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Unit 3 Review This version is for posting to the class web site. Genes are located on chromosomes and are the basic unit of heredity that is passed on from parent to child, through generations. A. Explain how a chromosome mutation could occur and why mutations are detrimental to the organism in which they take place. B. Explain why human males may suffer from having just one copy of the X chromosome, while females have two. 5 end O OH P –O O H2C Remember: phive-phosphate What do 3 and 5 stand for??? What does antiparallel mean? O O P –O O H2C OH O O A T O O O P –O O H2C O O O P –O O H2C 3 end Hydrogen bond G C C O CH2 O O– P O O G A 3 end Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings O CH2 O O– P O O O CH2 O O– P O O T OH Figure 16.7b O CH2 O O– P O O (b) Partial chemical structure 5 end Punnett Squares • Punnett squares help us visualize segregation of allelles: All F1’s show dominant trait 3:1 F2 ratio on eye color trait indicates heterozygous cross. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Operons are ___karyotic lac operon DNA lacl lacz 3 mRNA 5 lacA RNA polymerase mRNA 5' 5 mRNA -Galactosidase Protein Allolactose (inducer) lacY Permease Transacetylase Inactive repressor (b) Lactose present, repressor inactive, operon on. Allolactose, an isomer of lactose, derepresses the operon by inactivating the repressor. In this way, the enzymes for lactose utilization are induced. Why can’t this dad give hemophilia to his sons? How could a girl end up with hemophilia? Crossing Over Nonsister chromatids Prophase I of meiosis • Crossing over: Tetrad – when does this happen? – the closer two alleles are on a chromosome, the ____ chance of being separated by crossing over. Metaphase I Chiasma, site of crossing over Metaphase II Daughter cells Figure Copyright © 2005 Pearson Education, Inc. publishing as Benjamin13.11 Cummings Recombinant chromosomes How can you show that this pedigree is of an autosomal recessive trait? What’s the diff between autosomal & sex-linked? Multiple Alleles • Most genes exist in more than two allelic forms – Ex: ABO blood groups – Which type of blood cell(s) would be rejected by a person with type B blood? Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings mom dad A ? A ? A B B B What is Dad if 50% of offspring come out type A, and 50% come out type B? Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings If Y is a lethal allele, and it’s dominant, who will survive here? Can you think of an example of a lethal dominant that does persist in the population because it doesn’t kill till middle age? 12. Y y Y YY Yy y Yy yy Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Sporophytes produce __ploid ______s by ___osis. Gametophyes produce __ploid _______s by ___osis. Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings • What is Independent assortment? – pairs of maternal and paternal homologues sort into gametes independently of the other pairs Key Maternal set of chromosomes Possibility 1 Paternal set of chromosomes Two equally probable arrangements of chromosomes at metaphase I Possibility 2 Metaphase II Daughter cells Figure 13.10 Combination 1 Combination 2 Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Combination 3 Combination 4 W=waxy, w=dull; G=green, g=yellow •You can do a dihybrid Punnett for this problem, (WwGg x WwGg) or… – find probability of each character separately (dull and green) – then, multiply probabilities of dull and green together. – dull Green = wwGG or wwGg F1 W w W WW Ww w Ww ww x F1 G g G GG Gg g Gg Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings gg = ¼ x ¾ = 3/16 3/16 x 144 = 27 How are these virusus the same? How are they different? Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Same in all eukaryotes, from yeast to you. KNOW IT! Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings What are restriction enzymes? What are they used for? Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Movement of Transposons and Retrotransposons • Eukaryotic transposable elements are of two types – Transposons, which move within a genome by means of a DNA intermediate Transposon DNA of genome New copy of transposon Transposon is copied Insertion Mobile transposon (a) Transposon movement (“copy-and-paste” mechanism) – Retrotransposons, which move by means of an RNA intermediate Figure 19.16a, b Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Retrotransposon New copy of retrotransposon DNA of genome RNA Reverse transcriptase (b) Retrotransposon movement Insertion DNA Methylation • methylation of cytosines on the DNA strand lead to tight packing & reduces transcription • methylation patterns are copied during mitosis • what do we call it if methylation is passed to the next generation? Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings KNOW IT! Figure 17.26 Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings • Histone acetylation … – loosens chromatin structure and enhance transcription Unacetylated histones Figure 19.4 b Acetylated histones (b) Acetylation of histone tails promotes loose chromatin structure that permits transcription Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Lytic & Lysogenic phage infections: The Telemere problem in DNA Replication • The ends of eukaryotic chromosomes get shorter with each round of replication 5 Leading strand Lagging strand End of parental DNA strands 3 Last fragment Lagging strand Previous fragment RNA primer 5 3 Primer removed but cannot be replaced with DNA because no 3 end available 3 for DNA polymerase Telemerase can restore these ends Removal of primers and replacement with DNA where a 3 end is available 5 Second round of replication 5 New leading strand 3 New lagging strand 5 3 Further rounds of replication Figure 16.18 Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Shorter and shorter daughter molecules 5 DNA Amplification: Target sequence 3 Genomic DNA • PCR procedure: – use specific primers to bind to each end of the segment you want. – use a heat resistant DNA polymerase – after about 20 cycles, target DNA (white boxes) greatly outnumber longer strands. 3 1 Cycle 1 yields 2 molecules 2 Denaturation: Heat briefly to separate DNA strands 5 3 3 5 Annealing: Cool to allow primers to Cycle 3 yields 8 molecules; 2 molecules (in white boxes) match target sequence Primers hydrogen-bond. Extension: 3 DNA polymerase adds nucleotides to the 3 end of each primer Cycle 2 yields 4 molecules 5 New nucleotides Normal -globin allele • Restriction fragment analysis – Is useful for comparing two different DNA molecules, such as two alleles for a gene 201 bp 175 bp DdeI DdeI Large fragment DdeI DdeI Sickle-cell mutant -globin allele Large fragment 376 bp DdeI DdeI DdeI (a) DdeI restriction sites in normal and sickle-cell alleles of -globin gene. Normal allele Sickle-cell allele Large fragment 376 bp 201 bp 175 bp Figure 20.9a, b (b) Electrophoresis of restriction fragments from normal and sickle-cell alleles. How many cuts to get 9 pieces? Why would we want to “steal” a gene and put it in a bacterial plasmid? How could you do it? RNA Polymerase Binding and Initiation of Transcription • Promoters signal the initiation of RNA synthesis DNA RNA PROCESSING Pre-mRNA mRNA Ribosome TRANSLATION 1 Eukaryotic promoters Polypeptide Promoter 5 3 • Transcription factors TRANSCRIPTION TATA box – Help eukaryotic RNA polymerase bind to promoter sequences 3 5 T A T A A AA AT AT T T T Start point Template DNA strand Transcription factors 5 3 3 5 3 Additional transcription factors RNA polymerase II 5 3 Transcription factors 3 5 5 RNA transcript Figure 17.8 Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Transcription initiation complex • The ribosomal subunits – Are constructed of proteins and RNA molecules named ribosomal RNA or rRNA Figure 17.16a Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings • The DNA template (gene) determines the order of bases in the mRNA transcribed alongside. • mRNA determines the order of Amino Acids during translation Gene 2 DNA molecule Gene 1 Gene 3 DNA strand 3 5 A C C A A A C C G A G T (template) TRANSCRIPTION mRNA 5 U G G U U U G G C U C A Codon TRANSLATION Protein Figureas17.4 Copyright © 2005 Pearson Education, Inc. publishing Benjamin Cummings Trp Amino acid Phe Gly Ser 3 • Spliceosomes remove the __ons from pre mRNA RNA transcript (pre-mRNA) 5 Intron Exon 1 Exon 2 Protein 1 Other proteins snRNA snRNPs sn = small nuclear Spliceosome 2 5 Spliceosome components Figure 17.11 3 Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings mRNA 5 Exon 1 Exon 2 Cut-out intron Know how to use this table to determine which amino acids the mRNA is coding for! Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings Substitutions • A base-pair substitution can cause – missense or – nonsense Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings • DNA replication is semiconservative… – Each new daughter molecule has one old strand and one newly made strand Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings