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    Tarik Kader
    1. The DNA sequence codes for seven amino acids. The top strand is the coding strand, and the mRNA sequence would be AACCAUCAAGUGUAAUGU. The seven amino acids are Thr, Thr, Gln, Ser, Ser. 2. Segments E and I will be in the processed transcript. Segments CEG will be transcribed but not translated. A +1 insertion in segment D would likely prevent production of RNA and protein, while one in segment C may allow transcript detection but no functional protein. 3. Functional B-galactosidase will be produced in conditions where the P'O'Z' genotype is present and lactose is present.

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    Adobe Scan Nov 16, 2020

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    Tarik Kader
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    1. The DNA sequence codes for seven amino acids. The top strand is the coding strand, and the mRNA sequence would be AACCAUCAAGUGUAAUGU. The seven amino acids are Thr, Thr, Gln, Ser, Ser. 2. Segments E and I will be in the processed transcript. Segments CEG will be transcribed but not translated. A +1 insertion in segment D would likely prevent production of RNA and protein, while one in segment C may allow transcript detection but no functional protein. 3. Functional B-galactosidase will be produced in conditions where the P'O'Z' genotype is present and lactose is present.

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    Short micro bio homework

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    1. The DNA sequence codes for seven amino acids. The top strand is the coding strand, and the mRNA sequence would be AACCAUCAAGUGUAAUGU. The seven amino acids are Thr, Thr, Gln, Ser, Ser. 2. Segments E and I will be in the processed transcript. Segments CEG will be transcribed but not translated. A +1 insertion in segment D would likely prevent production of RNA and protein, while one in segment C may allow transcript detection but no functional protein. 3. Functional B-galactosidase will be produced in conditions where the P'O'Z' genotype is present and lactose is present.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    52 views4 pages

    Adobe Scan Nov 16, 2020

    Uploaded by

    Tarik Kader
    1. The DNA sequence codes for seven amino acids. The top strand is the coding strand, and the mRNA sequence would be AACCAUCAAGUGUAAUGU. The seven amino acids are Thr, Thr, Gln, Ser, Ser. 2. Segments E and I will be in the processed transcript. Segments CEG will be transcribed but not translated. A +1 insertion in segment D would likely prevent production of RNA and protein, while one in segment C may allow transcript detection but no functional protein. 3. Functional B-galactosidase will be produced in conditions where the P'O'Z' genotype is present and lactose is present.

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    © All Rights Reserved
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    of 4

    BIS 101 NAME

    Dr. Crowder ID#


    Homework 07

    BIS 101: Homework 07


    DUE Monday NOVEMBER 16th
    [submit using Homework 7 Quiz in Canvas]

    1. The DNA sequence is part of a gene. The sequence shown is AFTER the
    following
    it codes for seven amino acids of the polypeptide.
    start codon but within the sequence is the correct reading frame that
    Analyze the sequence to determine the what
    is not necessarily the start of
    codes for seven amino acids (note - the first nucleotide
    trial and error approach to
    the reading frame. For this question you will have to use
    frame is).
    determine what strand is the coding strand and what the codon reading

    3 caattgattagtcagtcaattgat5
    5 gtta actaatcagtc agttaacta3'

    Please see codon table on last page to answer the following questions:

    strand or the bottom strand?


    1A) Which DNA strand is the coding strand, the top

    Top
    Please
    that would be transcribed from this gene?
    1B) What is the mRNA sequence
    label the 5' &3' ends.
    aac
    Cag
    Ca gu4
    Cu 3

    would be translated from the open reading


    What the seven amino acids that
    1C) are

    frame within this segement of DNA?

    aSn
    Thr aSn gin Sor

    Marina Crowder, 2020


    Page 1 of 5
    All Rights Reserved
    BIS 101
    Homework 07
    Dr. Crowder

    2. The image shown below is the structure of a Drosophila gene, divided into 10
    segments, labelled A-J. The image represents the coding strand of the gene, which
    contains three exons, two introns, a promoter, and a site in segment J that
    corresponds to where the poly(A)-tail is added to the RNA transcript.

    Expn 1
    Exon 2 Lxon 3

    Promoter Intron1 Intron 2 AMAA poly (A)


    site
    ATGL
    41

    C
    EF GH
    2A) What segment or segments of the gene will be transcribed in the initial RNA
    transcript? List the appropriate letter or letters.

    E 6 /L
    28) What segment or segments of the gene will be found in the completely processed
    transcript?
    DE I

    2C) What segment or segments of the gene will be transcribed but not translated?

    CEG

    2D) A +1 insertion mutation occurred in region D, after the ATG sequence. How would
    this mutation likely affect the product produced? Would the RNA transcript be
    produced ? Would a functional polypeptide chain be made?
    nudd lezabey uleda
    Ahy mulal
    dop clin
    2E) A +1 insertion mutation occurred in region C, after the +1 but before the ATG
    Jence. How would this mutation likely affect the product produced? Would the
    transcript be detected? Would a functional polypeptide chain be made?

    hiy metatti vn ald ne Ahve

    iulalim
    Ahe
    lore
    ho
    Marina Crowder, 2020
    All Rights Reserved Page 2 of 5
    BIS 101
    Homework 07
    Dr. Crowder

    3. Predict whether functional B-galactosidase (product of LacZ gene) will be produced


    by the following E. coli strains under the conditions noted. The diploid genotypes
    represent partial diploid F (lac) strains. Use + to indicate that the enzyme is
    synthesized at greater than basal levels, and 0 to indicate that the enzyme is not
    synthesized.
    Gentotype No glucose Noglucosse
    No lactose lactose present
    3A) P o'Z
    3B) P 0" Z"
    3C)P O°Z
    3D) P 0 ° z
    3E) P o°Z
    3F) P 0 ' Z

    3G)P* O° Z/* P* O*Z*

    3H) P O* Z/t P 0° Z*

    31) PO* Z/t P 0* Z*

    glucose present glucose present


    Gentotype No lactosse lactose present
    33) P* 0* Z*

    3K)Pt O°Z*

    Marina Crowder, 2020


    All Rights Reserved Page 3 of 5
    BIS 101
    Homework 07
    Dr. Crowder

    4. Cystic fibrosis is a recessive human genetic condition that affects lung function.
    Cystic Fibrosis is caused by mutationsin the human CFTR gene that result in a
    defective CFTR protein. Taylor is a heterozygous carrier of a mutant CFTR allele.
    Below is the template strand DNA sequence for both of Taylor's CFTR alleles. Within
    this sequence is the first 5 codons of protein reading frame. The difference between the
    mutant allele and wildtype allele is underlined and bolded.

    Wildtype CFTR allele: 3 CCG TAC TGG GCC ATG TGGA 5

    mutant CFTR allele: 3' CCG TAC TGG ACC ATG TGGA 5"

    4A) What is the sequence of the first five amino acids of the wildtype CFTR protein
    made?

    me hr yn

    What is the anticodon of the tRNA that would pair with the 3rd codon of the
    4B)
    wildtype CFTR reading frame?

    4C) Taylor's mutant CFTR allele is what type of mutation?

    A) Silent mutation
    BFrameshift mutation
    C) Nonsense mutation
    D) Missense mutation
    E) Deletion mutation

    Marina Crowder, 2020


    All Rights Reserved Page 4 of 5

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