Genetics: Analysis and Principles
6th Edition
ISBN: 9781259616020
Author: Robert J. Brooker Professor Dr.
Publisher: McGraw-Hill Education
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Chapter 29, Problem 12CONQ
Summary Introduction
To review:
The lineup of the given DNA (deoxyribonucleic acid) sequence according to their homologous sequences.
Introduction:
A gene that is inheritedby two species from a common ancestor is known as a homologous gene. The term homology derives from the Greek word homos, which means equal and logos, which meansrelation. Homologous genes have the possibility to be similar in their sequences.
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Students have asked these similar questions
The bacterial gene shorty (sh) encodes for a small protein. The DNA sequence of the sh gene is shown below. The ORF is in CAPITAL LETTERS
5’-tataatgggcttaacaATGAGTAAAAGAGGTCCTTTACTCCGGTATCACCAATAGaaatattatttaa-3’ 3’-atattacccgaattgtTACTCATTTTCTCCAGGAAATGAGGCCATAGTGGTTATCtttataataaatt-5’
Answer the following questions:
Q1. Which is the coding strand? Which is the template strand? [10%]
Top-bottom.
Bottom-Top.
Both can be used as either coding or template for this gene.
The bacterial gene shorty (sh) encodes for a small protein. The DNA sequence of the sh gene is shown below. The ORF is in CAPITAL LETTERS
5’-tataatgggcttaacaATGAGTAAAAGAGGTCCTTTACTCCGGTATCACCAATAGaaatattatttaa-3’ 3’-atattacccgaattgtTACTCATTTTCTCCAGGAAATGAGGCCATAGTGGTTATCtttataataaatt
What is the length in AA’s of the Sh protein? Assume fMet is NOT CLEAVED
[10%]
39 AA
13 AA
12 AA
21 AA
For the following sequence design the forward and reverse primer... explain and justify your answer.
Gene of Interest:
a tgaaacaaca aaaacggctt tacgcccgat tgctgacgct gttatttgcg 61 ctcatcttct tgctgcctca ttctgcagca gcggcggcaa atcttaatgg gacgctgatg 121 cagtattttg aatggtacat gcccaatgac ggccaacatt ggaagcgttt gcaaaacgac 181 tcggcatatt tggctgaaca cggtattact gccgtctgga ttcccccggc atataaggga 241 acgagccaag cggatgtggg ctacggtgct tacgaccttt atgatttagg ggagtttcat 301 caaaaaggga cggttcggac aaagtacggc acaaaaggag agctgcaatc tgcgatcaaa 361 agtcttcatt cccgcgacat taacgtttac ggggatgtgg tcatcaacca caaaggcggc 421 gctgatgcga ccgaagatgt aaccgcggtt gaagtcgatc ccgctgaccg caaccgcgta 481 atttcaggag aacacctaat taaagcctgg acacattttc attttccggg gcgcggcagc 541 acatacagcg attttaaatg gcattggtac cattttgacg gaaccgattg ggacgagtcc 601 cgaaagctga accgcatcta taagtttcaa ggaaaggctt gggattggga agtttccaat 661 gaaaacggca actatgatta tttgatgtat gccgacatcg attatgacca tcctgatgtc 721 gcagcagaaa ttaagagatg gggcacttgg tatgccaatg…
Chapter 29 Solutions
Genetics: Analysis and Principles
Ch. 29.1 - Prob. 1COMQCh. 29.1 - Prob. 2COMQCh. 29.1 - 3. A pair of birds flies to a deserted island and...Ch. 29.1 - Prob. 4COMQCh. 29.2 - 1. Phylogenetic trees are based on
a. natural...Ch. 29.2 - Prob. 2COMQCh. 29.2 - An approach that is used to construct a...Ch. 29.2 - 4. Horizontal gene transfer is a process in which...Ch. 29.3 - Prob. 1COMQCh. 29.3 - Prob. 2COMQ
