1 / 34

Unit 3 Review

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.

garcianancy
Download Presentation

Unit 3 Review

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Unit 3 Review This version is for posting to the class web site.

  2. Genes are located on chromosomes and are the basic unit of heredity that is passed on from parent to child, through generations. • Explain how a chromosome mutation could occur and why mutations are detrimental to the organism in which they take place. • Explain why human males may suffer from having just one copy of the X chromosome, while females have two.

  3. O O O O– O– O– P P P O O O O O O 5 end O OH 3 end Hydrogen bond P Remember: phive-phosphate What do 3 and 5 stand for??? What does antiparallel mean? –O O OH H2C O A T O CH2 O O P O –O O– O P O H2C O O G C O O CH2 O P –O O H2C O C G O O CH2 O P –O O H2C O A T O CH2 OH 3 end Figure 16.7b 5 end (b) Partial chemical structure

  4. 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.

  5. lac operon DNA lacl lacz lacY lacA RNApolymerase 3 mRNA 5 mRNA 5' mRNA 5 -Galactosidase Permease Transacetylase Protein Inactiverepressor Allolactose(inducer) (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. Operons are ___karyotic

  6. Why can’t this dad give hemophilia to his sons? How could a girl end up with hemophilia?

  7. Nonsister chromatids Prophase I of meiosis Tetrad Chiasma, site of crossing over Metaphase I Metaphase II Daughter cells Recombinant chromosomes Figure 13.11 Crossing Over • Crossing over: • when does this happen? • the closer two alleles are on a chromosome, the ____ chance of being separated by crossing over.

  8. How can you show that this pedigree is of an autosomal recessive trait? What’s the diff between autosomal & sex-linked?

  9. Multiple Alleles • Most genes exist in more than two allelic forms • Ex: ABO blood groups • Which type of blood cell(s)would be rejectedby a person withtype B blood?

  10. mom dad What is Dad if 50% of offspring come out type A, and 50% come out type B?

  11. 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 populationbecause it doesn’t kill tillmiddle age?

  12. Sporophytes produce __ploid ______s by ___osis. Gametophyes produce __ploid _______s by ___osis.

  13. Key Maternal set of chromosomes Possibility 1 Possibility 2 Paternal set of chromosomes Metaphase II Daughter cells Combination 1 Combination 2 Combination 3 Combination 4 • What is Independent assortment? • pairs of maternal and paternal homologues sort into gametes independently of the other pairs Two equally probable arrangements of chromosomes at metaphase I Figure 13.10

  14. 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 dulland greentogether. • dull Green = wwGG or wwGg x = ¼ x ¾ =3/163/16 x 144 = 27

  15. How are these virusus the same? How are they different?

  16. Same in all eukaryotes, from yeast to you. KNOW IT!

  17. What are restriction enzymes? What are they used for?

  18. New copy of transposon Transposon DNA of genome Transposon is copied Insertion Mobile transposon (a) Transposon movement (“copy-and-paste” mechanism) New copy of retrotransposon Retrotransposon DNA of genome RNA Reverse transcriptase Insertion (b) Retrotransposon movement Movement of Transposons and Retrotransposons • Eukaryotic transposable elements are of two types • Transposons, which move within a genome by means of a DNA intermediate • Retrotransposons, which move by means of an RNA intermediate Figure 19.16a, b

  19. 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?

  20. KNOW IT! Figure 17.26

  21. Acetylated histones Unacetylated histones (b) Acetylation of histone tails promotes loose chromatin structure that permits transcription Figure 19.4 b • Histone acetylation … • loosens chromatin structure and enhance transcription

  22. Lytic & Lysogenic phage infections:

  23. 5 Leading strand Lagging strand End of parental DNA strands 3 Last fragment Previous fragment Lagging strand RNA primer 5 3 Removal of primers and replacement with DNA where a 3 end is available 5 3 Second round of replication 5 New leading strand 3 New lagging strand 5 3 Further rounds of replication Figure 16.18 Shorter and shorter daughter molecules The Telemere problem in DNA Replication • The ends of eukaryotic chromosomes get shorter with each round of replication Primer removed but cannot be replaced with DNA because no 3 end available for DNA polymerase Telemerase can restore these ends

  24. 3 2 1 3 5 Target sequence 3 Genomic DNA 5 3 5 Denaturation: Heat briefly to separate DNA strands 5 3 Annealing: Cool to allow primers to hydrogen-bond. Cycle 1 yields 2 molecules Primers Extension: DNA polymerase adds nucleotidesto the 3 end of each primer Newnucleo-tides Cycle 2 yields 4 molecules Cycle 3 yields 8 molecules; 2 molecules (in white boxes) match target sequence DNA Amplification: • 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.

  25. DdeI DdeI DdeI DdeI Normal  -globin allele 201 bp Large fragment 175 bp Sickle-cell mutant -globin allele Large fragment 376 bp DdeI DdeI DdeI (a) DdeIrestriction sites in normal and sickle-cell alleles of -globin gene. Sickle-cellallele Normalallele Largefragment 376 bp 201 bp175 bp (b) Electrophoresis of restriction fragments from normal and sickle-cell alleles. Figure 20.9a, b • Restriction fragment analysis • Is useful for comparing two different DNA molecules, such as two alleles for a gene

  26. How many cuts to get 9 pieces?

  27. Why would we want to “steal” a gene and put it in a bacterial plasmid? How could you do it?

  28. Eukaryotic promoters 1 Additional transcription factors 3 RNA Polymerase Binding and Initiation of Transcription • Promoters signal the initiation of RNA synthesis • Transcription factors • Help eukaryotic RNA polymerase bind to promoter sequences TRANSCRIPTION DNA Pre-mRNA RNA PROCESSING mRNA Ribosome TRANSLATION Polypeptide Promoter 5 3 A T A T A A A A T A T T T T 3 5 TATA box Start point Template DNA strand Transcription factors 5 3 3 5 RNA polymerase II Transcription factors 3 5 5 3 5 RNA transcript Transcription initiation complex Figure 17.8

  29. The ribosomal subunits • Are constructed of proteins and RNA molecules named ribosomal RNA or rRNA Figure 17.16a

  30. Gene 2 DNA molecule Gene 1 Gene 3 DNA strand (template) 5 3 A C C T A A A C C G A G TRANSCRIPTION A U C G C U G G G U U U 5 mRNA 3 Codon TRANSLATION Gly Phe Protein Trp Ser Figure 17.4 Amino acid • The DNA template (gene) determines the order of bases in the mRNA transcribed alongside. • mRNA determines the order of Amino Acids during translation

  31. 3 1 2 RNA transcript (pre-mRNA) 5 Intron Exon 1 Exon 2 Protein Other proteins snRNA snRNPs Spliceosome 5 Spliceosome components Cut-out intron mRNA 5 Exon 1 Exon 2 • Spliceosomes remove the __ons from pre mRNA sn = small nuclear Figure 17.11

  32. Know how to use this table to determine which amino acids the mRNA is coding for!

  33. Substitutions • A base-pair substitution can cause • missense or • nonsense

  34. DNA replication is semiconservative… • Each new daughter molecule has one old strand and one newly made strand

More Related