Biao Ding*, Myoung-Ok Kwon and Leif Warnberg
... kDa R-dextran(b). The photographwas taken 1 min after injection. (c and d) Thirty minutes after injection, the 20 kDa F-dextranwas clearlyvisible in ceils surrounding the injected cell (c), but the 40 kDa R-dextranstill remainedin the injected cell (d). The bar represents30 pm. ...
... kDa R-dextran(b). The photographwas taken 1 min after injection. (c and d) Thirty minutes after injection, the 20 kDa F-dextranwas clearlyvisible in ceils surrounding the injected cell (c), but the 40 kDa R-dextranstill remainedin the injected cell (d). The bar represents30 pm. ...
Cloning and characterisation of a cysteine proteinase gene
... amastigotes migrating around 30 kDa was completely inhibited after incubation of gelatin-coupled gels in the presence of E-64. 3.2. Cloning of a cysteine proteinase gene (Llacys1) from L. (L.) amazonensis By using genomic DNA of L. (L.) amazonensis amastigotes and a pair of primers derived from evol ...
... amastigotes migrating around 30 kDa was completely inhibited after incubation of gelatin-coupled gels in the presence of E-64. 3.2. Cloning of a cysteine proteinase gene (Llacys1) from L. (L.) amazonensis By using genomic DNA of L. (L.) amazonensis amastigotes and a pair of primers derived from evol ...
Activity: Invasion of the Snorks
... 1. Create the data charts in your lab book. Make sure to leave enough room to have all of the necessary information present. 2. Using the mRNA from the Snork, find the missing strand of DNA belonging to the Snork. From what we know about the Snorks, the base pairing rules are the same as us. 3. Code ...
... 1. Create the data charts in your lab book. Make sure to leave enough room to have all of the necessary information present. 2. Using the mRNA from the Snork, find the missing strand of DNA belonging to the Snork. From what we know about the Snorks, the base pairing rules are the same as us. 3. Code ...
Prophase II
... A special process of cell division that results in haploid sex cells The total number of chromosomes decreases to half the original number ...
... A special process of cell division that results in haploid sex cells The total number of chromosomes decreases to half the original number ...
meiosis_and_sexual_life_cycles
... In a literal sense, children do not inherit particular physical traits from their parents. It is genes that are actually inherited. ...
... In a literal sense, children do not inherit particular physical traits from their parents. It is genes that are actually inherited. ...
"Regulation of Prokaryotic Gene Expression". In: Microbial
... Prokaryotic gene expression is classically viewed as being controlled at two basic levels: DNA transcription and RNA translation. However, it will become apparent that mRNA degradation, modification of protein activity, and protein degradation also play important regulatory roles. This chapter deals ...
... Prokaryotic gene expression is classically viewed as being controlled at two basic levels: DNA transcription and RNA translation. However, it will become apparent that mRNA degradation, modification of protein activity, and protein degradation also play important regulatory roles. This chapter deals ...
Intra-genomic 16S rRNA gene heterogeneity in
... may, therefore, be a consequence of specialized organisms living in a uniform environment. 16S rRNA gene heterogeneity In total, 62.7% of all cyanobacterial genomes and 64.3% of filamentous forms contained more than one ribosomal operon (Table 1). Among these 37 cyanobacterial genomes with multiple ...
... may, therefore, be a consequence of specialized organisms living in a uniform environment. 16S rRNA gene heterogeneity In total, 62.7% of all cyanobacterial genomes and 64.3% of filamentous forms contained more than one ribosomal operon (Table 1). Among these 37 cyanobacterial genomes with multiple ...
Journal - International Journal of Systematic and Evolutionary
... The nearly complete 16S rRNA gene sequence and partial pmoA gene sequence of strain E10T were determined in a previous study (Ferrando & Tarlera, 2009). The phylogenetic analysis based on the 16S rRNA gene clearly showed that strain E10T represents a new distinct line of descent within the type I me ...
... The nearly complete 16S rRNA gene sequence and partial pmoA gene sequence of strain E10T were determined in a previous study (Ferrando & Tarlera, 2009). The phylogenetic analysis based on the 16S rRNA gene clearly showed that strain E10T represents a new distinct line of descent within the type I me ...
LWW PPT Slide Template Master
... 4. The process of body cell division is called: (a) separation ...
... 4. The process of body cell division is called: (a) separation ...
Section 11–4 Meiosis (pages 275–278) This section explains
... Both Mitosis and Meiosis begin with a ...
... Both Mitosis and Meiosis begin with a ...
1 The Diversity of Cells
... new single-celled organism. It has a cell wall, ribosomes, and long, circular DNA. Is it a eukaryote or a prokaryote cell? Explain. 8. Identifying Relationships You are looking at a cell under a microscope. It is a single cell, but it also forms chains. What characteristics would this cell have if t ...