Ch. 29.3 - When the chromosomes of closely related species...Ch. 29 - 1. Discuss the two principles on which evolution...Ch. 29 - 2. Evolution, which involves genetic changes in a...Ch. 29 - Prob. 3CONQCh. 29 - Prob. 4CONQCh. 29 - 5. Would each of the following examples of...Ch. 29 - Distinguish between anagenesis and cladogenesis....Ch. 29 - 7. Describe three or more genetic mechanisms that...Ch. 29 - Explain the type of speciation (allopatric,...Ch. 29 - Prob. 9CONQCh. 29 - Prob. 10CONQCh. 29 - Discuss the major differences among allopatric,...Ch. 29 - Prob. 12CONQCh. 29 - Prob. 13CONQCh. 29 - Would the rate of deleterious or beneficial...Ch. 29 - 15. Which would you expect to exhibit a faster...Ch. 29 - Prob. 16CONQCh. 29 - 17. Plant seeds contain storage proteins that are...Ch. 29 - Take a look at the -globin and -globin amino acid...Ch. 29 - Compare and contrast the neutral theory of...Ch. 29 - Prob. 20CONQCh. 29 - 21. As discussed in Chapter 27, genetic variation...Ch. 29 - Prob. 22CONQCh. 29 - Two populations of snakes are separated by a...Ch. 29 - 2. Sympatric speciation by allotetraploidy has...Ch. 29 - 3. Two diploid species of closely related frogs,...Ch. 29 - A researcher sequenced a portion of a bacterial...Ch. 29 - F1hybrids between two species of cotton,Gossypium...Ch. 29 - 6. A species of antelope has 20 chromosomes per...Ch. 29 - Prob. 7EQCh. 29 - 8. Prehistoric specimens often contain minute...Ch. 29 - From the results of the experiment of Figure...Ch. 29 - InChapter 23, a technique called fluorescence in...Ch. 29 - Prob. 11EQCh. 29 - 12. Discuss how the principle of parsimony can be...Ch. 29 - 13. A homologous DNA region, which was 20,000 bp...Ch. 29 - Prob. 14EQCh. 29 - Prob. 1QSDCCh. 29 - 2. Compare the forms of speciation that are slow...Ch. 29 - 3. Do you think that Darwin would object to the...
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- The following are DNA fragments containing a small gene. The top strand is the coding strand. Transcribe all groups and translate. FIND THE POSSIBLE MUTATIONS Group B - MUTATION 5’-GGCAATGGGTTTGTGAAATTCTAAAAGTTTTTAATTC-3’ 3’-CCGTTACCCAAACACTTTAAGATTTTCAAAAATTAAG-5’ Group C- 5’-GGCAATGGGTTTGTGCAATTCTAAGAGTTTTTAATTC-3’ 3’-CCGTTACCCAAACACGTTAAGATTCTCAAAAATTAAG-5’arrow_forwardTGAGGATGAAACTCACACCGGGGCGCAGTTTGGCACTTAGATTCTTGTACACGACCTAGTATAACACAGTT Compare this mutated sense sequence given below to the original one given above and identify and classify all mutations that can be found in this new DNA sequence? TGAGCATGAAACTCACACCGGGGGCAGTTTCGCACTTAGGATTCTTGTACAGGACCTAGTATAACAAGTTarrow_forwardThe DNA sequence of one strand of a gene from threeindependently isolated mutants is given here (5′ endsare at left). Using this information, what is the sequence of the wild-type gene in this region?mutant 1 ACCGTAATCGACTGGTAAACTTTGCGCGmutant 2 ACCGTAGTCGACCGGTAAACTTTGCGCGmutant 3 ACCGTAGTCGACTGGTTAACTTTGCGCGarrow_forward
- A portion of the sequence from the DNA coding strand of the chick ovalbumin gene is shown. Determine the partial amino acid sequence of the encoded protein. CTCAGAGTTCACCATGGGCTCCATCGGTGCAGCAAGCATGGAA-(1104 bp)-TTCTTTGGCAGATGTGTTTCCCCTTAAAAAGAA Enter the 3-letter abbreviation for each amino acid in sequence, separated with dashes, and no spaces (example: xxx-xxx-XXX-XXX...) The amino acid sequence is .1104bp..