Concept explainers
Videos
_______________ make(s) up the nucleus of an atom.
- a. Protons and electrons
- b. Protons and neutrons
- c. DNA and RNA
- d. Neutrons and electrons
- e. DNA only
Introduction: Atom is the smallest structural unit of any matter. The nucleus is the centermost part of the atom. The atom consists of proton, electron, and neutron. Multiple numbers of atoms combine together and form the molecules.
Answer to Problem 1TY
Correct answer: The “protons and neutrons” forms the nucleus of an atom. Hence, the correct answer is option b.
Explanation of Solution
Reason for correct answer:
The nucleus of the atom is formed by the “neutrons and protons”. These two particles provide the mass to the atom and both the particles are heavier than the electron. The charge possessed by the proton is positive in nature. However, in the case of neutrons, the charge is not present. Both the particles are present in the nucleus of an atom and the electrons circulate around them in their orbitals.
Option b. is given as “Protons and neutrons”.
The “protons and neutrons” are found in the nucleus of an atom. Hence, the correct answer is option b.
Reason for incorrect answer:
Option a. is given as “Protons and electrons”.
The electrons are found in the orbitals present outside the nucleus. The protons and neutrons are present inside the nucleus. Hence, option a. is incorrect.
Option c. is given as, “DNA and RNA”.
DNA and RNA are made up of a number of atoms and they are found in the nucleus of the cell, not in the nucleus of the atom. So, the possibility of the presence of DNA and RNA inside the atom nucleus is negligible. Hence, option c. is incorrect.
Option d. is given as “Neutrons and electrons”.
Electrons are present on the orbitals and they surround the nucleus of an atom in their respective energy levels. The protons and neutrons are present inside the nucleus. Hence, option d. is incorrect.
Option e. is given as, “DNA only”.
The DNA stands for “deoxyribonucleic acid”. Presence of DNA molecule is found in case of the cellular nucleus, not in the atom nucleus. Hence, option e. is incorrect.
Hence, the options a., c., d., and e. are incorrect.
Inside the nucleus of an atom, the “protons and neutrons” are found.
Want to see more full solutions like this?
Chapter 2 Solutions
BIOLOGY (LL)
Additional Science Textbook Solutions
Biological Science (6th Edition)
Campbell Essential Biology (7th Edition)
HUMAN ANATOMY
Biology: Life on Earth with Physiology (11th Edition)
Genetics: From Genes to Genomes
General, Organic, and Biological Chemistry: Structures of Life (5th Edition)
- Please indentify the unknown organismarrow_forwardPlease indentify the unknown organismarrow_forward5G JA ATTC 3 3 CTIA A1G5 5 GAAT I I3 3 CTIA AA5 Fig. 5-3: The Eco restriction site (left) would be cleaved at the locations indicated by the arrows. However, a SNP in the position shown in gray (right) would prevent cleavage at this site by EcoRI One of the SNPs in B. rapa is found within the Park14 locus and can be detected by RFLP analysis. The CT polymorphism is found in the intron of the Bra013780 gene found on Chromosome 1. The Park14 allele with the "C" in the SNP has two EcoRI sites and thus is cleaved twice by EcoRI If there is a "T" in that SNP, one of the EcoRI sites is altered, so the Park14 allele with the T in the SNP has only one EcoRI site (Fig. 5-3). Park14 allele with SNP(C) Park14 allele with SNPT) 839 EcoRI 1101 EcoRI 839 EcoRI Fig. 5.4: Schematic restriction maps of the two different Park14 alleles (1316 bp long) of B. rapa. Where on these maps is the CT SNP located? 90 The primers used to amplify the DNA at the Park14 locus (see Fig. 5 and Table 3 of Slankster et…arrow_forward
- From your previous experiment, you found that this enhancer activates stripe 2 of eve expression. When you sequence this enhancer you find several binding sites for the gap gene, Giant. To test how Giant interacts with eve, you decide to remove all of the Giant binding sites from the eve enhancer. What results do you expect to see with respect to eve expression?arrow_forwardWhat experiment could you do to see if the maternal gene, bicoid, is sufficient to form anterior fates?arrow_forwardYou’re curious about the effect that gap genes have on the pair-rule gene, evenskipped (eve), so you isolate and sequence each of the eve enhancers. You’re particularly interested in one of the enhancers, which is just upstream of the eve gene. Describe an experimental technique you would use to find out where this particular eve enhancer is active.arrow_forward
- For short answer questions, write your answers on the line provided. To the right is the mRNA codon table to use as needed throughout the exam. First letter U บบบ U CA UUCPhe UUA UCU Phe UCC UUG Leu CUU UAU. G U UAC TV UGCys UAA Stop UGA Stop A UAG Stop UGG Trp Ser UCA UCG CCU] 0 CUC CUA CCC CAC CAU His CGU CGC Leu Pro CCA CAA Gin CGA Arg CUG CCG CAG CGG AUU ACU AAU T AUC lle A 1 ACC Thr AUA ACA AUG Mot ACG AGG Arg GUU GCU GUC GCC G Val Ala GAC Asp GGU GGC GUA GUG GCA GCG GAA GGA Gly Glu GAGJ GGG AACASH AGU Ser AAA1 AAG Lys GAU AGA CAL CALUCAO CAO G Third letter 1. (+7) Use the table below to answer the questions; use the codon table above to assist you. The promoter sequence of DNA is on the LEFT. You do not need to fill in the entire table. Assume we are in the middle of a gene sequence (no need to find a start codon). DNA 1 DNA 2 mRNA tRNA Polypeptide C Val G C. T A C a. On which strand of DNA is the template strand (DNA 1 or 2)?_ b. On which side of the mRNA is the 5' end (left or…arrow_forward3. (6 pts) Fill in the boxes according to the directions on the right. Structure R-C R-COOH OH R-OH i R-CO-R' R R-PO4 R-CH3 C. 0 R' R-O-P-OH 1 OH H R-C-H R-N' I- H H R-NH₂ \H Name Propertiesarrow_forward4. (6 pts) Use the molecule below to answer these questions and identify the side chains and ends. Please use tidy boxes to indicate parts and write the letter labels within that box. a. How many monomer subunits are shown? b. Box a Polar but non-ionizable side chain and label P c. Box a Basic Polar side chain and label BP d. Box the carboxyl group at the end of the polypeptide and label with letter C (C-terminus) H H OHHO H H 0 HHO H-N-CC-N-C-C N-C-C-N-GC-OH I H-C-H CH2 CH2 CH2 H3C-C+H CH2 CH2 OH CH CH₂ C=O OH CH2 NH2arrow_forward
Concepts of BiologyBiologyISBN:9781938168116Author:Samantha Fowler, Rebecca Roush, James WisePublisher:OpenStax College
Biology Today and Tomorrow without Physiology (Mi...BiologyISBN:9781305117396Author:Cecie Starr, Christine Evers, Lisa StarrPublisher:Cengage Learning