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Information Flow Lecture 1: The flow of biological information starts with the genetic code. PRE-CLASS WORK: READ: Ch. 3.6 Nucleic acids store, transmit, and help express hereditary information. 1. Complete the sketch of the DNA nucleotide below by labeling the following: ● Deoxyribose sugar ● Nitrogenous base ● Phosphate group ● Number each of the carbons on the sugar (1′, 2′, 3′, 4′, 5′) ● The free -OH group on the sugar ● The location on the sugar that makes deoxyribose different from ribose. 2. Nitrogenous bases: ● The purine bases are ______________________ and _______________ ● The pyrimidine bases are ________________ and ________________. ● In a molecule of DNA, a nucleotide with an adenine (A) base will always pair with a nucleotide with a ____________________ base. ● A nucleotide with a guanine (G) base will always pair with a nucleotide with a _____________ base. READ: Ch. 13.1 DNA is the genetic material. 3. Frederick Griffith’s classic experiment: ● What was Griffith’s central research goal? What was he studying at the time? ● Which of the four experimental results was unexpected? Explain. ● What hypothesis would explain the unexpected result? What did Griffith call this phenomenon? 1
4. DNA structure: ● 1. Using the last nucleotide in the chain below, label the three components of a DNA nucleotide: nitrogenous base, deoxyribose sugar, and phosphate group. ● 2. Circle a phosphodiester bond. (Your textbook calls this a sugar-phosphate covalent bond) ● 3. Label the 5’ and 3’ ends of the chain. ● 4. What chemical group is attached to the 5’ carbon of the deoxyribose sugar? ______________________ ● 5. What chemical group is attached to the 3’ carbon of the deoxyribose sugar? ______________________ ● 6. What makes up the backbone of a DNA molecule? ________________________ ● 7. The two sides of a DNA molecule are held together by ________________ bonds. 2. DNA has an antiparallel structure, with the ends of the two strands identified as 5’ and 3’. Make a brief simplified sketch below that demonstrates the antiparallel structure of DNA, using 5’ and 3. (Figure 13.8 has a nice review of the different ways that a DNA molecule may be drawn from detailed to simplified.) END OF PRE-CLASS 2
C ricket evolution: Identify mechanism of information flow at each level. Flatwing male normal normal Gene Mutant gene involved in hormone signaling RNA/Protein Cellular Action Phenotype Behavior Environment Identify mechanism of information flow at each level. 3
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Was Griffith’s “transforming principle” protein or DNA? Frederick Griffith (1928) “ transforming principle” Alfred Hershey and Martha Chase (1952) DNA from a virus enters the cell, not protein. Oswald Avery Maclyn McCarty (b. South Bend, IN 1911) Colin MacLeod DNA is responsible for bacterial transformation. (1944) The structure of DNA is essential for understanding its function in information flow. James Watson and Francis Crick (1953) Maurice Wilkins Raymond Gosling Rosalind Franklin Photo 51 4
Fidelity: the driving force of why information flow works. The “telephone game” - no fidelity checks ANALYSIS: Identify at least 3 ways in which the structure of DNA provides stability and fidelity of the genetic code. How does the complementarity of DNA help ensure fidelity? 5
"It has not escaped our notice that the specific pairing we have postulated immediately suggests a possible copying mechanism for the genetic material." - Watson and Crick, Nature, 1953 T A G C phosphate group deoxyribose sugar nitrogenous base 5’ end 3’ end phosphodiester bond nucleotide REPLICATE the DNA: DRAW the complementary strand including the correct number of hydrogen bonds, label the 5’ and 3’ ends 6
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DNA polymerase catalyzes the addition of new nucleotides. Formation of phophodiester bond Technique connection: Polymerase chain reaction (PCR) follows the basic rules of DNA replication. (Photo credit: Wikipedia, Creative Commons Licence) 7
REPLICATION FORK CHALLENGE PROBLEM 1. Label the ends of the existing strands as 5’ or 3’. 2. On both forks, draw in helicase and single stranded binding proteins to show where these proteins would be found. 3. ON THE LEFT HAND FORK: One of the strands is replicated continuously in the 5’to 3’ direction. Use the term leading strand template to label this strand. 4. The other strand is replicated in fragments. Use the term lagging strand template to label this strand. 5. Draw an RNA primer (synthesized by enzyme primase) on the leading strand template. If available, use a different color or style to differentiate RNA from DNA. 6. Add DNA polymerase III and add an arrow to represent the addition of new nucleotides by the enzyme. Label the 5’ and 3’ ends of the growing strand. 7. Draw the FIRST RNA primer that would form along the lagging strand template, and label as (1). Add an arrow to represent the new nucleotides that would be added by DNA polymerase III. Label 5’ and 3’ ends of the growing strand. 8. Repeat this process for the SECOND and THIRD RNA primers for the lagging strand, labeling as (2) and (3) . 9. REPEAT STEPS 3-7 ON THE RIGHT HAND FORK. 10. What enzyme would remove RNA nucleotides and replace them with DNA nucleotides? _________________________________. 11. Add arrows and labels to show locations where the enzyme DNA ligase would be active. 8
POS7-C+ECK 48ES7,ONS Question Cluster: The figure to the right is of a DNA nucleotide. Identify the following statements as either True or False: 1. This nucleotide contains a pyrimidine base. TRUE/FALSE 2. Alternating sugars and phosphates make up the backbone of a DNA molecule. TRUE/FALSE 3. This nucleotide would form 3 hydrogen bonds when base-paired with an adenine. TRUE/FALSE 4. During replication, a new nucleotide would be added at the 5’ carbon. TRUE/FALSE Multiple Choice: 5. In his work with pneumonia-causing bacteria and mice, Griffith found that: A. the protein coat from pathogenic cells was able to transform nonpathogenic cells. B. heat-killed pathogenic cells caused pneumonia. C. some substance from pathogenic cells was transferred to nonpathogenic cells, making them pathogenic. D. the polysaccharide coat of bacteria caused pneumonia. 6. A sample of DNA is found to have 35% thymine. What percent guanine would be expected? A. 10 B. 15 C. 35 D. 65 7. What is meant by the description "antiparallel" regarding the two strands of nucleic acids that make up DNA? A. The twisting nature of DNA creates nonparallel strands. B. The 5' to 3' direction of one strand runs counter to the 5' to 3' direction of the other strand. C. Base pairings create unequal spacing between the two DNA strands. D. One strand contains only purines and the other contains only pyrimidines. 8. Which of the following contribute to the fidelity of the genetic message stored in DNA? Select all TRUE statements. A. The complementary strands of DNA provide a template to make new DNA strands. B. The complementary strands of DNA provide a template to repair DNA. C. The bases of DNA nucleotides are protected in the middle of the helix. D. A relatively simple code using 4 letters can store the information to produce complex proteins. 9. If one strand of a DNA molecule has the sequence of bases 5’ATTGCA3’, the complementary strand would have the sequence: A. 5’TAACGT3’ B. 5’TGCAAT3’ C. 5’UAACGU3’ D. 5’UGCAAU3’ 9
