html
http://biology.about.com/library/bldnamodels.htm
3. DNA Extraction experiments
Along with model building students can very easily perform DNA extraction experiments at little
cost using household items. The Genetic Science Learning Center website provides a simple and
straight forward protocol for extracting DNA from wheat germ. In addition it provides
information as to why various items are used in the protocol. For example it explains the role of
detergent in extracting the DNA from the wheat germ. This site also has a virtual DNA
extraction lab.
To review the virtual DNA extraction lab at the Genetics Science Learning Center, go to
http://www.teachersfirst.com/getsource.cfm?id=5346
4. DNA fingerprinting activity
To understand that our genes are passed from parent to offspring the teacher could have the
students look at the role of DNA fingerprinting in helping to identify individuals. At this Nova
Online site students can play the role of detective and use genetic fingerprinting to solve a crime.
To review the DNA fingerprinting lab at Nova Online, go to
http://www.pbs.org/wgbh/nova/sheppard/analyze.html