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Oogenesis
The formation of the ovum (mature female gamete) from undifferentiated germ cells is called oogenesis. This process takes place in the ovaries (female gonads). Oogenesis consists of three stages known as the multiplication phase, growth phase, and maturation phase.
Cell Division
Cell division involves the formation of new daughter cells from the parent cells. It is a part of the cell cycle that takes place in both prokaryotic and eukaryotic organisms. Cell division is required for three main reasons:
choices:
purine or pyrimidine
ribose or deoxyribose
1-9?
hydrogen or phosphodiester
1-9?
an -OH or no -OH
1-9?
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- Below is a sequence of 540 bases from a genome. What information would you use to find the beginnings and ends of open reading frames? How many open reading frames can you find in this sequence? Which open reading frame is likely to represent a protein- coding sequence, and why? Which are probably not functioning protein-coding sequences, and why? Note: for simplicitys sake, analyze only this one strand of the DNA double helix, reading from left to right, so you will only be analyzing three of the six reading frames shown in Figure 19.4.In the following gel showing stained bands of the Alu insertion sequence, what is the genotype of individual 2? 941 bp 641 bp->>> 1 2 3 4 5 6 Homozygous for the 641 bp sequence that does not contain in the Alu insertion Heterozygous, containing one 941 bp sequence and one 641 bp sequence O Homozygous for the 941 bp sequence containing the Alu insertion5' GTGCTAGCGGGAATGAGCTGGGATACTAGTAGGGCT 3' 3' CACGATCGCCCTTACTCGACCCTATGATCATCCCGA 5' Template Strand: 9. Using the template strand, transcribe the DNA above, Be sure you write your sequence 5 - 5 a indicate the 5' and 3' ends of any nucleic acid molecule(s). 10. Use the codon chart below to translate your mRNA into an amino acid sequence. Begin at the first codon. Third First position (5' end) Second position position (3'end) UGU Cys UAU Tyr Cc UGC Cys UGA Stop UGG Trp UCU Ser -Y UAC Tyr UAA Stop UAG Stop UUU Phe - F UUC Phe UUA Leu UUG Leu FL UCC Ser -- UCA Ser UCG Ser CGU Arg CGC Arg ER CGA Arg CGG Arg CCU Pro CAU His CUU Leu CUC Leu -- CAC His CAA Gln CAG Gln CCC Pro -P A - CUA Leu CUG Leu CCA Pro CCG Pro AAU Asn AAC Asn AGU Ser AGC Ser AGA Arg ACU Thr AUU lle AUC lle AUA lle AUG Met M ACC Thr -T ACA Thr ACG Thr A. AAA Lys K AAG Lys -R AGG Arg A. GAU Asp -D GAC Asp GGU Gly GGC Gly GCU Ala GUU Val GUC Val GCC Ala A -G GGA Gly GGG Gly A -V GUA Val GUG Val GCA Ala GCG Ala GAA Glu -E…
- In a standard procedire, when writing and reading base sequences for nucleic acids (both DNA and RNAs) always to specify base sequence in 5' > 3' direction unless otherwise directed 1. From the base sequence 5' A-T-G-C-C-A 3' in a DNA template strand, determine the base sequence in hnRNA synthesized from the DNA template strand 2. From the base sequence 5' T-A-A- C-C-T 3' in a DNA template strand, determine the base sequence in hnRNA synthesized from the DNA template strandBelow is a sequence of DNA. 5'-ttaccgataattctctctcccctcttccatgattctgattaaagaaggcgagaacgaaactatttgttaatacc-3' Using the one letter code for Amino Acids, what is the predicted AA sequence of the shortest ORF (from N to C-terminal end)? Using the one letter code for Amino Acids, what is the predicted AA sequence of the longest ORF (from N to C-terminal end)?You are sequencing the genome of newly discovered bacterium, and know nothing of its sequence except that it is one single circular chromosome about 6,000,000 bp long. Your raw sequencing data, from two reactions, are given: 5'-ACCGTCGGTTACGCTTAGA-3' 5'-GTTACGCTTAGATAACACAAG-3' Based on this data, give the sequence of one sequence read: Based on this data, give the sequence of one sequence contig: C. So far, the researchers have assembled all the data they have into three sequence contigs. Have they sequenced the whole genome? Briefly explain, in one or two sentences.
- Using the figure below, what is molecule "A" (type a 1, 2 or 3 in the blank) nuclease ligase DNA polymerase What is the function of molecule "A"? to separate the double helix into two to piece together the Okazaki segments to copy the new DNA strand to the old strand by complementary base pairing Using the figure below, what is molecule "G" (type a 1, 2 or 3 in the blank) nuclease ligase DNA polymerase What is the function of molecule "G"? to separate the double helix into two to piece together the Okazaki segments to copy the new DNA strand to the old strand by complementary base pairing Which of the following statements best describes why one of the daughter strands is synthesized in pieces? the enzymes that synthesize DNA are slower that the enzymes that unwind the double helix and this produces 'lagging time' the enzymes that synthesize DNA can only do so in a 5' --->3' direction this figure illustrates a eukaryotic cell since prokaryotic cells do not synthesize DNA…Using the figure below, what is molecule "A" (type a 1, 2 or 3 in the blank) nuclease ligase DNA polymerase What is the function of molecule "A"? to separate the double helix into two to piece together the Okazaki segments to copy the new DNA strand to the old strand by complementary base pairing Using the figure below, what is molecule "G" (type a 1, 2 or 3 in the blank) nuclease ligase DNA polymerase What is the function of molecule "G"? to separate the double helix into two to pieceGiven the sequence shown below, write the complementary DNA sequence, using the base-pairing rules, as well as the directionality of the strands: 5'- CGAGGCTAGGTTAACCTG-3'
- Design 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'If the recognition sequence of the restriction enzyme HindiIl is AAGCTT, then how many covalent bonds will be broken by the enzyme in the following DNA molecule? 5'-T-C-A-A-G-C-T-T-C-G-A-A-G-C-T-T-G-A-3 3-A-G-T–T-C-G-A-A-G-C -T-T-C-G-A-A-C-T-5 А. 1 В. 2 С.3 D. 4Sanger sequencing originally used 4 lanes in gels. These lanes represented sequences of different lengths obtained by adding: O All of the 4 dideoxynucleotides (ddATP; ddGTP; ddCTP; ddTTP) to the reaction vials; together with one of the 4 deoxynucleotides (DATP, DGTP, DCTP and DTTP), one for each lane, in each vial. O One of the 4 dideoxynucleotides (ddATP; ddGTP; ddCTP; ddTTP) to the reaction vials; one for each lane O All of the 4 dideoxynucleotides (ddATP; ddGTP; ddCTP; ddTTP), together with all of the 4 deoxynucleotides (DATP, DGTP, dCTP and dTTP), to all of the reaction vials O One of the 4 dideoxynucleotides (ddATP; ddGTP; ddCTP; ddTTP) to the reaction vials; one for each lane, together with all of the 4 deoxynucleotides (dATP, DGTP, dCTP and ATTP) in each vial