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10. You are conducting a PCR experiment in the lab. In this image, Primer 1 and Primer 2 have annealed to the antiparallel strands of the template DNA during the PCR reaction. Given your understanding of the structure and directionality of DNA, which of the following statements about the template DNA is TRUE? A. A & D are 3 ′ ends B. C & B are 5 ′ ends C. C & A are 3 ′ ends D. A & D are 5 ′ ends 10
Information Flow Lecture 2: How does the cell maintain fidelity of the message? PRE-CLASS WORK READ: Section 13.2 Many proteins work together in DNA replication and repair. WATCH : DNA replication video 1. What do we mean that DNA replication is “semi-conservative”? Define the roles of the following in DNA replication: 2. Helicase: 3. Single-stranded binding proteins: 4. RNA primers: (make sure you know why RNA primers are needed) 5. Primase: 6. DNA polymerase III: 7. Okazaki fragments: (make sure you know why Okazaki fragments are needed) 8. DNA polymerase I: 9. DNA ligase: 11
10. Looping of the lagging strand (your textbook calls this the “trombone model”): 11. NOTE: Include a picture of your completed replication fork problem with today’s pre-class submission. WATCH: What happens when your DNA is damaged ? READ: Ch. 13.2 Proofreading and repairing DNA. 12. Given that initial base pairing errors during replication occur at a rate of 1 in 10 5 , why is the error rate in completed DNA much lower (1 in 10 10 )? 13. What is a nuclease? 14. What reaction does ligase carry out? What bond does it form in DNA? (google this answer). 15. What type of DNA damage is repaired by the processes of non-homologous end joining and homologous recombination? 16. THOUGHT QUESTION: What would happen if a cell never made a mistake during replication? Would this be beneficial for the organism? Why or why not? REMINDER: INCLUDE COMPLETED REPLICATION FORK WITH YOUR PRE-CLASS SUBMISSION END OF PRE-CLASS 12
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Replication in bacteria begins at a single origin of replication. Replication starts at multiple origins in eukaryotes. Why? Fig. 13.15 TEM of Replication bubbles 13
How can DNA be replicated in two different directions by ONE complex of DNA polymerase? What happens when DNA polymerase makes a mistake? How is fidelity of the message maintained? Brainstorm: What key steps would need to be involved in a repair process? Pre-class: What if you had a perfect repair system that allowed no mutation? 14
Two classes of DNA repair mechanisms “Single-strand” repairs: 1. DNA proofreading 2. DNA mismatch repair 3. Nucleotide excision repair “Double-strand” repairs: 1. Non-homologous end-joining (NEHJ) 2. Homology-directed repair (HDR) DNA repair terminology: Endonucleases vs exonucleases DNA backbone is cut (nicked) at an internal site Nucleotides are removed from DNA (chewed back) from the cut end. 15
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How are errors fixed during replication?: DNA proofreading Step 2: Remove mismatched nucleotide Step 3: Resynthesize DNA Is this ENDONUCLEASE or EXONUCLEASE activity? How are errors repaired immediately after replication?: DNA mismatch repair 1. Why is it essential to be able to identify the parental from the newly synthesized strand? 2. What type of nuclease (blue) breaks the phosphodiester bond between C-A? a. Endonuclease b. Exonuclease 3. What type of nuclease (purple) removes the mismatched nucleotide? a. Endonuclease b. Exonuclease 4. What enzyme (orange) would synthesize and fill in the missing nucleotides? 5. What enzyme (not shown) would form a phosphodiester bond to seal the backbone? Step 1: Detect mismatch Step 2: Remove mismatch Step 3: Resynthesize Step 4: Seal backbone 16
How does the cell identify the parental strand of DNA? Parental strand Newly synthesized strand DNA can be damaged by exposure to environmental factors like UV light. 17
How is UV damage repaired?: Nucleotide excision repair Step 1: Detect damage. Step 2: Cut out damage. Step 3: Resynthesize Step 4: Seal backbone 1. Specialized proteins scan the genome for UV damage. Knowing what you know about the damage caused by UV exposure, what would the proteins look for? 2. What type of nuclease is used to remove the damaged DNA in this mechanism? a. Endonuclease b. Exonuclease c. Both an endonuclease and an exonuclease 3. What enzyme (orange) would synthesize and fill in the missing nucleotides? 4. What enzyme (not shown) would form a phosphodiester bond to seal the backbone? Xeroderma pigmentosum: What happens when nucleotide excision repair is lost? 18
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High energy radiation and specialized cellular endonucleases can cause double-stranded breaks. Double stranded breaks repaired by non-homologous end joining (NHEJ) OR homology directed repair (HDR). =homology directed repair non homologous end joining= Double stranded break DNA polymerase nucleases DNA ligase 19
Two classes of DNA repair mechanisms “Single-strand” repairs: 1. DNA proofreading 2. DNA mismatch repair 3. Nucleotide excision repair “Double-strand” repairs: 1. Non-homologous end-joining (NEHJ) 2. Homology-directed repair (HDR) What is the order of enzyme activity in these repair mechanisms? 20
REVIEW OF DNA REPAIR MECHANISMS DNA Repair Type Single or Double Stranded Break When does process occur? Nuclease? (Endo or Exo) DNA Polymerase? DNA Ligase? DNA proofreading DNA mismatch repair Nucleotide excision repair Nonhomologous end joining (NHEJ) Homology directed repair (HDR) 21
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POST-CHECK 4UESTIONS: 4uestions 1-±: The figure below shows an active replication fork. Evaluate the following statements as either True or )alse. 1. The strand labeled & is the leading strand template. TRUE±)ALSE 2. The end labeled A is a 3¶ end. TRUE±)ALSE 3. A molecule of helicase would be active at d. TRUE±)ALSE 4. The seTuences shown in red at the beginning of each new segment are eventually replaced. TRUE±)ALSE 5. Which seTuence is the correct order of proteins (enzymes) used in DNA replication: a. RNA primase, DNA helicase, DNA polymerase I, DNA polymerase III, DNA helicase b. DNA helicase, DNA polymerase I, DNA polymerase III, RNA primase, DNA ligase c. RNA primase, DNA polymerase III, DNA ligase, DNA polymerase I, DNA helicase d. DNA helicase, RNA primase, DNA polymerase III, DNA polymerase I, DNA ligase 6. In eukaryotes, an origin of replication: a. uses the same strand of DNA as the leading strand at both forks. b. starts at one end of a chromosome and replicates to the other end c. is present at multiple locations along a chromosome to replicate DNA more Tuickly. d. contains two forks moving in the same direction. 7. Which of the following statements about the synthesis of DNA during DNA replication is true? Select all that apply. a. New DNA must be synthesized in the 5¶ to 3¶ direction. b. DNA polymerase adds a new nucleotide to the 5¶ carbon of the existing chain of nucleotides. c. RNA primers are only needed to initiate synthesis of the lagging strand of DNA. d. DNA ligase is reTuired to link together fragments of the lagging strand. e. Replication uses a DNA polymerase complex that synthesizes simultaneously on both leading and lagging strands 8. In DNA proofreading during replication, DNA polymerase III enzyme acts as a: Select all that apply. a. exonuclease b. endonuclease c. ligase d. polymerase 9. In mismatch repair (00R), the cell is able to distinguish the newly synthesized DNA from the original template DNA because the new DNA is: a. methylated 22
