Sample Exam II
... 8. In the example above, what offspring would be expected if the two genes are 10 map units apart and the heterozygote has the dominant alleles on one chromosome and the recessive alleles on the other? 1. 45% of the offspring will exhibit A and B, 45% will exhibit a and b, 5% will exhibit A and b, a ...
... 8. In the example above, what offspring would be expected if the two genes are 10 map units apart and the heterozygote has the dominant alleles on one chromosome and the recessive alleles on the other? 1. 45% of the offspring will exhibit A and B, 45% will exhibit a and b, 5% will exhibit A and b, a ...
Unit 3 Practice Exam
... a. the age of selected fossils is calculated. b. organisms with traits well suited to their environment survive and reproduce at a greater rate than less well-adapted organisms in the same environment. c. acquired traits are passed on from one generation to the next. d. All of the above 11. The proc ...
... a. the age of selected fossils is calculated. b. organisms with traits well suited to their environment survive and reproduce at a greater rate than less well-adapted organisms in the same environment. c. acquired traits are passed on from one generation to the next. d. All of the above 11. The proc ...
Plant Nuclear Genome Size Variation
... Class 2 TEs are excised during transposition and may undergo “cut and paste” transposition with no duplication or “gap repair” where the gap is filled with a copy of the transposon ...
... Class 2 TEs are excised during transposition and may undergo “cut and paste” transposition with no duplication or “gap repair” where the gap is filled with a copy of the transposon ...
GENE EXPRESSION CHAPTER 11
... the enzyme lactase. Lactase will only be made if necessary. This will save the bacteria energy. If lactose, the inducer, is not present, than transcription of the mRNA that is translated into lactase is not made. Once enough mRNA is produced, it is immediately translated, even if the entire mRNA str ...
... the enzyme lactase. Lactase will only be made if necessary. This will save the bacteria energy. If lactose, the inducer, is not present, than transcription of the mRNA that is translated into lactase is not made. Once enough mRNA is produced, it is immediately translated, even if the entire mRNA str ...
GO ON TO THE NEXT PAGE. -28- 8. In 1952 Alfred Hershey and
... 8. In 1952 Alfred Hershey and Martha Chase used radioactive phosphorus and radioactive sulfur to selectively label the DNA and proteins of bacteriophage T2, a virus that infects bacteria. After incubating the labeled bacteriophage particles with Escherichia coli and separating extracellular phage pa ...
... 8. In 1952 Alfred Hershey and Martha Chase used radioactive phosphorus and radioactive sulfur to selectively label the DNA and proteins of bacteriophage T2, a virus that infects bacteria. After incubating the labeled bacteriophage particles with Escherichia coli and separating extracellular phage pa ...
lecture notes
... Sex cells are ‘haploids’ having a single chromosome and produced by meiosis Gene expression All cells have the same chromosome. Genes express themselves to perform the cells respective function Question : is the genome analogous to a computer program? Genome is too complex and gene expression ...
... Sex cells are ‘haploids’ having a single chromosome and produced by meiosis Gene expression All cells have the same chromosome. Genes express themselves to perform the cells respective function Question : is the genome analogous to a computer program? Genome is too complex and gene expression ...
Final Review
... 13. What organisms did Mendel use for his experiments? Why did he choose these organisms? What traits did he observe? 14. Explain the difference between genotype and phenotype. 15. Complete the following crosses and give the genotype & phenotype ratios: a. Qq x Qq ...
... 13. What organisms did Mendel use for his experiments? Why did he choose these organisms? What traits did he observe? 14. Explain the difference between genotype and phenotype. 15. Complete the following crosses and give the genotype & phenotype ratios: a. Qq x Qq ...
The entire human genome consists of 23 pairs of chromosomes
... The entire human genome consists of 23 pairs of chromosomes – long, coiled strands of DNA. Each chromosome may contain thousands of genes – sections of DNA that encode the information needed to create the proteins that are essential for all biological functions. NUCLEOSOME ...
... The entire human genome consists of 23 pairs of chromosomes – long, coiled strands of DNA. Each chromosome may contain thousands of genes – sections of DNA that encode the information needed to create the proteins that are essential for all biological functions. NUCLEOSOME ...
Recombinant DNA and Cloning The Impact of Biotechnology
... GloFish, marketed as the world’s first GM-pet ...
... GloFish, marketed as the world’s first GM-pet ...
Construction of Reporter Luciferase Genes to Assess NOC4
... Vectors have four features they are able to replicate they have selectable markers foreign DNA can be inserted in them they often carry a reporter gene ...
... Vectors have four features they are able to replicate they have selectable markers foreign DNA can be inserted in them they often carry a reporter gene ...
Biology 303 EXAM III
... Two types of post-transcriptional modifications that take place in the mRNA of eukaryotes are 1. the addition of a poly T sequence at the 5' end of the gene and the addition of a poly U tail at the 3' end. 2. addition of a poly A sequence at the 5' end and the addition of a “cap” at the 3' end of th ...
... Two types of post-transcriptional modifications that take place in the mRNA of eukaryotes are 1. the addition of a poly T sequence at the 5' end of the gene and the addition of a poly U tail at the 3' end. 2. addition of a poly A sequence at the 5' end and the addition of a “cap” at the 3' end of th ...
