Concept explainers
To review:
In the given table, there are two DNA and polypeptide sequences shown with alleles for a hypothetical locus that generates different polypeptides of five amino acids. In both cases, the lower DNA strand is the template strand:
Based on this data of DNA and polypeptide sequence, is it possible to determine the dominant and recessive allele. Explain with reason.
Introduction:
Genotype is the organism’s genetic makeup while
The dominance or recessive terms state to the inheritance design of certain trait. The dominant alleles will produce dominant phenotype while recessive alleles produce recessive phenotype.
The resultant protein decides whether the trait is dominant or recessive.
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Genetic Analysis: An Integrated Approach (2nd Edition)
- (i) For the chromatogram below, what is the sequence of the template DNA from base 115 to 125? CTGTGTGAAATTGT TA T CCGC T CA CA AT T C CACA CA A CATA CGAGC CGGAAG CA TA A 110 120 130 140 150 160 (ii) An allele of a gene has the following change in it's sequence ATG GTG CÁC CTG ACT CCT GTG GAG AAG TCT compared to the wild type ATG GTG CAC CTG ACT CT GAG GAG AAG TCT With reference to the sequence; there is a codon, resulting in a change from is a mutation in the to which mutation.arrow_forwardSupercoiled DNA is slightly unwound compared to relaxed DNA and this enables it to assume a more compact structure with enhanced physical stability. Describe the enzymes that control the number of supercoils present in the E. coli chromosome. How much would you have to reduce the linking number to increase the number of supercoils by five?arrow_forwardThe restriction endonuclease NciI recognizes and cuts the five-base-pair sequence 5’- CC(G/C)GG-3’ [where (G/C) means either G or C will work at that position]. (1) How often, on average, would this sequence occur in random DNA? Assume the DNA contains 25% each of A, G, T & C. (2) After digestion, Nci1 leaves a one-base 5’ overhang. Write/draw the cut site/digested products.arrow_forward
- A certain section of the coding (sense) strand of some DNA looks like this: 5'- ATGGGCCACTCATCTTAG-3' It's known that a very small gene is contained in this section. Classify each of the possible mutations of this DNA shown in the table below. I Don't Know mutant DNA 5'- ATG GGCCACAGTTCTTAG-3' 5'- ATG GG CTCATCTTAG - 3' 5'- ATG GGCCACGCATCTTAG-3' Submit type of mutation (check all that apply) ооооо O point O silent O noisy ооооо insertion deletion insertion O deletion Opoint Osilent noisy insertion O deletion ооооо Opoint silent O noisy X S Ⓒ2023 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center Accessibilityarrow_forwardDesign primers that will amplify the following region of DNA (assume this is one strand from a double stranded region of DNA). The primers should be 15 bases in length. Indicate the 5' and 3' ends of the primers. 5' GGATCGATCAAGAACAATGACAGGATCGAGGAATTCAGCCTACGCAGCCCGTAGCTGGAGGGA 3'arrow_forwardYou have the following DNA coding sequence of a wild-type allele: 5’-ATG TTC CAG CTA GAT GAT ATG CTG GTA ATT GGG GAA CGC GCG CGG TAA-3’ 1. For the second, third, fourth, and fifth codons, write all possible anticodon sequences (left-to-right, 5’- 3’), including anticodons with wobble and inosine.arrow_forward
- For the following short sequence of double stranded DNA and the given primers, there will be one major duplex DNA product after many cycles (imagine 10 cycles) of PCR. Provide the sequence of this one major duplex product and label the 5’ and 3’ ends of each strand. Sequence to be amplified: 5’- GGTATTGGCTACTTACTGGCATCG- 3’ 3’- CCATAACCGATGAATGACCGTAGC- 5’ Primers: 5’-TGGC-3’ and 5’-TGCC-3’arrow_forwardDraw the structure of the double Holliday junctionthat would result from strand invasion by both ends of thebroken duplex into the intact homologous duplex shownin Figure Q5–3. Label the left end of each strand in the Hol-liday junction 5ʹ or 3ʹ so that the relationship to the paren-tal and recombinant duplexes is clear. Indicate how DNAsynthesis would be used to fill in any single-strand gaps inyour double Holliday junction.arrow_forwardConsider the following segment of DNA, which is part ofa much longer molecule constituting a chromosome:5′.…ATTCGTACGATCGACTGACTGACAGTC….3′3′.…TAAGCATGCTAGCTGACTGACTGTCAG….5′If the DNA polymerase starts replicating this segmentfrom the right,a. which will be the template for the leading strand?b. Draw the molecule when the DNA polymerase ishalfway along this segment.c. Draw the two complete daughter molecules.d. Is your diagram in part b compatible with bidirectional replication from a single origin, the usual modeof replication?arrow_forward
- The table shows the partial sequences of a wild type polypeptide and three mutant polypeptides as well as the type of single nucleotide mutation that produced each mutant polypeptide. Peptide sequence Met - Leu - Arg - Ile - ... Type of mutation Wild type Met - Leu - Arg - Met - ... Met - Leu - [STOP] Mutant 1 transition Mutant 2 transversion Mutant 3 Met - Phe - Arg - Ile - ... transition Determine the mRNA sequence for the wild type polypeptide by identifying the codons that correspond to each amino acid. The first codon has been filled in for you. Codon information can be found in the codon access table. Met Leu Arg Ile Answer Bank CỦA AUA CGU CGA AUG AUC AGA UUG CÚCarrow_forwardThe beginning of the hexose kinase gene's sequence can be found below, the +1 nucleotide is underlined and bolded. It also contains an origin of replication (ORI) which is found at position 30. 1 20 ORI 40 60 5'..TTCGAGCTCTCGTCGTCGAGATACGCGATGATATTACTGGTAATATGGGGATGCACTATC...3’ 3'...AAGCTCGAGAGCAGCAGCTСТАТGCGCTАСТАТААTGACCATTATАССССТАСGTGATAG...5' promoter 2a. Assume that replication has been initiated at that ORI. Provide the sequence of the primer that is complementary to the DNA in each of the following positions. Site A - binding to the top strand of the DNA at position 20 – 30 5' 3' Site B - binding to the top strand of the DNA at position 31 – 41 5' 3' 2b. Replication is occurring normally in these cells; would you expect to find a primer in both positions? Why or why not?arrow_forwardThe beginning of the hexose kinase gene's sequence can be found below, the +1 nucleotide is underlined and bolded. It also contains an origin of replication (ORI) which is found at position 30. 20 ORI 40 60 5'..TTCGAGCTCTCGTCGTCGAGATACGCGATGATATTACTGGTAATATGGGGATGCACTATC...3' 3'...AAGCTCGAGAGCAGCAGCTСТАТGCGCTАСТАТААTGACСАТТАТАССССТАСGTGATAG...5" promoter a. Assume that replication has been initiated at that ORI. Provide the sequence of the primer that is complementary to the DNA in each of the following positions. Site A - binding to the top strand of the DNA at position 20 – 30 5' 3' Site B - binding to the top strand of the DNA at position 31 – 41 5' 3'arrow_forward
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