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b. not methylated c. oriented in the 5¶ to 3¶ direction d. oriented in the 3¶ to 5¶ direction 10. Once a mismatch has been detected in 00R, which of the following best describes the mechanism that removes the mismatched nucleotide: a. a DNA polymerase removes the mismatched nucleotide. b. an endonuclease cuts out the segment of DNA containing the mismatch. c. an exonuclease cuts out the segment of DNA containing the mismatch. d. an endonuclease nicks the DNA backbone, and then an exonuclease removes the mismatched nucleotide. 11. In nucleotide excision repair of thymine dimers: a. a single nucleotide is removed by an exonuclease b. a segment of nucleotides are removed by an exonuclease c. a single nucleotide is removed by endonucleases d. a segment of nucleotides are removed by endonucleases 12. When a double-strand DNA break occurs in the genome: (Select all correct answers) a. the correct seTuence from the opposite DNA strand could be used as a template for repair. b. the correct seTuence from a homologous chromosome could be used to repair the damage. c. insertions or deletions of DNA may occur during repair of the DNA. d. the DNA mismatch repair process could repair the break. 13. In which order do the necessary enzymes act to repair mismatched or damaged DNA? a. nuclease, DNA polymerase, RNA primase b. helicase, DNA polymerase, DNA ligase c. DNA ligase, nuclease, DNA polymerase d. nuclease, DNA polymerase, DNA ligase 23
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Information Flow Lecture 3: How is genetic information “expressed” to make a protein? (Part I: Transcription) PRE-CLASS WORK What is the difference between a gene and a genome? WATCH: An overview of the “central dogma”. READ: Concept 3.6 Nucleic acids store, transmit, and help express hereditary information. The figure below is of a generic RNA nucleotide. Label the main parts of the nucleotide and highlight two main ways in which RNA nucleotides are different from DNA nucleotides. Concept 14.0 The flow of genetic information What is “gene expression”? Concept 14.1 Genes specify proteins via transcription and translation Define the following terms: 1. Transcription 2. Translation 3. primary transcript 4. template strand 24
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5. codon 6. reading frame What is the connection between the nucleotide sequence of DNA and a sequence of amino acids in a protein? How are AUG, UAA, UAG, and UGA different from the other codons in the code? Concept 14.2 Transcription is the DNA directed synthesis of RNA Define the following terms: 1. RNA polymerase 2. Promoter 3. Transcription factors The figure below shows the start of transcription: 1. Label the template strand, the non-template strand, and the primary transcript. 2. Draw the RNA polymerase at its correct location in the diagram. 3. The RNA molecule is synthesized in the ______________________ direction. 25
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14.3 Eukaryotic cells modify RNA after transcription Define the following terms and label each on the mRNA molecule below 1. 5’ cap 2. poly-A tail 3. intron 4. exon mRNA END OF PRECLASS 26
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THE CENTRAL DOGMA: Information stored in the genome is expressed to make a protein. Transcription Translation DNA mRNA Protein Ribosome Transcription begins at a promoter and ends at a terminator. Contains “consensus” sequences 27
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How does transcription begin? How is RNA polymerase recruited to a promoter? RNA transcript Template strand RNA polymerase synthesizes the RNA transcript in the 5’ to 3’ direction. 28
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A primary transcript is processed to make a mature mRNA. Open Reading Frame (ORF) Start transcription ATG* start translation TAG, TAA, TGA* stop translation Intron 1 Intron 2 Intron 3 Protein coding region interrupted by introns Overview of eukaryotic gene structure 3′ untranslated region 5′ untranslated region *refers to sequence on non-template/coding strand 29
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Draw the primary transcript: include 5’ and 3’ labels Draw the mature mRNA: include 5’ and 3’ labels Intron 1 Intron 2 Intron 3 UTR UTR Differential gene expression: How can cells have the same DNA but express different proteins? 30
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Initiation of transcription requires presence of both general and regulatory transcription factors. ● General transcription factors present in all cells. ● Regulatory transcription factors only present in specific cells. Differential gene expression involves two key elements: specific binding sites (DNA) and regulatory transcription factors (proteins) 31
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Regulatory transcription factors bound to enhancer sequences PROMOTE formation of a transcriptional complex. General transcription factors Promoter (ACTIVATOR) Protein Regulatory transcription factors bound to silencer sequences INHIBIT formation of a transcriptional complex. General transcription factors Promoter Protein (REPRESSOR) 32
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ANALYSIS: A given gene has two enhancer sites and one silencer site “upstream” of the promoter. Each sequence is recognized and bound by a specific regulatory transcription factor. The diagram shows this gene and its regulatory transcription factors in 3 different types of cells. 1. Predict expression of the gene in each cell type (Yes or No) 2. Given that the DNA is the same in all 3 cell types, how would you explain the differences in gene expression? 3. Why are there different transcription factors present in each cell type? 33
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Transcription and translation application problem Use the space below to transcribe and translate a gene starting with the DNA. TRANSCRIPTION: 1. Below is a DNA sequence of a simple gene. 2. Show attachment and direction of synthesis of RNA polymerase enzyme. 3. Label the template and non-template strand of the DNA . 4. Write out the sequence of the mRNA transcribed from this gene and label 5’ and 3’ ends. _________promoter__________ ↓transcription start 5’ TCGAGCG TATAAA GTGGCAA|AGTATTTAAAGGTAACCCGATGCCTTGGCAGTTAAGGTCAATTTGACCAT 3’ 3’ AGCTCGC ATATTT CACCGTT|TCATAAATTTCCATTGGGCTACGGAACCGTCAATTCCAGTTAAACTGGTA 5’ ANNOTATE YOUR mRNA TRANSCRIPT : 5. Scan the mRNA for a start codon (AUG) and highlight. (HINT: don’t break into triplets, just look for an AUG) 6. Find the stop codon (UAG, UAA, UGA) and highlight. 7. Underline and label the 5’ untranslated region before the start codon. 8. Underline and label the 3’ untranslated region after the stop codon. 9. Draw a box around the coding region of the mRNA sequence. TRANSLATION: 10. How many amino acids will be in the translated protein? 11. Use the codon table to generate the sequence of amino acids that make up the protein. 34
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POST CHECK 48EST,ONS 4uestion 1-4: Researchers Walters et al. ±2²12³ were interested in studying the effect of regulatory transcription factors on the level of transcription in a gene mPGES-1, which codes for a prostaglandin involved in tissue inflammation. They generated a series of deletions, each removing one of 3 potential binding sites for regulatory transcription factors, and measured the level of transcription of an associated reporter gene. The results of the experiment are shown below. Evaluate the following statements as either TR8E or )ALSE: 1. Each regulatory binding site on the DNA would be recognized by a specific regulatory transcription factor. 2. The 3 binding sites have the same nucleotide sequence. 3. Whenever a binding site was deleted, the level of transcription of the reporter gene decreased. 4. The experiment suggests that binding site 3 is an enhancer, and binding sites 1 and 2 are silencers. 4uestion ± - ²: Evaluate the following about the process of transcription as either TR8E or )ALSE. 5. General transcription factors help recruit RNA polymerase to the promoter. 6. RNA polymerase synthesizes the primary transcript in the 3’ to 5’ direction. ´. RNA polymerase requires a primer to begin transcription. 4uestion ²-12: Evaluate the following about eukaryotic genes and transcription as TR8E or )ALSE. µ. The promoter for a gene would be located “upstream” of the coding region. ¶. Exons are non-coding sections of DNA that are transcribed into RNA but removed during RNA processing. 1². The 5’cap and poly-A tail are transcribed into RNA when the primary transcript is made. 11. Transcription starts at the AUG start codon. 35
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Information Flow Lecture 4: How is genetic information “expressed” to make a protein? (Part II: Translation) PRE-CLASS WORK READ: Concept 14.4 Translation is the RNA-directed synthesis of a polypeptide. Briefly describe the function of the following RNA molecules in the process of translation: 1. Messenger RNA (mRNA) 2. Transfer RNA (tRNA) 3. Ribosomal RNA (rRNA) Label the schematic of the ribosome below and provide a brief description of the function of each component. Concept 14.5 Mutations of one or a few nucleotides can affect protein structure and function. Define: 1. Missense mutation 2. Nonsense mutation 3. Silent mutation 36
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4. Frameshift mutation Concept 3.5: Proteins include a diversity of structures, resulting in a wide range of functions. Define: 1. Protein 2. Polypeptide 3. Amino acid 4. Peptide bond How are amino acids classified or grouped? END OF PRECLASS 37
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How is genetic information “expressed” to make a protein? Transcription Translation DNA mRNA Protein Ribosome The universal genetic code : RNA → Amino Acids 38
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TRANSLATION: Is the RNA world alive and well? 3 different types of RNA are involved in translation: 1. messenger (mRNA) 2. transfer (tRNA) 3. ribosomal (rRNA) How do tRNAs link amino acids to codons during translation? anticodon amino acid attachment site tRNA synthetase 39
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Base-pairing plays an essential role throughout the process of protein synthesis. For the triplet on the template strand, fill in the: a. codon on the mRNA b. anti-codon on the tRNA c. code for the amino acid on the tRNA Template strand Non-template strand mRNA mRNA tRNA E P A Large subunit Small subunit The ribosome coordinates synthesis of a polypeptide. EXIT PROTEIN AMINO ACID (peptidyl) (amino acyl) 40
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What components are required for initiation of translation? 5′ 3′ 5′ 3′ Initiation factors 5′ UTR 41
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42
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What terminates translation? GLY ARG VAL THR 3′ UTR 43
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What happens to a synthesized string of amino acids? Nonpolar side chains; hydrophobic Glycine (Gly or G) Alanine (Ala or A) Valine (Val or V) Leucine (Leu or L) Isoleucin e ( I le or I ) Methionin e (Met or M) Polar side chains; hydrophilic Phenylalanin e (Phe or F) Tryptopha n (Trp or W) Proline (Pro or P) Serine (Ser or S) Threonin e (Thr or T) Cysteine (Cys or C) Tyrosine (Tyr or Y) Asparagin e (Asn or N) Glutamin e (Gln or Q) Electrically charged side chains; hydrophilic Aspartic acid (Asp or D) Glutamic acid (Glu or E) Lysine (Lys or K) Arginine (Arg or R) Histidine (His or H) What dictates protein folding? 44
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Functional Non-functional 45
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Mutation analysis problem Consider a gene for which there are several known mutants. The normal mRNA transcript and each mutant transcript are listed below. Mutations are indicated in bold and underlined . 1. Use your codon table to translate the transcripts into amino acid sequences. 2. Classify the type of mutation: missense, nonsense, silent, or frameshift 3. Predict whether the mutation would likely alter the structure (and therefore function) of the protein. mRNA transcript Type of mutation Alter protein structure? Normal: 5’AUGAAGUUUGGCUAA3’ NA NA Mutant 1: 5’AUGAAGUU C GGCUAA3’ Mutant 2: 5’AUGAAGUUUG A CUAA3’ Mutant 3: 5’AUGAAGUUUG U CUAA3’ Mutant 4: 5’AUG U AGUUUGGCUAA3’ Mutant 5: 5’AUGAAGUUGGCUAAC3’ U deleted 46
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P26T C+(C. 48(6T,2N6 48(6 1- ±: The diagram below shows one stage in the process of translation. EYaluate the following statements as either TR8E or FALSE. 1. The large subunit is composed of protein and the small subunit is composed of RNA. TR8E±FALSE 2. The mRNA shown should be labeled as 3¶A8**8AA*A ²¶. TR8E±FALSE 3. The small subunit and a charged Met tRNA scan the mRNA for a start codon. TR8E±FALSE 4. The tRNA with the anticodon 3¶ 8AC ²¶ can only carry a methionine (Met) amino acid. TR8E±FALSE ². Methionine and Yaline (9al) amino acids are Moined by a peptide bond. TR8E±FALSE ³. The ribosome will directly linN the incoming arginine (Arg) to the methionine (Met) amino acid. TR8E±FALSE ´. A polypeptide has the amino acid seTuence met-phe-pro-lys-met. Which of the following seTuences in the coding (non-template) strand of the DNA could code for this polypeptide? a. 3µ AT*-TTT-CCC-AAA-AT* b. 3µ A8*-888-CCC-AAA-A8* c. ²µ AT*-TTT-CCC-AAA-AT* d. ²µ TAC-AAA-***-TTT-TAC ·. Which of the following mutations would be most liNely to haYe a negatiYe effect on protein structure and function? a. A deletion of 3 nucleotides near the middle of that gene. b. A deletion of a single nucleotide in the middle of an intron of a euNaryotic gene. c. A deletion of a single nucleotide near the end of the coding seTuence. d. A deletion of a single nucleotide after (downstream) the start codon. ¸. <ou are studying Protein ;. An aspartic acid at a Ney position in Protein ; acts to promote protein folding and stability. ,f this aspartic acid is changed to a different amino acid¹ which one of the following amino acid substitutions is most liNely to allow the protein to fold correctly? (See amino acid table in te[tbooN or online.) a. threonine b. glycine c. arginine d. glutamic acid e. serine 1º. The mutations in the gene encoding Protein ; described in the preYious Tuestion are e[amples of a. nonsense mutations b. silent mutations c. missense mutations 47