Repeated DNA sequences - lecture 1
... during meiosis. When unequal crossing over is combined with a bit of gene conversion (see next lecture) then it can account for variation in copy number, and homogeneity of sequence, between rRNA genes (and more generally in other types of repeat sequence). ...
... during meiosis. When unequal crossing over is combined with a bit of gene conversion (see next lecture) then it can account for variation in copy number, and homogeneity of sequence, between rRNA genes (and more generally in other types of repeat sequence). ...
Genetics review sheet VOCABULARY- on the test, the vocabulary
... VOCABULARY- on the test, the vocabulary section will be fill in the blank with a word bank 1. gene ...
... VOCABULARY- on the test, the vocabulary section will be fill in the blank with a word bank 1. gene ...
Welkin`s Presentation on Assigning and Correctly
... number. – Most of these will be supported by HHpred and BLAST on NCBI. ...
... number. – Most of these will be supported by HHpred and BLAST on NCBI. ...
BIOLOGY 207 - Dr.McDermid Lecture #1: DNA is the Genetic Material
... Figure 8-3 Bacteriophage (bacterial virus) T2 Radioisotope 32P to follow DNA; P not found in protein 35S labels protein; S not found in DNA Results 35S protein -> 32P DNA -> Conclusion: If DNA is the hereditary material then: 1) How do cells replicate their DNA? 2) How is genetic information stored? ...
... Figure 8-3 Bacteriophage (bacterial virus) T2 Radioisotope 32P to follow DNA; P not found in protein 35S labels protein; S not found in DNA Results 35S protein -> 32P DNA -> Conclusion: If DNA is the hereditary material then: 1) How do cells replicate their DNA? 2) How is genetic information stored? ...
Key Concepts File - Northwest ISD Moodle
... cells). Eukaryotic multicellular organisms reproduce sexually by combining two gametes containing homologous chromosomes (one set of chromosomes from each parent) during fertilization. Crossing over during meiosis allows for the reshuffling of genetic combinations between individual homologous chrom ...
... cells). Eukaryotic multicellular organisms reproduce sexually by combining two gametes containing homologous chromosomes (one set of chromosomes from each parent) during fertilization. Crossing over during meiosis allows for the reshuffling of genetic combinations between individual homologous chrom ...
PowerPoint 簡報
... only 6kb (kilobases) up to more than 2000kb, with the human version being some 16kb in length. In animals and plants, mitochondria are maternally inherited through the egg cytoplasm. • mtDNA does not appear to undergo recombination and in mammals evolves about tenfold faster than nuclear DNA, make i ...
... only 6kb (kilobases) up to more than 2000kb, with the human version being some 16kb in length. In animals and plants, mitochondria are maternally inherited through the egg cytoplasm. • mtDNA does not appear to undergo recombination and in mammals evolves about tenfold faster than nuclear DNA, make i ...
Cow DNA: How DNA Controls the Workings of the Cell
... this case, the sequence contains the gene to make the protein insulin. Insulin is necessary for the uptake of sugar from the blood. Without insulin, a person cannot use digest sugars the same way others can, and they have a disease called diabetes. Instructions: 1. Using the DNA sequence, make a com ...
... this case, the sequence contains the gene to make the protein insulin. Insulin is necessary for the uptake of sugar from the blood. Without insulin, a person cannot use digest sugars the same way others can, and they have a disease called diabetes. Instructions: 1. Using the DNA sequence, make a com ...
Section 1.3 Name:
... • In order to prepare for protein synthesis in the cytoplasm, DNA must copy its genetic instructions into messenger RNA (mRNA) molecules. This process is known as ____________________ (see Figure 10-7 on page 191). Transcription: ...
... • In order to prepare for protein synthesis in the cytoplasm, DNA must copy its genetic instructions into messenger RNA (mRNA) molecules. This process is known as ____________________ (see Figure 10-7 on page 191). Transcription: ...
Worksheet – DNA and Protein Synthesis Biology 11 Name: DNA
... 2. What is the main difference between the structure of chromatin and the structure of chromosomes? ...
... 2. What is the main difference between the structure of chromatin and the structure of chromosomes? ...
Genetics Study Guide Answers
... A particular triplet of bases in the template strand of DNA is 5' AGT 3'. The corresponding codon for the mRNA transcribed is A) 3' UCA 5'. B) 3' UGA 5'. C) 5' TCA 3'. D) 3' ACU 5'. E) either UCA or TCA, depending on wobble in the first base. Which of the following does not occur in prokaryotic euka ...
... A particular triplet of bases in the template strand of DNA is 5' AGT 3'. The corresponding codon for the mRNA transcribed is A) 3' UCA 5'. B) 3' UGA 5'. C) 5' TCA 3'. D) 3' ACU 5'. E) either UCA or TCA, depending on wobble in the first base. Which of the following does not occur in prokaryotic euka ...
Presentation
... from their genes where others will not based on histone modifications One twin may express a trait or get a disease that the other does not, despite same genes Schizophrenia Some cancers Etc. ...
... from their genes where others will not based on histone modifications One twin may express a trait or get a disease that the other does not, despite same genes Schizophrenia Some cancers Etc. ...
Biology 105
... An RNA primer is needed at the beginning point of DNA replication. This primer is synthesized by DNA primase The RNA primer is later filled in with DNA ...
... An RNA primer is needed at the beginning point of DNA replication. This primer is synthesized by DNA primase The RNA primer is later filled in with DNA ...