For example, as shown below, suspect A has one allele with 5 repeats, and one allele with 3 repeats, giving a DNA profile for the TH01 locus of 5-3. Suspect A Genotype for TH01 locus Suspect B Genotype for TH01 locus CCC AAA CCC CCC AAA CCC AAA AAA Number of [TCAT] repaats
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Please answer these two questions regarding PCR
a) First question you must use the image: What is the DNA profile for suspect B for the TH01 locus?
b) What determines the target sequence amplified in PCR?
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- The TPOX locus is one of the genetic loci used for forensic analysis of DNA. The TPOX locus has the repeat structure [TAGA]n. Based on this information, you can infer that the TPOX allele with the structure [TAGA]10 will be ______ nucleotides longer than the allele with the structure [TAGA]5. Type the appropriate number to fill in the blank.For the following sequence please design an 18 base pair forward primer. ATG TCA AAA GCT GTC GGT ATT GAT TTA GGT ACA ACA TAC TCG TGT GTT GCT CAC TTT GCT TAAIn addition to the standard base-paired helical structures, DNA can form X-shaped hairpin structures called cruciforms in which most bases are involved in Watson–Crick pairs. Such structures tend to occur at sequences with inverted repeats. Draw the cruciform structure formed by the DNA sequence TCAAGTCCACGGTGGACTTGC.
- Using figure 1 and the following background information answer the following questions. Identification of the genetic cause of hornlessness in cattle has been the subject of intensive genetic and genomic research, culminating in the nomination of two different candidate neomutations on cattle chromosome 1 that are predicted to have arisen 500-1,000 years ago: a complex allele of Friesian origin (PF), an 80,128 base pair (bp) duplication (1909352–1989480 bp), and a second, simple allele of Celtic origin (PC) corresponding to a duplication of 212 bp (chromosome 1 positions 1705834–1706045) in place of a 10-bp deletion (1706051–1706060)We report the use of genome editing using transcription activator-like effector nucleases (TALENs) to introgress the putative PC POLLED allele into the genome of bovine embryo fibroblasts to try and produce a genotype identical to what is achievable using natural mating, but without the attendant genetic drag and admixture. In our previous studies, we…#2.) Mary wants to further confirm if she is a carrier for this lethal gene. She was told that a new SNP marker was identified RFPL marker. She requested to do SNP test for all the DNA samples. The results are: her parents are heterozygous for SNPs (that is to say, one chromosome is wild-type and another one contains the SNP). The sister who passed away is homozygous for the SNP. Mary does not have this SNP. So now what is the probability of Mary as a carrier of the death gene?Why are the first 20 bases typically ignored from a Sanger sequencing .ab1 file?
- The letter R isconventionally used to represent anypurine base (A or G) and Y to represent anypyrimidine base (T or C). In double-strandedDNA, what is the relation between %Rand %Y?The DNA of every individual in the pedigree shown in image B (below) has been sequenced at the causative locus, all the non-shaded individuals are wild type apart from III.1 and III.6. III.1 and III.6 have both been proven to have the causative allele for the condition but they do not exhibit any of the phenotypic signs or symptoms. Based on this pedigree, what is the level of penetrance for the condition? Please give your answer as a percentage to one decimal place, give the number only, no percentage symbol. Given the information above I calculate the level of penetrance seen in image B to be "Blank" 1 percent.The DNA of every individual in the pedigree shown in image B (below) has been sequenced at the causative locus, all the non- shaded individuals are wild type apart from III.1 and III.6. III.1 and III.6 have both been proven to have the causative allele for the condition but they do not exhibit any of the phenotypic signs or symptoms. Based on this pedigree, what is the level of penetrance for the condition? Please give your answer as a percentage to one decimal place, give the number only, no percentage symbol. ANSWER: Given the information above I calculate the level of penetrance seen in image B to be Blank 1 percent. A KEY Homozygous Homozygous Heterozygous Heterozygous Wild Type Male Female Male Female Male Note: Completely red symbol denotes an individual exhibiting the phenotype of interest CI || III IV V 1/4 1/2 1/2 1/2 1/2 Wild Type Female 1/4 1/2 Affected Known carrier Affected female Normal female Affected male Normal male D ●●●
- Please answer all parts along with the reason. I'll definitely give a like. Thank you in advance! 1A) From the cross Ab/aB x ab/ab, what is the recombination frequency if the progeny numbers are 17 AB/ab, 72 Ab/ab, 68 aB/ab, and 21 ab/ab? 1B)In human gene mapping, a LOD score is calculated to see if a gene causing a rare disease is linked to a known SNP. The LOD score is -4. This means that 1C) A three-point testcross is used to determine the order of three linked genes. The following crossover progeny result: single crossovers, double crossovers, and no crossovers. To determine the order, the no-crossover progeny must be compared to what other class of progeny?(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.Table I CACGT A GA CTGAGG ACTC CACGTAGACTGAG G ACAC Wild-type beta-globin gene fragment Sickle-cell beta-globin gene fragment > Circle the mutation in DNA of the sickle-cell beta-globin gene fragment Compare fragments of DNA the wild-type and mutant beta-globin genes in the Table I above, what are the similarities and differences you observe?