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Information Flow Lecture 5: What’s in a genome? PRE-CLASS WORK 1. What is the difference between the terms “gene” and “genome”? READ: CONCEPT 13.3 A chromosome consists of a DNA molecule packed together with proteins. 2. Define the following terms: Histones Nucleosome Chromatin READ: CONCEPT 15.2 Eukaryotic gene expression is regulated at many stages. 3. Define Histone acetylation DNA methylation Epigenetic inheritance END OF PRE-CLASS 48
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A genome is the complete set of DNA of an organism. Publication of the first draft of the human genome. February 2001 Is there a relationship between genome size and organism complexity? 49
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Is there a relationship between the size of a genome and the number of genes? Prokaryotes Eukaryotes What else is in the genome besides “genes”? Estimates for the human genome. 50
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Transposon (TE) content varies across eukaryotic genomes. Gaut and Ross-Ibarra (2008) Transposons are “mobile elements” that can move around the genome. Barbara McClintock Cold Spring Harbor Lab - 1947 Nobel Prize 1983 51
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Friend or foe of the genome? Transposon insertion can disrupt the coding region of a gene. (Kazazian and Moran 2017) Friend or foe of the genome? Transposon insertion can coordinate expression of sets of genes. 52
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Transposon-mediated rewiring of gene regulatory networks contributed to the evolution of pregnancy in mammals. (Lynch et al. 2015) Mammals Vincent Lynch University of Buffalo How is the DNA of the genome organized? 53
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The nucleosome is the basic structural unit of DNA packaging in eukaryotes. 8 histone proteins Segment of DNA Nucleosomes along a DNA molecule. Chromatin coils into a condensed chromosome. 54
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The human genome is arranged into 23 chromosomes. (University of Washington Medical School) Gametes have 1 copy the genome; 23 chromosomes ( haploid ) Body cells have 2 copies of the genome; 46 chromosomes ( diploid ) The location of chromosomes in the nucleus is spatially organized. Genes switched off in this cell type Repressed/intermediate genes Transcriptionally active genes 55
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How do “epigenetic” modifications influence gene expression? ● “Epi” (above, over) → modifications that do not change nucleotide sequence. ● Examples include methylation/acetylation of histones and methylation of DNA nucleotides. ● Reversible in response to internal or external signals. ● Potentially passed to offspring. Chromatin can be “remodeled” by chemical modification of histones. 56
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Disruption of histone acetylation implicated in human disease. Pathological histone acetylation in Parkinson’s disease (Harrison et al 2018) AcH3-Lys9 ePD = early Parkinson’s lPD = late Parkinson’s Ian Harrison Univ. College London 57
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POSTC+ECK 48EST,ONS 1. The “C paradox” refers to the fact that: A. Large genomes have more genes B. More complex organisms have larger genomes C. There is no clear relationship between between genome size and organisms complexity 2. Which of the following makes up the largest proportion of the human genome? A. Exons of genes B. Introns of genes C. Transposons 3. The number of genes is positively correlated with genome size in: A. Prokaryotic genomes B. Eukaryotic genomes C. Both prokaryotic and eukaryotic genomes 4. Which of the following statements about transposons is TRUE? A. Transposons insertion is primarily targeted in the exons of genes. B. Most transposons in the genome remain active and “jumping”. C. Transposons insertion may influence the expression of nearby genes. D. Transposons rely on enzymes from the host genome to be copied and inserted. 5. In a nucleosome, a length of DNA is wound around 8 positively charged : A. DNA polymerases B. RNA polymerases C. Helicases D. Histones 6. Which of the following correctly represents the levels of DNA packaging in a eukaryotic chromosome? A. DNA helix, nucleosome fiber, chromatin fiber, condensed chromosome B. DNA helix, chromatin fiber, nucleosome fiber, condensed chromosome C. Chromatin fiber, nucleosome fiber, condensed chromosome, DNA helix D. Nucleosome fiber, DNA helix, condensed chromosome, chromatin fiber 7. Which of the following statements about the arrangement of chromosomes in the nucleus is TRUE? (Select all that apply.) A. During the majority of the life of a cell, chromatin is coiled to form a condensed chromosome. B. Chromosomes are anchored in a specific territory within the nucleus. C. Actively transcribed genes are oriented towards the center of the nucleus. D. The arrangement of chromosomes is the same in all types of cells. 8. Which of the following is true of epigenetic modifications but NOT of a DNA mutation? A. Can be inherited by offspring B. Influences the phenotype of an individual C. Involved the substitution of a DNA nucleotide D. Can be reversed in response to internal or external signals 9. Which of the following epigenetic modifications would likely DECREASE expression of a gene? (Select all that apply). A. Addition of methyl groups (methylation) of histone tails near the gene. B. Addition of acetyl groups (acetylation) of histone tails near the gene. C. Removal of methyl groups (demethylation) of histone tails near the gene. D. Removal of acetyl groups (deacetylation) of histone tails near the gene. 10. Which of the following epigenetic modifications would likely INCREASE expression of a gene? (Select all that apply). A. Addition of methyl groups (methylation) of histone tails near the gene. B. Addition of acetyl groups (acetylation) of histone tails near the gene. C. Removal of methyl groups (demethylation) of histone tails near the gene. D. Removal of acetyl groups (deacetylation) of histone tails near the gene. 58
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Information Flow Lecture 6: How does mitosis produce genetically identical cells? PRE-CLASS WORK READ: Section 9.1 Most cell division results in genetically identical daughter cells. Define the following: 1. chromatin 2. sister chromatids (also draw two sister chromatids in a chromosome) 3. centromere 4. homologous chromosome (see section on meiosis) Section 9.2 The mitotic phase alternates with interphase in the cell cycle. Define the following: 1. M phase 2. G1 phase 3. S phase 4. G2 phase 5. Microtubules 6. Kinetochore 59
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DNA replication and cell division occur in a highly regulated cycle. Growth and normal metabolic roles Growth and prep for mitosis Chromosomes replicated How do chromosomes change as they move through the cell cycle? Gene 1 Gene 1 60
