Please use the "Talk to Ed and Mike" Forum in Moodle if you wish to discuss the assignment further.
Don't overlook the links to web resources and references to text pages below that might be useful in answering these questions. Activities in lecture, discussion, and laboratory classes will also address these questions.
Completion of these questions will help you achieve most of the objectives for lectures #14 and #15.
Beta
Thalassemia is a common blood disorder. The product of
the gene (HBB) is the beta globin chain of hemoglobin.
Hemoglobin consists of 4 protein chains, two alpha globin and two beta
globin chains. The HBB gene is located on chromosome 11. The
mutated, recessive allele results in a reduction of beta globin
protein production. Children who are homozygous recessive are unable to
produce beta globin, reducing the amount of complete hemoglobin they
can produce.
This inhibits the development of functional red blood cells and leads
to severe anemia and other medical problems.
The following links may be helpful as background information, but are not necessary to answer the question on this assignment.
Map of chromosome #11 Follow the directions to enlarge the upper end of the chromosome. Note the beta thalassemia locus listed 5th from the top on the right hand list of gene loci.
More information on beta thalassemia from the National Library of Medicine, National Institutes of Health
Beta
Thalassemia is a common blood disorder. The product of
the gene (HBB) is the beta globin chain of hemoglobin.
Hemoglobin consists of 4 protein chains, two alpha globin and two beta
globin chains. The HBB gene is located on chromosome 11. The
mutated, recessive allele results in a reduction of beta globin
protein production. Children who are homozygous recessive are unable to
produce beta globin, reducing the amount of complete hemoglobin they
can produce.
This inhibits the development of functional red blood cells and leads
to severe anemia and other medical problems.
The following links may be helpful as background information, but are not necessary to answer the question on this assignment.
Map of chromosome #11 Follow the directions to enlarge the upper end of the chromosome. Note the beta thalassemia locus listed 5th from the top on the right hand list of gene loci.
More information on beta thalassemia from the National Library of Medicine, National Institutes of Health
Prepare a simple diagram and include labels and text to explain the expression of the dominant (normal) allele for the beta hemoglobin gene locus (HBB). Include and BOLD or HIGHLIGHT the terms in the table below in your diagram and text.
Your diagram should contain enough visual detail and/or textual explanation to convince your instructor that you truly do understand the relationship of each of the terms to the process of expression of the dominant beta hemoglobin allele.
Note: This MUST be YOUR OWN drawing, NOT a
copy of a drawing from the web or your text book.
We will deduct major points if it is clear that your drawing is a
version of this one from Access Excellence - or from any
other
print or web source, for that matter.
One half point for each term that is correctly illustrated and explained.
| dominant (normal) allele for the hemoglobin beta gene (HBB) |
DNA | mRNA |
| tRNA | rRNA | ribosome |
| mRNA processing + introns & exons |
codon | anticodon |
| protein folding | cytoplasm | transcription |
| translation | the beta chain of the hemoglobin protein |
Complete hemoglobin molecule |
| capillaries | nucleus | red blood cells |
| amino acid | RNA polymerase |
The dominant (normal) allele of the HBB gene, produces the beta protein chain of the hemoglobin molecule. Two beta chains combine with two alpha protein chains to form a complete and functional hemoglobin molecule. These hemoglobin molecules are produced and reside in the red blood cells. Red blood cells with normal hemoglobin are round and smooth. They flow easily in single file through the smallest blood vessels in your body, the capillaries where most of the nutrient and gas exchange occurs between the blood and the surrounding tissues.
When the HBB gene is mutated to the Thalassemia
form - and a
person is homozygous recessive, the beta hemoglobin molecules are
produced in insufficient number to produce normal hemoglobin.
You might also remember that the HBB gene is
activated in bome marrow cells in response to the protein hormone, EPO
produced in the cells of the kidney in response ot low blood oxygen
levels. See Take-Home Assignment #2.
| .....ACT | CCT | GAG | GAG | AAG | TCT..... | |
| .....TGA | GGA | CTC | CTC | TTC | AGA..... | Transcribed strand (template strand) |
The DNA sequence above is an actual bit of the dominant hemoglobin beta gene (HBB). Note that the preceding and following "......s" represent preceding and following nucleotides. The letters included here are a very small part of the much larger gene.
Use this sequence to answer parts a, b, c, & d.
HINT: Typing your answer in a word processor and using copy/paste to edit the subsequent parts of this question will save you time and make scoring your answers more efficient and accurate.
a. Use the mRNA codon chart on the web or in your book, pg. 246 of the Life text to decode this bit of the gene. Write out the mRNA sequence that would be produced when the bottom DNA strand is transcribed (from left to right). Write the sequence of amino acids that would be produced when the resulting mRNA strand is decoded from left to right.
For parts (b, c, and d), include the following information:
1. the original DNA
2. your mutated DNA
3. the new mRNA strand
4. the new sequence of amino acids
5. a sentence or two naming the type of mutation and describing how the change affects the sequence of amino acids and the resulting structure and function of the protein (include this bit whether the amino acid sequence is changed or not)
b. Delete a single nucleotide base pair from the original DNA (Highlight or Bold the deleted section in the original DNA).
c. Change one nucleotide base pair someplace in the original DNA (Highlight or Bold the changed base pair).
d. Randomly insert one or more nucleotide base pairs in the original DNA (Highlight or Bold your inserted base pairs).
