Practice set 8 lecs 14-19

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Jan 9, 2024

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Practice set 8: Covers lectures 14-19 Complete each set of questions as we advance through the course material. Every question in this exam is connected! PsbS gene is a gene in plants which codes for a protein called “pigment binding protein S that is essential for proper photosystem II functioning. When PsbS is mutated, plants have a much slower rate of growth than is typical. They are “poor growing”. PSBS denotes the normal allele for this gene. psbs denotes the mutated. recessive allele for this gene. Pineapples have 25 chromosomes. PsbS is located on chromosome 22. Also on chromosome 22 and located right next to PsbS. is the gene for spikes. where R represents the allele for sharp spikes and r represents the allele for blunt spikes. Chromosome 21 contains the skin gene, where F represents the allele for tough skin and f represents the allele for soft skin. Questions 1-6 covers lectures 14-15 1. The wild ancestors of pineapples had two copies of each of their 25 chromosomes. Commercially grown pineapples of present day have three copies of each of their 25 chromosomes. What is the correct scientific notation for commercial pineapple chromosome number? a) 25.75 b) 75.25 ¢ 253 d) 325 2. Kiara Jackson PhD is a plant geneticist working for a major university in southern California. She has been tasked with assessing a local farmer’s pineapple crops to find out why 20% of them displayed poor growth this season. Kiara takes samples of the well-growing pineapples and the poor-growing pineapples. Kiara extracts an egg from one well-growing pineapple plant and performs genetic analysis. The egg’s genotype is PSBS/R/f. How did the plant’s chromosomes line up at the metaphase plate? (Assume the well-growing plant genotype is PSBSpsbs. Rr. Ff. In other words. it is heterozygote for all genes.)

A B PS%“S pg}{shs PS%SBSI p% R R B: p: sbs R r R r r r %f %F f% %’z D ¢ | ps! fosbs PSBr 7 sbs R t | | ‘Which two alleles will most likely be inherited in a different combination than in their parent’s chromosomes (recombination)? a) PSBS/psbs and R/t b) PSBS/psbs and F/f Remember that the extracted egg’s genotype is PSBS/R/f. Kiara fertilizes this egg from a random sampling of pollen (sperm) from a male pineapple plant with the genotype PSBS/psbs R/r f/f. In a punnet square of this cross, how many gametic combination of the female’s must be represented? How many gametic combinations of the male must be represented? a) Female: 1. Male: 4 b) Female: 2. Male: 2 c) Female: 4. Male 4: Kiara takes a sample from a different well-growing plant. It also has sharp spikes and tough skin. She doesn’t know its genotype. She mates this plant with one that is poor-growing, has blunt spikes and soft skin. Half of the resulting offspring are well-growing with sharp spikes and tough skin. The other half is poor-growing with sharp spikes and tough skin. What is the genotype of the unknown parent plant? a) PSBSpsbs. RR. FF b) PSBSpsbs. Rr. Ff c) psbspsbs. RR. ff Like humans. pineapple “males™ contain X and Y chromosomes in their diploid cells while “females” will contain two X chromosomes in their diploid cells. There is a gene located on the X

chromosome for skin glow. with B denoting the allele for bright and b denoting the allele for pale. Kiara mates a pale female with a bright male. What kind of glow will their offspring have? a) Half of the females will be bright. half will be pale. All of the males will be pale. b) All the females will be pale and all the males will be bright. c) All the males will be pale and all the females will be bright. Questions 7-20 covers lectures 16-19 7. The normal PSBS gene sequence is 3’ | ATCATGCGCTAATATACG TCGTAATACTGCTCGTTAATCTCGATCTCTAATTCTGCGCGCGCGCG J +1 At what point does RNA polymerase begin transcription (making the mRNA)? a) ATC (far left) b) GCG (far right) ¢ TCG (+1) d) TAC (middle left) 8. Which region of the gene is the termination signal? a) Atthe far 3" end. ATCTAT. b) Atthe far 5° end. CGCGCG. 9. Is the termination signal transcribed info mRNA? Is it eventually translated into amino acids? a) It is not transcribed into mRNA. It is not translated into amino acids. b) It is transcribed into mRNA. It is not translated into amino acids. c) Itis transcribed into mRNA. It is translated into amino acids. 10. What will be the first several ribonucleotides in the sequence of the mRNA that is made? a) 57 AGCAUUAUG 3° b) 3" AGCAUUAUG 5’ c) 5 AUGACGAGC 3’ d) 37 AUGACGAGC 5’ e) 5 UAGUACGCG 3’ f) 3'UAGUACGCG 5’ 11. After the mRNA is made. what occurs? a) Transcription factors binds. b) Methyl groups are added. while aceyl groups are removed. c) Introns are removed. Exons are fused together.