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Sister chromatids are synthesized during the S phase. DNA replication REPLICATED CHROMOSOMES UNREPLICATED CHROMOSOMES A d Pair of homologous chromosomes for individual with genotype AaDd Trace changes in chromosomes through the cell cycle. 61
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Terminology review: haploid (1n) vs diploid (2n) (University of Washington Medical School) Human gametes have 1 copy the genome; 23 chromosomes ( haploid ) Body cells have 2 copies of the genome; 46 chromosomes ( diploid ) 62
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Mitosis ensures daughter cells have the same genetic information as the parent cell. ● How many chromosomes are in a diploid (2n) cell of this organism? __________ ● How many homologous pairs of chromosomes are present in this cell? ____________ ● How many chromatids make up each chromosome at this stage of the cell cycle? ___________ ● Are sister chromatids exact copies of each other? Why or why not? Prophase key process: Prometaphase key process: Mitotic spindles Centrosome Nuclear envelope Chromatin fibers Nuclear envelope starts to break down. Sister chromatids Chromosome (Assume starting with a diploid cell) Spindle pole Metaphase key process: Anaphase key process : Telophase key process: ● How many chromosomes do the daughter cells have compared with the original parent cell? ○ The same number of chromosomes ○ Twice as many chromosomes ○ Half the number of chromosomes ● How much DNA is in the daughter cell compared with the original parent cell at the start of mitosis? ○ The same amount of DNA ○ Twice as much DNA ○ Half the amount of DNA ● Are the two daughter cells genetically identical to each other? To the parent cell? Why or why not? 63
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What would be checked at each cell cycle checkpoints? ● DNA replication complete ● DNA damage repaired ● Microtubules correctly attached to chromatids Consider your rationale for each placement. Cell cycle checkpoints are critical for proper development and growth. Loss of cycle control in cancer cells results in genomic instability. Karyotype (Chromosome composition) of tumor and normal tissue. Bladder Cancer. Credit: Image courtesy of University of California – Berkeley. TUMOR CELL NORMAL CELL 64
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Lack of fidelity in cell division causes mitotic non-disjunction and aneuploidy in somatic cells. Fidelity in mitosis: How does the cell ensure that each daughter cell gets the appropriate chromosomes? Prof. Kevin Vaughan works on the regulation of cell division Metaphase checkpoint 65
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Proper connection of chromosomes to microtubules is required to progress through metaphase checkpoint. 66
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POST-C+ECK 48EST,O1S 4uestions 1-±: 7he picture to the right is of a cell of a diploid organism with condensed chromosomes at the beginning of mitosis. 1. 7he diploid number (2n) of the organism is 6. 758(±)$/S( 2. ,n its current state² the cell contains 12 chromatids. 758(±)$/S( 3. 7he cell contains 6 pairs of homologous chromosomes. 758(±)$/S( 4. 7he sister chromatids of a chromosome are e[act copies of each other. 758(±)$/S( 5. $ human somatic cell contains 46 chromosomes. MarN all that are 758( about these chromosomes: a. ,n mitosis² homologous chromosomes are not present. b. 7he D1$ double heli[ is wound around proteins called histones. c. &hromosomes alwa\s looN liNe ³;´ shaped structures. d. &hromosomes condense to undergo cell diYision. 6. $ Nar\ot\pe can proYide information about the genome of an organism. :hich of the following can a Nar\ot\pe 127 proYide" a. the number of chromosome pairs in an organism b. the se[ of the organism c. whether mitotic nondisMunction occurred d. the presence of mutant genes on chromosomes µ. ,f the D1$ content of a diploid cell in the G1 phase of the cell c\cle is represented b\ x ² then the D1$ content of the same cell is: (Select all correct answers) a. 0.5x at the end of the S phase of the cell c\cle b. x at the end of the S phase of the cell c\cle c. 2x at the end of the S phase of the cell c\cle d. 4x at the end of G2 phase of the cell c\cle e. [ at the end of telophase and c\toNinesis of mitosis ¶. 7he daughter cells formed b\ a mitotic cell diYision of diploid cell: a. haYe half the number of chromosomes as the parent cell. b. contain 1 chromosome from each homologous pair. c. are geneticall\ identical to each other but not to the parent cell. d. are geneticall\ identical to the parent cell and to each other. ·. 7he G2 checNpoint in the cell c\cle: a. checNs for D1$ damage before the S phase. b. checNs for complete D1$ replication before the cell diYides b\ mitosis. c. replicates the D1$ so cells can diYide. d. checNs for proper attachment of microtubules to sister chromatids. 1¸. Short answer: $ cell cannot moYe through the metaphase checNpoint without proper attachment of microtubules to sister chromatids. ,n 2¹3 sentences² briefl\ e[plain wh\ this is an essential process to maintain fidelit\ of the genetic code during mitosis. 67
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Information Flow Lecture 7: How does meiosis produce genetic diversity? PRE-CLASS WORK READ: Section 10.2 Fertilization and meiosis alternate in sexual life cycles. 1. Define the following: a. homologous chromosome (homologs): b. ”n” and “2n”: c. diploid cell: d. haploid cell: 2. If a pollen (male gamete) of a pea plant contains seven chromosomes, what is the diploid number in this species? 68
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Section 10.3 Meiosis reduces the number of chromosome sets from diploid to haploid. 3. Starting with the cell below (2N = 6) in Prophase I, draw the remaining stages: metaphase I, anaphase I, telophase I and cytokinesis, metaphase II, anaphase II, telophase II and cytokinesis. Your end product should be four haploid daughter cells. END OF PRE-CLAS 69
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Diploid (2n) Diploid (2n) Diploid (2n) Diploid (2n) Haploid (1n) Haploid (1n) Haploid (1n) Haploid (1n) Haploid (1n) Haploid (1n) Mitosis Meiosis Genetically identical Genetically variable Meiosis makes haploid (1n) gametes in sexual life cycle. 71
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Terminology review: haploid (1n) vs diploid (2n) (University of Washington Medical School) Human gametes have 1 copy the genome; 23 chromosomes ( haploid ) Body cells have 2 copies of the genome; 46 chromosomes ( diploid ) Summary of diploid and haploid cells 72
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Like mitosis, the starting point for meiosis is a diploid cell (2N) with replicated chromosomes. Two homolgous chromosomes Each chromosome with two chromatids Metaphase I Anaphase I Telophase I and cytokinesi s Metaphase II Anaphase II Telophase II and cytokinesis Meiosis generates haploid (1n) gametes from a diploid (2n) precursor cell. When do the cells become haploid? Prophase I (later stage) Prophase II 73
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Draw the four possible gametes formed during meiosis in a “dihybrid” individual (AaBb) b b B B A A a a Parent cell at end of prophase I Arrangements of chromosomes in Metaphase I OR Meiosis II Gametes formed Gametes formed Recombination is a unique event during prophase I. In this example, the A, B and C alleles are linked ( they are on the same chromosome ). So are the a, b, and c alleles. 74