…........arrow_forwardFor the following sequence design the forward and reverse primer... explain and justify your answer. Full sequence would be: 1 tctagagtca tgaaacaaca aaaacggctt tacgcccgat tgctgacgct gttatttgcg 61 ctcatcttct tgctgcctca ttctgcagca gcggcggcaa atcttaatgg gacgctgatg 121 cagtattttg aatggtacat gcccaatgac ggccaacatt ggaagcgttt gcaaaacgac 181 tcggcatatt tggctgaaca cggtattact gccgtctgga ttcccccggc atataaggga 241 acgagccaag cggatgtggg ctacggtgct tacgaccttt atgatttagg ggagtttcat 301 caaaaaggga cggttcggac aaagtacggc acaaaaggag agctgcaatc tgcgatcaaa 361 agtcttcatt cccgcgacat taacgtttac ggggatgtgg tcatcaacca caaaggcggc 421 gctgatgcga ccgaagatgt aaccgcggtt gaagtcgatc ccgctgaccg caaccgcgta 481 atttcaggag aacacctaat taaagcctgg acacattttc attttccggg gcgcggcagc 541 acatacagcg attttaaatg gcattggtac cattttgacg gaaccgattg ggacgagtcc 601 cgaaagctga accgcatcta taagtttcaa ggaaaggctt gggattggga agtttccaat 661 gaaaacggca actatgatta tttgatgtat gccgacatcg attatgacca tcctgatgtc 721 gcagcagaaa ttaagagatg gggcacttgg…arrow_forwardIn the copies of each sequence below, divide the sequences into codons (triplets) by putting a slash between each group of three bases. Sequence ATCTTCCCTCCTAAACGTTCAACCGGTTCTTAATCCGCCGCCAGGGCCCCGCCCCTCAGAAGTTGGTSequence BTCAGACGTTTTTGCCCCGTAACAACTTGTTACAACATGGTCATAAACGTCAGAGATGGTCAATCTCTTAATGACTSequence CTACAAACATGTAAACACACCCTCAGTGGACCAACTCCGCAACATAAACCAAACACCGCTCGCGCCGAAAAAGATATGGarrow_forward
- For the following sequence please design an 18 base pair forward primer. ATGGCTGATAAGATAGAGAGGCATACTTTCAAGGTCTTCAATCAAGATTTCGAAAAAGAGCTGGAGTTTGGATTAGATAGAAAATATTTTTAGarrow_forwardhe Sequence below comes from the alpha-2 globin of the human hemoglobin gene cluster found in chromosome 16. The globin region of the hemoglobin protein itself consists of 2 alpha chains and 2 beta chains. 1 actcttctgg tccccacaga ctcagagaga acccaccatg gtgctgtctc ctgccgacaa 61 gaccaacgtc aaggccgcct ggggtaaggt cggcgcgcac gctggcgagt atggtgcgga 121 ggccctggag aggatgttcc tgtccttccc caccaccaag acctacttcc cgcacttcga 181 cctgagccac ggctctgccc aggttaaggg ccacggcaag aaggtggccg acgcgctgac 241 caacgccgtg gcgcacgtgg acgacatgcc caacgcgctg tccgccctga gcgacctgca 301 cgcgcacaag cttcgggtgg acccggtcaa cttcaagctc ctaagccact gcctgctggt 361 gaccctggcc gcccacctcc ccgccgagtt cacccctgcg gtgcacgcct ccctggacaa 421 gttcctggct tctgtgagca ccgtgctgac ctccaaatac cgttaagctg gagcctcggt 481 agccgttcct cctgcccgct gggcctccca acgggccctc ctcccctcct tgcaccggcc 541 cttcctggtc…arrow_forwardA 210-bp sequence within the CFTR gene on human chromosome 7 is shown below. The three bold underlined nucleotides are deleted in a common cystic fibrosis (CF) mutation, removing a phenylalanine amino acid from the CFTR protein. 1 AGAGGGTAAA ATTAAGCACA GTGGAAGAAT TTCATTCTGT TCTCAGTTTT 51 CCTGGATTAT GCCTGGCACC ATTAAAGAAA ATATCATCTT TGGTGTTTCC 101 TATGATGAAT ATAGATACAG AAGCGTCATC AAAGCATGCC AACTAGAAGA 151 GGTAAGAAAC TATGTGAAAA CTTTTTGATT ATGCATATGA ACCCTTCACA 201 CTACCCAAAT PCR primers have been designed to amplify fragments within this sequence: Forward: GGATTATGCCTGGCACCATT Reverse: AGTGTGAAGGGTTCATATGC DNA from a CF patient is tested with a PCR assay using a pair of these primers, and the PCR product is found to be 3 bp shorter than that expected from the sequence shown above. What length PCR products (in bp) would you expect in the mother of the CF patient? A. 95 and 92 B. 149 C. 133 and 130 D. 149 and 146 E. 146arrow_forward
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