... new single-celled organism. It has a cell wall, ribosomes, and long, circular DNA. Is it a eukaryote or a prokaryote cell? Explain. 8. Identifying Relationships You are looking at a cell under a microscope. It is a single cell, but it also forms chains. What characteristics would this cell have if t ...
Rate of Gene Transfer From Mitochondria to Nucleus
... than in plants, implying a higher transfer rate of animals. The evolution of gene transfer may have been affected by an intensity of intracellular competition among organelle strains and the organelle inheritance system of the organism concerned. This article reveals a relationship between those fac ...
... than in plants, implying a higher transfer rate of animals. The evolution of gene transfer may have been affected by an intensity of intracellular competition among organelle strains and the organelle inheritance system of the organism concerned. This article reveals a relationship between those fac ...
biol 4469 – molecular biology - School of Biological Sciences
... 2) Most lectures use outside sources (provided via Web) in addition to the textbook. 3) If you have previously taken BIOL 4469 (Mol. Biology) or BIOL 7668 (Euk. Mol. Genetics - graduate), you cannot get additional credit for BIOL 4668. 4) Exam will be given only during class hours on the dates speci ...
... 2) Most lectures use outside sources (provided via Web) in addition to the textbook. 3) If you have previously taken BIOL 4469 (Mol. Biology) or BIOL 7668 (Euk. Mol. Genetics - graduate), you cannot get additional credit for BIOL 4668. 4) Exam will be given only during class hours on the dates speci ...
BI 112 Instructor: Waite Exam #4 Study Guide Cell Membrane
... Understand how RNA is processed after transcription before it is translated; know the difference between introns and exons, which one codes for protein, and the 2 main advantages of storing coding information in this way ...
... Understand how RNA is processed after transcription before it is translated; know the difference between introns and exons, which one codes for protein, and the 2 main advantages of storing coding information in this way ...
Expansion of tandem repeats and oligomer
... interest due to their role in genome organization and evolutionary processes [1–11]. It is known that SSR constitute a large fraction of noncoding DNA and are relatively rare in protein coding sequences. SSR are of considerable practical and theoretical interest due to their high polymorphism [7]. T ...
... interest due to their role in genome organization and evolutionary processes [1–11]. It is known that SSR constitute a large fraction of noncoding DNA and are relatively rare in protein coding sequences. SSR are of considerable practical and theoretical interest due to their high polymorphism [7]. T ...
Bdellovibrio
... differentiation. This recombination is catalyzed by the xisF gene which acts at two directly repeating 5 bp sequences within the fdxN gene (shown as green triangles above). As a result, the nifB-fdxN-nifS-nifU operon can then be expressed properly. * a 10.6 kb element is excised from the hupL gene, ...
... differentiation. This recombination is catalyzed by the xisF gene which acts at two directly repeating 5 bp sequences within the fdxN gene (shown as green triangles above). As a result, the nifB-fdxN-nifS-nifU operon can then be expressed properly. * a 10.6 kb element is excised from the hupL gene, ...
igcse biology (double award) year 11 learning objectives for the first
... Flowering plants 3.3 describe the structures of an insect-pollinated and a wind-pollinated flower and explain how each is adapted for pollination Students will be assessed on their ability to: 3.4 understand that the growth of the pollen tube followed by fertilisation leads to seed and fruit formati ...
... Flowering plants 3.3 describe the structures of an insect-pollinated and a wind-pollinated flower and explain how each is adapted for pollination Students will be assessed on their ability to: 3.4 understand that the growth of the pollen tube followed by fertilisation leads to seed and fruit formati ...
Module 6: Enzymatic Function
... MetaCyc provides an abundance of information. Your protein will only participate in one of these pathways if it has an enzymatic function. If you have determined that your protein is NOT an enzyme you can skip this module (state that fact in your notebook). 2. Navigate to MetaCyc at http://metacyc.o ...
... MetaCyc provides an abundance of information. Your protein will only participate in one of these pathways if it has an enzymatic function. If you have determined that your protein is NOT an enzyme you can skip this module (state that fact in your notebook). 2. Navigate to MetaCyc at http://metacyc.o ...
Chloroplast DNA and Molecular Phylogeny
... restriction-site mutations found among a given group of cpDNAs. The tree shown in Fig. 3 is based on analysis of the 10 cpDNAs shown in Fig. 2, plus five other DNAs, with25 different restriction endonucleases.I3 Among the 15 DNAs, a total of only 40 independent restriction site mutations (Fig. 3) we ...
... restriction-site mutations found among a given group of cpDNAs. The tree shown in Fig. 3 is based on analysis of the 10 cpDNAs shown in Fig. 2, plus five other DNAs, with25 different restriction endonucleases.I3 Among the 15 DNAs, a total of only 40 independent restriction site mutations (Fig. 3) we ...