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Meiosis and sexual reproduction shuffles existing genetic variation. Independent assortment Recombination Fertilization 75
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MEIOSIS and LIFE CYCLE SIMULATION Start the cycle with fertilization for an organism with two pairs of chromosomes. Begin with a haploid (1n) egg and haploid (1n) sperm. There are _______chromosomes in each gamete cell. Each chromosome has _______ chromatid(s). Form a diploid (2n) zygote from haploid egg and sperm. There are _______chromosomes in the zygote cell. Each chromosome has _______ chromatid(s). Show a diploid (2n) cell from this organism following the S phase of the cell cycle. There are ______ chromosomes in the cell. Each chromosome has _______ chromatid(s). This diploid cell is a germline cell and now enters meiosis. Simulate the process of synapsis that occurs during prophase I What other meiotic event occurs at this time? _____________________ Line up the chromosomes as they would appear in metaphase I. The chromosomes are arranged __________________________ (single file / homologous pairs) along the metaphase plate. How many different ways could the chromosomes be arranged in metaphase? __________ This is the basis for the law of ____________________________. Complete the process of meiosis I (anaphase I, telophase I, cytokinesis). When the chromosomes separate in anaphase I, this is the basis for the law of _________________. There are _______ chromosomes in each daughter cell after telophase I. 76
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Each of these chromosomes has _______ chromatid(s). The daughter cells are now _____________ (haploid 1n / diploid 2n). Demonstrate the arrangement of chromosomes in metaphase II. The chromosomes are arranged ____________________________ (single file / homologous pairs) along the metaphase plate. This is similar to the arrangement of chromosomes during metaphase of ______________________. Complete the process of meiosis II (anaphase II, telophase II, cytokinesis). There would be a total of __________ cells formed from the original diploid germline cell. There are _______ chromosomes in each daughter cell after telophase II. Each of these chromosomes has _______ chromatid(s). The daughter cells are _____________ (haploid 1n / diploid 2n). Are the daughter cells all genetically identical? ______________ What happens to the daughter cells formed in meiosis? The daughter cells will become _____________ , either egg or sperm. The diploid state will be regained in the next generation by the process of ______________________. 77
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POS7-C+ECK 48ES7,ONS 1. 7wo homologous chromosomes: (Select all that are 758() $. are identical copies of each other. %. c arry the same genes in the same order, but the alleles for each trait may be different. &. are inherited, one from each parent D. separate from each other in anaphase II (. are present in cells that initiate mitosis and meiosis I 2. :hich two cells together demonstrate the principle of independent assortment. a. &ell 1 and &ell 2 b. &ell 1 and &ell 3 c. &ell 2 and &ell 3 3. 7he movement of homologous chromosomes to opposite poles during BBBBBBBBB is the basis for the principle of segregation. a. $naphase I b. $naphase II c. 0etaphase I d. 0etaphase II 4uestions ± - ²: $nswer the following Tuestions about the cell in Prophase I of meiosis. +omologous pairs of chromosomes are circled by a dashed line. ±. :hat is the diploid (2n) number of this cell? a. 3 b. 6 c. 12 ². +ow many pairs of homologous chromosomes are found in the cell? a. 3 b. 6 c. 12 6. (ach chromosome at this stage of meiosis has: a. 1 chromatid b. 2 chromatids c. ± chromatids 78
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³. $t the end of meiosis I, the two daughter cells will be: a. diploid with 6 chromosomes in each cell. b. diploid with 3 chromosomes in each cell. c. haploid with 6 chromosomes in each cell. d. haploid with 3 chromosomes in each cell. ´. $t the end of meiosis II, each chromosome will have: a. 1 chromatid b. 2 chromatids c. ± chromatids µ. In the same cell above, consider any single gene on any chromosome. +ow many copies of that gene would the cell have at the start of meiosis¶ at the end of meiosis? a. 2¶ 2 b. 2¶ 1 c. ±¶ ± d. ±¶ 2 e. ±¶ 1 1·. :hich of the following statements about the phases of meiosis are 758(? (Select all that apply.) a. +omologous chromosomes separate from each other during anaphase I. b. 5ecombination (crossing over) occurs during prophase II. c. Sister chromatids separate from each other during anaphase II. d. &ells first become haploid at the end of telophase II and cytokinesis. 11. 7he diploid cell to the right is in metaphase I of meiosis. *ene $¸a and *ene D¸d are located on the same chromosome. :hich of the following gene combinations would 2N/Y be found in gametes formed if a recombination event (dashed line) occurred ? (Select all that apply.) $. $ ¹ D %. a ¹ d &. $ ¹ d D. a ¹ D 79
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MITOSIS and MEIOSIS COMPARISON TABLE Use this table to review concepts of mitosis and meiosis. Mitosis Meiosis Chromosomes replicated prior to process Chromosomes condense at the beginning of process. Starting chromosome number of parent cell (2n, 1n) Number of cell divisions Number of daughter cells formed Final chromosome number of daughter cells (2n,1n) Daughter cells genetically identical to parent cell? Daughter cells identical to each other? Homologous chromosome pair along metaphase plate Recombination between homologous chromosomes Chromatids of replicated chromosome separate during anaphase Purpose for division 80
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Information Flow 8: How is genetic information passed to offspring? PRE-CLASS WORK READ: Concept 11.1 Mendel used the scientific approach to identify two laws of inheritance. Reminder: Many of the terms from this section were already defined in the pre-class work for Evolution Lecture 3, but would still be relevant here. (dominant, recessive, genotype, phenotype, heterozygous, etc.) 1. What does P, F 1 and F 2 refer to in genetic terminology? 2. Use the Punnett square to the right to perform a monohybrid cross for seed color in peas, where Y is the dominant yellow allele and y is the recessive green allele. (A monohybrid cross refers to a cross where both parents are heterozygous for a single trait.) 3. What are the genotypes of the parents? 4. What is the expected ratio of genotypes in the offspring of this cross? 5. What is the expected ratio of phenotypes in the offspring of this cross? Concept 12.2 Sex-linked genes exhibit unique patterns of inheritance. 1. For an x-linked trait, how many copies of the gene will a human male have? How many copies will a female have? 2. What does the term “hemizygous” mean? 81
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3. Fathers will pass an X-linked allele to all of their daughters, but none of their sons. WHY? BRAINSTORM: List 3-4 facts that you already know about dominant and recessive traits. END OF PRECLASS 82
