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- D Question 23 What is predictive DNA Phenotyping? O a method using known phenotypes to predict a suspect's DNA profile an STR profile a method using DNA sequence information to predict identifying phenotypes O a genotypic profile based on combined mitochondrial and Y-chromosome STR genotypes Question 24Question 3 Review genome size, gene number, and gene density. Select the false statement. Number of genes is not correlated to genome size O Genomes of most bacteria and archaea range from 1 to 6 million base pairs (Mb) O Eukaryotic genomes tend to be larger. O Free-living bacteria and archaea have 1,500 to 7,500 genes. O Humans and other mammals have the highest gene density, or number of genes, in a given length of DNAQuestion 12 What does this illustration show? O cladogram dichotomous key DNA barcode binomial nomenclature branches
- QUESTION 1 You want to perform PCR on the CDNA of the spike gene from a SARS CoV-2 sample so that you can sequence it. Based on the sequence below, which of the following primer pairs would probably work for PCR of this gene? Spike gene Sequence: 5' ATGTTTATTTTCTTATTATTTTTTACTCTCACTAGTGGTAGTGACCTTGACCGGTGCACCACTTTTGATG ATGTTCAAGCTCCTAATTACACTCAACATACTTCATCTATGAGGGGGGTT TACTATCCTGATGAAATTTT. .. (it's really long so didn't post the whole thing.).TCTTGCTTTGTTGCATGACTAGTTGTTGCAGTTGCCTCAAGGGTGCATGCTCTTGTGGTTCTTGCTGCAA GTTTG ATGAGGATGACTCTGAGCCAGTTCTCAAGGGTGTCAAATTACATTACACATAA 3' Forward primer: 5' - CTC TCA CTA GTG GTA GTG ACC - 3' (Tm = 60.5 °C in a standard qPCR mix) Reverse Primer: 5' GGG TGT CAA ATT ACA TTA CAC ATA - 3' (Tm= 59.6 °C in a standard QPCR mix) Forward Primer: 5'- ATG TTT ATT TTC TTA TTA TT -3' (Tm=D 47.2 °C in a standard qPCR mix) Reverse Primer: 5'- GCA AGA ACC ACA AGA GCA TGC ACC -3' (Tm= 68 °C in a standard qPCR mix) Forward primer: 5' - CTC TCA CTA GTG GTA GTG ACC -3'…Preview File Edit View Go Tools Window Help )) 100% Thu Apr 29 10:44 PM A BI-201 Syllabus Virtual Spring 2021.pdf (page 1 of 19) A BioLab Fly_Group_F.pdf (page 1 of 4) Q Search DOCX v BioLab Fly Group_F.pdf HIST111 signment1 Mutation: Lobe eye shape P generation Phenotypes: Normal Lobe DOCX _female X male W1 - Puffins F. generation Phenotype Females Males Total Ratio Normal 8. 8. Lobe Screen Shot 021-04..3.03 AM F, XF, Phenotypes: Normal Normal female X male DOCX F2 generation HIST111 Phenotvoe Females Males Total Ratio Normal 6 6 Lobe 2 2 F. Punnett square F2 Punnett square DOCX HIST111 2 2 Atomic This mutation is inherited as: uctur...pdf .pdf dominant autosomal (da) recessive autosomal (ra) DOCX recessive sex-linked (rs) History homework4 APR 29 X FENEQUESTION 2 Variable expression of the MERRF syndrome arises from recombination between nuclear and mtDNA O homoplasmic cells mitotic segregation nuclear genes environmental factors
- Jekyll-Hyde Afflicted Spider Curse Afflicted Jekyll-Hyde/Spider Curse I O Human 1 II 1 III 4. Jekyll-Hyde disease is characterized by transformation into an unfeeling, aggressive, alter ego at night. While the "Spider Curse" is a disease that causes affected individuals to grow extra arms and extra eyes during a full moon. The genes responsible for these diseases sort independently and both traits run in one family. Based on the above pedigree answer the following questions: What is the mode of inheritance for Jekyll-Hyde disease? (Hint: look at individuals III-1 & III-3) O a. Autosomal Dominant O b. Autosomal Recessive O c. X-Linked Dominant O d. X-linked Recessive What evidence supports your hypothesis? Give at least 2 pieces of information from the pedigree that support your answer to 3a. Answer: 2. 3. 2. 2.Nullisomy is a genome mutation where a pair of homologous chromosomes that would normally be present is ______.Preview File Edit View Go Tools Window Help )) 100% Thu Apr 29 10:46 PM Q A BI-201 Syllabus Virtual Spring 2021.pdf (page 1 of 19) A BioLab Fly_Group_F.pdf (page 4 of 4) Q Search DOCX v BioLab Fly_Group_F.pdf HIST111 Mutation: Star eye shape signment1 P generation Phenotypes: Normal Star 2 _female X male DOCX W1 - Puffins Fi generation Phenotvne Females Males Total Ratio Normal 8 8 Star Screen Shot 021-04..3.03 AM F, XF, Phenotypes: Normal female X Normal male DOCX F2 generation HIST111 Phenotvoe Females Males Total Ratio Normal 4 Star 4 F. Punnett square F2 Punnett square DOCX HIST111 2 Atomic This mutation is inherited as: uctur...pdf .pdf dominant autosomal (da) recessive autosomal (ra) recessive sex-linked (rs) DOCX History homework4 APR 29 X HENE
- Extra Question Chapter 4 1. Consider the following cross concerning 4 different gene loci: AaBbCcDd (x) AabbCcdd From this cross, what is the probability of getting a progeny (offspring) with genotype AABbccdd? b. From this cross, what is the probability of getting a progeny (offspring) with genotype AabbCcDd? c. From this cross, what is the probability of getting a male progeny (offspring) with genotype aaBbCcdd? 2. Your neighbor has twelve children. One is blue eye color and short. Two are brown eye color and short. Two are blue eye color and tall. Seven look just like the parents; brown eye color with tall. What can you discover about the genetics of eye color and height of the children? a. How many traits are you dealing with? Each trait has phenotypes: Specify the phenotypes. b. What is the probability of the height of the children? What is the probability of the eye color of the children? (Refer to monohybrid punnett square slides 17-19) c. What are recessive traits based on the…Word AaBbCcI AaBb CcD AaB AaBbCсI AaBbCcL 4 1 Normal Subtitle Title 1 No Spaci... Subtle Em..... Create and Share Adobe PDF Styles Adobe Acro P 1 1 2 4 A 3. A AL T L 4 E 5 E Question 6: The largest chromosome in Mischievous gremlinus has a length of 9 x 107 base pairs. The rate of DNA synthesis is 3,000 bp per minute at each replication fork. a. How long does it take to replicate the entire chromosome from a single origin of replication located exactly in the middle of the chromosome? Assume no pauses. I b. Experiments performed by Professor Trink reveal that it takes only 20 minutes for a growing cell to replicate this chromosome. What is the minimum number of replicons present on the chromosome? Florian Fischer Find v Replace Select Editing 153 23Dogs T G A F C M Murgia et al., 2006 Cell E 100 H HC ·Α F ABIG ED L M B Tumors Observe this phylogeny of DNA derived of normal cells of eleven dogs (circles around tip names), and DNA derived from tumor cells of those same eleven dogs (no circles around tip names). Dog individuals are named A to M, and both the normal cell samples and the tumor cell samples carry the letter-names corresponding to the dog individual they were sampled from. What can we conclude from this phylogenetic tree? D The evolutionary distance among tumor cells of different dog individuals is greater than the evolutionary distance among normal cells of different dog individuals. The tumors are derived from an infectious dog cell: The tumor cells of each dog are not the descendants of normal cells of the corresponding individual dogs. Rather, the tumor cells of different dogs all share a common ancestor. As in all forms of tumorous cancer, the tumor cells of each dog are evolutionary descendants of normal cells of…