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12. The mutated version of the gene. psbs. is 3 ATCATGCGCTAATATACGTCGTAATACCGCTCGTTAATCTCGATCTCTAATTCTGCGCGCGCGCG I +1 Locate the point mutation directly following the start codon. What kind of change will that create in the 2 b) B b resulting amino acid chain? CYS in the normal amino acid chain will become ARG in the mutated amino acid chain. This will have a mild effect on the resulting protein. CYS in the normal amino acid chain will become ARG in the mutated amino acid chain. This will have a severe effect on the resulting protein. THR in the normal amino acid chain will become ALA in the mutated amino acid chain. This will have a mild effect on the resulting protein. THR in the normal amino acid chain will become ALA in the mutated amino acid chain. This will have a severe effect on the resulting protein. 13. During translation of the mutated psbs gene. the ribosome catalyzes the peptide bond between . via Amino acids: IRNA Amino acids: Protein 1RNA: amino acids Protein: amino acids 14. During translation of the mutated psbs gene, what would be the result if there was not any protein release factor available? a) b) 0 The ribosome would recruit the amino acid associated with the stop codon. The ribosome would skip the stop codon and attempt to recruit the amino acid for the following codon. The ribosome would completely stop translation. 15. During Kiara’s investigations into the poor-growing pineapple plants. she discovers that some of the plants are infected with a bacteria called Dickeye zaea. Dickeye zaea has an operon system which operates according to the diagrams below:

H\GH TRYPTOPHAN: 7. P polymenase Teykophown . / No transadption onA ? o [ mpe [ wpD || teC || #pB || TrpA | iAo Tryptophan from the environment binds to the trp repressor protein making it active. Then, the trp repressor binds to the promoter, thus preventing RNA polymerase from binding. LOW TRYPTOPHAN: repressor L-m::*wo.) Trauscripion ONA _[&® ol e [[awpp [ mec [ #pB [ mpa [ [ PN polywerase If there is no tryptophan, the trp repressor stays inactive. The trp repressor does not bind to the promoter. RNA polymerase can bind. Transcription of genes trpE-trpA occurs when a) The repressor is active. b) The repressor is inactive. 16. What kind of transcriptional control is this? a) Positive control b) Negative control 17. Pineapples also have a set of genes related to coping with heat. If a pineapple that is living in a cool environment suddenly experiences a sharp increase in temperature, what is the fastest way (of the options below) that it can respond? a) Evolve a change in the sequence of one of the enhancers. b) Destroy the mRNA right after its transcribed. c) Cut the amino acid chains in half while its traveling through the golgi apparatus. 18. Under extreme heat stress, a pineapple (“female” XX) plant grew fewer leaves than is typical but grew a standard sized fruit. The gene responsible for this response is HT8. After analysis. it was found that HT8 had lowered expression in the plant’s diploid cells. The plant was propagated via sexual reproduction. If transgenerational epigenetic modification is operating in this system. what you expect to occur in the offspring? (SELECT ALL) a) Grow fewer leaves than is typical and a standard sized fruit even if it never experiences extreme heat stress.

b) Grow fewer leaves than is typical and a standard sized fruit only if it also experiences extreme heat stress. c) HT8 will be methylated. d) HTS8 will be acetylated. 19. The R/r gene mentioned earlier is actually expressed slightly differently in the pineapple skin cells compared to the pineapple phloem cells. In the skin cells. the protein that is made from the R/r gene is 98 amino acids long, but in the phloem cells its only 79 amino acids long. What feature/process accounts for this discrepancy? a) Different transcription factors b) RNA interference c) Folding difference d) Alternative splicing 20. Overall. a skin cell and a phloem cell should have the same exact sequences of (SELECT ALL) a) mRNA b) tRNA c) 1RNA Key 1. Answer: D. 3 copies of 25 chromosomes is 3.25. 2. Answer: B. To end up in the same gamete, the genes must have lined up on the same side of the metaphase plate. 3. Answer: B. Because they are far apart on different chromosomes. Independent assortment will break up the parental allelic combination. 4. Answer: A. The egg’s genotype is already known, PSBS/R/f. That’s the only gametic combination that needs to be considered because we know that’s the only one that’s being used for this fertilization. The male sperm is chosen randomly. The possible gametic combinations are PRf Prf pRf prf (P=PSBS).

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5. Answer: A. Look at each gene individually. Half the offspring are well growing and half are poor growing. They each must have inherited one poor growing allele from the poor growing parent. because thats the only allele the poor growing parent can provide. So the offspring that is poor growing must have inherited the second poor growing allele from the well growing parent. Therefore, you know the well growing parent is PSBS/psbs. Do the same thing for the other genes. Then check the punnet square: prf PRF PpRIFf PRFE ppRIFf 6. Answer: C. X* X X XEX° XEX® Y XY XY 7. Answer: C. That’s the +1 site. 8. Answer: B. The termination signal is on the DNA and it is composed of Cs and Gs. 9. Answer: B. It is transcribed into mRNA. It’s that part of the transcript that folds back to make the hair pin. It is not translated into the amino acid chain because the stop codon comes before it. 10. Answer: A. The mRNA is complementary to the template DNA strand. Transcription starts at the +1 site. 11. Answer: C. Processing occurs (eukaryotic cell). 12. Answer: D. Make the mRNA and then look it up in the codon chart. 13. Answer: A. The IRNA composes the active site. So it is responsible for catalyzing the bond between amino acids. 14. Answer: B. The stop codons do not code for an amino acid. So the ribosome will skip it and attempt to recruit amino acids for the next section of mRNA. 15. Answer: B. When the repressor is inactive it is not bound to DNA. so RNA polymerase can bind. 16. B. Because the regulatory molecule is stopping transcription. 17. Answer: C. Post translation modification is fastest way to adjust gene expression. 18. Answer: A and C. The response is inherited even if the environment changes. Methylation causes reduced expression

19. Answer: D. Makes slightly different proteins out of the same mRNA by cutting out some exons. 20. Answer: B and C. Because they perform translation in the same way.

Practice set 8 lecs 14-19

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