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What did Mendel mean by dominant and recessive? 83
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Normal phenotype Homozygous dominant AA Normal phenotype Heterozygous Aa Loss of function phenotype Homozygous recessive aa li li Case 1- a recessive “loss of function” mutant phenotype Normal phenotype Gain of function phenotype Homozygous recessive bb Heterozygous Bb Case 2 – dominant “gain of function” mutant phenotype li 84
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Enzyme Allele for purple flowers Locus for flower- color gene Pair of homologous chromosomes Allele for white flowers No enzyme C T A A A T C G G T G A T T T A G C C A A T T A A A T C G G T A T T T A G C C A What is the molecular basis for dominance in Mendel’s flowers? Heterozygous (Pp) Molecular analysis clarifies dominant vs. recessive inheritance of human genetic disease. Cystic fibrosis Achondroplasia Duchenne muscular dystrophy 85
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Cystic fibrosis: an autosomal recessive condition ● Sticky mucus in lungs and other organs ● Mutation in CFTR gene ( cystic fibrosis transmembrane conductance regulator); identified in 1989 ● CFTR codes for chloride channel protein CFTR gene: chromosome 7 Q arm band 31.2 (NIH, Genetics Home Reference) Normal Cystic fibrosis There are different ways for the CFTR protein to “lose function”. Reduced channel activity (R347P) Normal “wild-type” CFTR protein Misfolded protein (F508del) Nonfunctional channel (G551D) No protein made (G542X) Golgi apparatus nucleus What type of mutation? 86
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ANALYSIS: What makes CFTR mutant alleles “recessive”? (adapted from Derichs 2013) Background: A high concentration of chloride in sweat is a phenotype associated with cystic fibrosis. A “sweat test” is a common diagnostic test for CF. Homozygous wildtype 1. Review the axes and identify the major trend: What is the relationship between CFTR protein function % and the concentration of sweat chloride? 2. Compare the approximate CFTR protein function among homozygous wildtype, heterozygous carriers, and homozygous mutant individuals. homozygous wildtype = ______% heterozygous carriers = ________% homozygous mutant = _______ % 3. Identify key points: Based on the standard diagnostic cutoff for the CF sweat test, which individuals would identified as CF patients? a. Homozygous F508del b. Homozygous R347P c. Heterozygotes 4. Analyze trends: What do these results suggest about the reason that these CFTR mutant alleles are recessive to the wildtype allele? 87
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CFTR mutant alleles result from different changes in nucleotide sequence. F508del (misfolded protein) R347P (channel with reduced activity) Normal sequence DNA 5’ XXX XXX C G G XXX XXX 3’ Protein Arg Position 347 Mutant sequence DNA 5’ XXX XXX C C G XXX XXX 3’ Protein Pro Position 347 Inheritance patterns of cystic fibrosis (CF) 88
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Achondroplasia: an autosomal dominant condition ● Defect in long bone development ● Can’t convert cartilage to bone (ossification) ● Caused by mutation in FGFR3 gene ( fibroblast growth factor receptor 3); identified in 1994 FGFR3 gene: chromosome 4 P arm band 16.2 (NIH, Genetics Home Reference) cytoplasmic domain extracellular domain FGF receptor 3 (adapted from Mason 2007) Irregular bone growth in achondroplasia (Biomarin, Inc.) Develop a hypothesis: 1. How could a “gain of function” mutation in FGF receptor result in the phenotype of achondroplasia? 2. Why would the mutant allele for FGFR be dominant to the normal allele? triggers pathway that inhibits bone growth Normal FGF receptor and growth factor (FGF) dimer forms after binding growth factor 89
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Two different nucleotide changes in the FGFR3 gene result in the same change in the protein. Normal sequence DNA 5’ XXX XXX GGG XXX XXX 3’ Protein Gly Position 380 Mutant sequence 1 DNA 5’ XXX XXX A GG XXX XXX 3’ Protein Arg Position 380 Mutant sequence 2 DNA 5’ XXX XXX C GG XXX XXX 3’ Protein Arg Position 380 The G380R mutation accounts for 99% of achondroplasia cases X-linked traits have unique patterns of inheritance. Female Male 90
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Patterns of inheritance for x-linked traits: Duchenne muscular dystrophy (DMD): an X-linked recessive condition ● Progressive wasting of skeletal and cardiac muscle ● Symptoms begin in early childhood; wheelchair dependent by adolescence ● Caused by mutation in DMD gene; codes for dystrophin protein stabilizes muscle fibers band 21.2 and 21.1 (NIH, Genetics Home Reference) 91
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POST-CHECK QUESTIONS 1. Which of the following statements might explain why a mutation is a “loss of function” mutation? (Select all that apply.) a. No mutant protein product is produced. b. The mutant protein is produced but is nonfunctional. c. The mutant protein has increased activity relative to the normal protein. 2. If one parent is a carrier of a CF mutation and the other parent is homozygous normal, what is the chance that an offspring has the disease? a. 75% b. 50% c. 25% d. 0% 3. If both parents are carriers for a CF mutation, what is the chance that an offspring has the disease? a. 75% b. 50% c. 25% d. 0% 4. If both parents are carriers for a CF mutation, what is the chance that an offspring is a carrier? a. 75% b. 50% c. 25% d. 0% 5. A couple would like to know the chances of their child having Achondroplasia. Parent #2 has Achondroplasia; parent #1 does not. What is the chance of having a child with Achondroplasia? (Note that being homozygous for an Achondroplasia mutation is normally lethal.) a. 100% b. 50% c. 25% d. 0% 6. Can two parents who are heterozygous for an FGFR mutation have a typical height child? If yes, what percent of the time? a. Yes; 75% of the time. b. Yes; 50% of the time c. Yes; 25% of the time d. No, they cannot have a typical height child. 7. Two unaffected parents have a son with DMD. What are the genotypes of the parents? a. X d X d and X d Y b. X d X d and X D Y c. X D X D and X d Y d. X D X D and X D Y e. X D X d and X D Y 8. How likely is it for a female to have DMD? a. Females will never have this condition. b. One-fourth of the female children of an affected father would have this condition. c. One-half of the female children of an affected father and a carrier mother would be expected to have this condition. 9. In cats, black fur color is caused by an X-linked allele; the other allele at this locus causes orange color. The heterozygote is a tortoiseshell. What kinds of offspring would you expect from the cross of a black female 92
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and an orange male? (Hint: Write out the genotypes of the parents and set up a Punnett square to solve this question.) a. tortoiseshell females; tortoiseshell males b. black females; orange males c. orange females; orange males d. tortoiseshell females; black males 10. Short answer: The majority of loss of function mutations are inherited as recessive traits. In general, why is a normal allele able to mask a recessive loss of function allele? 93
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