There are many different ways to answer these questions. The important thing is that your mRNA, amino acid sequence, name of the mutation type, and description of the possible changes in the structure and function of the protein are consistent with your mutations.
a. Write out the mRNA sequence that would be produced when
the
bottom DNA strand is transcribed (from left to right). Write the
sequence of amino acids that would be produced when the resulting mRNA
strand is decoded from left to right.
| .....ACT | CCT | GAG | GAG | AAG | TCT..... | |
| .....TGA | GGA | CTC | CTC | TTC | AGA..... | Transcribed strand (template strand) |
| .....ACU | CCU | GAG | GAG | AAG | UCU..... | |
| Thr | Pro | Glu | Glu | Lys | Ser..... | Amino Acid Sequence |
b. Delete a nucleotide from the original DNA (Highlight or Bold the deleted section in the original DNA).
| .....ACT | CCT | GAG | GAG | AAG | TCT..... | |
| .....TGA | GGA | CTC | CTC | TTC | AGA..... | Transcribed strand (template strand) |
| .....ACT | CCT | GAG | G_G | AAG | TCT..... | |
| .....TGA | GGA | CTC | C_C | TTC | AGA..... | MUTATED DNA - Transcribed strand (template strand) |
| .....ACU | CCU | GAG | GGA | AGU | CU..... | |
| Thr | Pro | Glu | Gly | Ser | Leu..... | Amino Acid Sequence |
The deletion of the single DNA nucleotide causes a frame shift mutation that will result in a potential change in the amino acid controlled by the codon that was initially changed by the deletion and all of the following amino acids. All the following amino acids will be changed. This will undoubtedly change the structure and functionality of the protein.
c. Substitute one nucleotide for another someplace in the original DNA (Highlight or Bold the new nucleotide).
| .....ACT | CCT | GAG | GAG | AAG | TCT..... | |
| .....TGA | GGA | CTC | CTC | TTC | AGA..... | Transcribed strand (template strand) |
| .....ACT | CCT | GAG | GAG | AAT | TCT..... | |
| .....TGA | GGA | CTC | CTC | TTA | AGA..... | MUTATED DNA - Transcribed strand (template strand) |
| .....ACU | CCU | GAG | GAG | AAU | UCU..... | |
| Thr | Pro | Glu | Glu | Asn | Ser..... | Amino Acid Sequence |
The substitution mutation changed the third nucleotide in the fifth codon, which in this case resulted in a the substitution of an Asn amino acid for Lys. This is called a missence mutation.
This change might result in a change in the protein.
A missense mutation, or a change in one amino acid, might radically change the protein if it affects the shape of the protein. Sickle cell disease is the result of a change in one amino acid. If the amino acid that is changed is in a portion of the protein that is not important in determining its shape or chemical properties it might not affect the function of the protein. A missense mutation might change the shape and functionality of the protein slightly, making it perform slightly worse or even slightly better.
Other single nucleotide changes could result in a "silent mutation" that would produce no change in the amino acid encoded by that codon.
A substitution like this might produce a stop codon that would stop translation at the ribosome, even though there was more RNA to be translated. The resulting protein would be shorter than it should be and would probably not fold properly, loosing its function. A misplaced stop codon is called a "nonsense mutation".
d. Randomly insert one or more nucleotides in the original DNA (Highlight or Bold your inserted sequence of nucleotides).
| .....ACT | CCT | GAG | GAG | AAG | TCT..... | ||
| .....TGA | GGA | CTC | CTC | TTC | AGA..... | Transcribed strand (template strand) | |
| .....ACT | CCT | GAG | GAG | AGG | GAG | TCT..... | |
| .....TGA | GGA | CTC | CTC | TCC | CTC | AGA..... | MUTATED DNA - Transcribed strand (template strand) |
| .....ACU | CCU | GAG | GAG | AGG | GAG | UCU..... | |
| Thr | Pro | Glu | Glu | Arg | Glu | Ser..... | Amino Acid Sequence |
Adding three nucleotides into a codon caused a temporary frame shift mutation. The first amino acid at that point was changed to Arg. The mutation changed the next codon and adding a Glu amino acid that was not originally in the protein. After the addition of a new amino acid, the reading frame was regained and the rest of the amino acids were the same as in the original DNA strand. Depending on where the extra amino acid was added in the protein, this mutation might or might not change the structure and function of the protein.
If one or two or some number other than three or a multiple of three nucleotides were added, a frame shift would persist and drastically change the structure and function of the protein.
Mutations are the way that new alleles are formed. When we talk about the normal, dominant allele in a recessively inherited genetic disorder we are talking about the allele that produces the normal protein and the normal phenotype. When we talk about a person having two recessive alleles for a genetic disease (homozygous recessive), we are talking about two of, perhaps several mutated forms of the normal allele. The mutated alleles produce mutant mRNAs, which assemble mutant proteins that do not function properly. The result is a phenotype that exhibits the diseased condition. Depending on the types of the two recessive alleles inherited from the parents, the person my be mildly or severely affected. There may be several intermediate degrees of expression of the diseased condition. If a person has one dominant (good) allele and one recessive (bad) allele (heterozygous) the good allele can make enough normal protein to keep the person healthy.
With dominant disorders the reverse is true. The recessive allele is the "normal" form and produces the normal protein. The dominant allele is the mutant form that produces the "bad" protein that makes the person sick. In dominantly inherited diseases it only takes one copy of the mutated gene to have the disease.