RECAP 17.1
Before sequencing, a long DNA molecule is cut into many fragments. These cuts are random (i.e., they are made at different random locations). The overlaps are important in arranging the final sequence. If there were only one type of cut, aligning the final sequence would not be possible. The sequences of the fragments are arranged by computer using the overlaps as a guide.
Open reading frames are recognized by the presence of a promoter sequence, triplet start and stop codons, transcription termination sequence, and recognition sequences at the beginning and end of introns. The reading frame gives the promoter sequence (signal for when and where the gene is expressed) and the sequence of amino acids of the protein (an indication of its function).
Comparative genomics is the study of DNA sequences across different organisms. Similar sequences detected in different DNA samples may help reveal the function of a protein (when a sequence matches that of a known protein in another organism), the identity of a species known only by its DNA (metagenomics), and evolutionary relationships. The gene that influences the relative sizes of dogs, for example, was identified by comparative genomics.
RECAP 17.2
The following characteristics of prokaryotic genomes are essential for organisms that exist in a rapidly changing environment and must adapt rapidly: (1) Small genomes can show large changes with a few mutations. (2) Plasmids allow individual cells to share genomic information rapidly. (3) Transposons shuffle gene locations and therefore the rate of gene expression, and they cause mutations that make evolution possible.
Transposons either splice out of the host chromosome directly and move to a different location, or they make a DNA copy that is inserted at a different location, leaving the original transposon in place.
One could take a sample of belly-
RECAP 17.3
Genes for tissue formation are present in C. elegans but not in yeast.
Gene families allow for evolutionary “tinkering.” If there are several copies of a gene in the genome, mutations in one copy of the gene may temper a harmful effect of the wild-
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In some tissues, such as the egg cell, there is a need for massive protein synthesis (after fertilization) and therefore a need for a lot of ribosomes. The rate of transcription of the rRNA genes is not sufficient to meet the cell’s need for ribosomes, so there are multiple copies of the gene.
The four plants exist in different environments. For example, corn grows in drier and cooler regions than rice. Mutations of the basic plant genome allowed plants that carried them to adapt to the different environment to survive and reproduce.
RECAP 17.4
Each human gene has several introns. Alternative splicing of pre-
The human genome sequence has haplotypes that are linked to differential sensitivity to a drug. If the patient’s DNA is isolated and amplified, a classification with regard to the haplotypes can be made, and whether the patient will respond to a particular drug can be determined.
RECAP 17.5
The proteome is analyzed by identifying large numbers of proteins using two-
Patterns of metabolites correlate with different physiological states. The simplest example of why it would be useful to have a database of primary and secondary metabolites involved in human metabolism might be the high level of blood glucose that is associated with diabetes. Most diseases are more complex than this. Diagnosis may be helped by a pattern of metabolites in the metabolome.
WORK WITH THE DATA, P. 363
There was a 4.4 percent change in DNA sequence between the two cat genomes in 10.8 million years. This indicates a rate of change of 0.4 percent per million years. By comparison, the rate of change between the human and gorilla genomes was 0.6 percent per million years, or 50 percent faster.
14,425 gene families are shared by all the mammalian genomes examined.
103 are unique to the tiger and domestic cat genomes; 231 are unique to the human and mouse genomes.
Over 90 percent of the mammalian genome is common to all mammals. Relatively few gene families are unique to each kind of mammal examined.
The tiger is a hunter, and its genome reflects this, with genes for smell, signaling and digestion.
FIGURE QUESTIONS
Figure 17.4 A retrovirus infects a cell and makes a cDNA copy of its genome. The cDNA is inserted into the host cell by the action of viral integrase. The cDNA is carried along with the host chromosome as the cell divides. Through recombination, the cDNA adds a gene for excision. It also adds adjacent genes and is able to move about the genome.
Figure 17.6 No. Protein-
Figure 17.10 In both processes a protein structure is moving along a nucleic acid. In the polysome, the ribosome moves along mRNA; in rRNA synthesis, RNA polymerase moves along DNA. In both processes a polymer product is made: in the polysome the product is a polypeptide, and in the rRNA the product is RNA. The two processes are similar also in having a “cafeteria” system in which multiple polymers are made at the same time.
APPLY WHAT YOU’VE LEARNED
For SNP 5689, individuals with genotype AA have the highest concentrations (and thus the slowest metabolism) of Calm. Individuals with GG have the lowest concentrations (fastest metabolism) of Calm. Heterozygotes are intermediate. For SNP 8835, AA homozygotes have the lowest concentrations (fastest metabolism), CC homozygotes have the highest concentrations (slowest metabolism), and heterozygotes are intermediate. For SNP 11286, the heterozygotes have the highest concentrations (slowest metabolism). The two homozygotes have roughly similar concentrations and thus similar metabolisms.
Because AA individuals at this SNP have a slower metabolism of the drug than do GG individuals, the AA individuals should receive a lower dose.
AA homozygotes have the slowest metabolism and GG homozygotes have the fastest. Thus one would expect that the enzyme encoded by the A allele of the gene would be less active than the enzyme encoded by the G allele.
AA homozygotes have the fastest metabolism and CC homozygotes have the slowest. Thus one would expect that the inhibitor encoded by the A allele of the gene would be less active than the inhibitor encoded by the C allele.
Give the treatment mice the drug and leave others as the control. Prepare liver tissue from the treatment and the control mice. Use two-