ND STONY BROOK UNIVERSITY LOOSELEAF GENETICS: FROM GENES TO GENOMES
6th Edition
ISBN: 9781260406092
Author: HARTWELL, Leland, HOOD, Leroy, Goldberg, Michael
Publisher: Mcgraw-hill Education/stony Brook University
expand_more
expand_more
format_list_bulleted
Concept explainers
Textbook Question
Chapter 15, Problem 16P
Suppose you are examining a newly found plant species, and you want to determine whether the inheritance of mtDNA is maternal, paternal, or biparental. You find that in the population two variants of mtDNA exist that can be distinguished by size differences of PCR amplification products made with a particular pair of primers. You first perform PCR analysis on DNA isolated from a leaf on each of two individual plants. Then you cross eggs from plant 1 with pollen from plant 2, obtain four seedlings, and perform PCR analysis with DNA from each of the whole seedlings. The results are shown below.
a. | Assuming plant 2 is homoplasmic, do these results exclude any of the three possible models for the inheritance of mtDNA in this species (maternal, paternal, or biparental)? |
b. | What experiment could you perform to distinguish between the models that remain? |
c. | What experiment(s) could you perform to check your assumption that plant 2 is in fact homoplasmic? Why is such an experiment necessary to make a conclusion about the mode of inheritance of mtDNA in this species? |
d. | Explain the differing proportions of the two forms of mtDNA in the four seedlings. Be as specific as possible. |
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solutionStudents have asked these similar questions
A region of the genome is amplified by PCR to analyze two closely linked SNPs. PCR products from individual A were labelled with a red-fluorescing dye and those from individual B labelled with a green-fluorescing dye. These PCR products were mixed and hybridized to an oligonucleotide microarray as shown in the figure below. What can we conclude about the data shown?
Select all true:
a. individual A is heterozygous for the M1 and M2 haplotypes, individual B is heterozygous for M4 and M5
b. individual A is heterozygous for the M1 and M2 haplotypes, individual B is heterozygous for M3 and M6
c. Sanger sequencing traces of these PCR products would show the same double peaks at the same positions for these two individuals
d. analyzing this PCR amplified region with conventional Sanger sequencing would be more accurate than the microarray analysis
e. individuals A and B could be siblings who share a parent that is homozygous for one of the haplotypes
{urgent}
ou are studying a new gene “X” that you think controls skin color in Bearded Dragons. In order to determine what gene X does, you need lots of gene X DNA to work with. So, you decide to amplify it through the process known as PCR. You have a forward and a reverse primer that have been designed to gene Xf – a gene in frogs that is similar to the one found in the Bearded Dragon. You are pretty sure that these two primers will work in Bearded Dragons, so you decide to go ahead with the PCR. In frogs, the Forward Xf primer binds gene Xf between nucleotides 32 and 47. The Reverse Xf primer binds between nucleotides 1110 and 1135.
a. What size PCR product would these two primers create if they amplified the frog gene Xf??
Transgenic tobacco plants were obtained in which the vector Ti plasmid was designed to insert the gene of interest plus an adjacent kanamycin-resistance gene. The inheritance of chromosomal insertion was followed by testing progeny for kanamycin resistance. Two plants typified the results obtained generally. When plant 1 was backcrossed with wild-type tobacco, 50 percent of the progeny were kanamycin resistant and 50 percent were sensitive. When plant 2 was backcrossed with the wild type, 75 percent of the progeny were kanamycin resistant and 25 percent were sensitive. What must have been the difference between the two transgenic plants? What would you predict about the situation regarding the gene of interest?
Chapter 15 Solutions
ND STONY BROOK UNIVERSITY LOOSELEAF GENETICS: FROM GENES TO GENOMES
Ch. 15 - Match each numbered item with the most closely...Ch. 15 - Assuming human cells have on average 1000...Ch. 15 - Reverse translation is a term given to the process...Ch. 15 - The human nuclear genome encodes tRNAs with 32...Ch. 15 - The human mitochondrial genome includes no genes...Ch. 15 - How do you know if the halibut you purchased at...Ch. 15 - Is each of these statements true of chloroplast or...Ch. 15 - Suppose you are characterizing the DNA of a...Ch. 15 - An example of a gene-targeting DNA plasmid vector...Ch. 15 - Which of the following characteristics of...
Ch. 15 - The Saccharomyces cerevisiae nuclear gene ARG8...Ch. 15 - The so-called hypervariable regions HV1 and HV2 of...Ch. 15 - Suppose a new mutation arises in a mitochondrial...Ch. 15 - Describe at least two ways in which the...Ch. 15 - Why are severe mitochondrial or chloroplast gene...Ch. 15 - Suppose you are examining a newly found plant...Ch. 15 - A form of male sterility in corn is inherited...Ch. 15 - Plant breeders have long appreciated the...Ch. 15 - A mutant haploid strain of Saccharomyces...Ch. 15 - Prob. 20PCh. 15 - What characteristics in a human pedigree suggest a...Ch. 15 - The first person in the family represented by the...Ch. 15 - In 1988, neurologists in Australia reported the...Ch. 15 - If you were a genetic counselor and had a patient...Ch. 15 - Kearns-Sayre syndrome KSS, Pearson syndrome, and...Ch. 15 - Many clinically relevant mitochondrial diseases...Ch. 15 - Leigh syndrome is characterized by psychomotor...Ch. 15 - All mutations in mitochondrial genes ultimately...Ch. 15 - How could researchers have determined that the...
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, biology and related others by exploring similar questions and additional content below.Similar questions
- You are using the restriction enzyme HAEIII to digest different samples of the taster gene isolated from cheek cells of different people and amplified by PCR. When viewing the bands on the electrophoresis gel, one would expect that a taster (homozygote) would have---------band(s), whereas a carrier (heterozygote) would show--------band(s), and a non-taster would show------band(s).arrow_forwardThe temperature at which the primers and target DNA hybridize may be changed to influence the stringency of PCR amplification. What effect will changing the hybridization temperature have on the amplification? Let's say you have a certain yeast gene A and want to check whether it has a human equivalent. How might managing the hybridization's rigor benefit you?arrow_forwardThe gel image below shows 7 alleles, let’s call them 1-7 in order of size, with 1 being the largest and 7 being the smallest. In this sample of 11 individuals, how many times does allele 5 appear?arrow_forward
- Polymerase Chain Reaction (PCR) was invented by Kary Mullis in 1983. This technique had indeed facilitated research in various areas of molecular biology and genetics. You would like to amplify a particular gene fragment from the yeast genome using Polymerase Chain Reaction (PCR). What are the THREE (3) main cycles in PCR? Discuss the processes at each PCR cycle mentioned.arrow_forwardAfter much work the researcher maps the mutation in the mut1 plant to a locus on chromosome 2. The gene associated with the mut1 mutation is called COB1 (i.e. mutations in COB1 cause the mut1 phenotype.) The researcher extracts DNA and RNA from developing ears of wild type, mut1 and mut2 plants. From the DNA samples COB1 PCR primers are used to amplify the coding region of the COB1 gene (see DNA gel below). An RNA gel is used to examine the expression of COB1 all three plants (see RNA gel below). results in photo: Which of the following is TRUE about the COB1 gene. a. There is a deletion in the COB1 gene in the mut1 plants. b. COB1 is not transcribed in the mut2 plant. c. There is a deletion in the COB1 gene in the mut2 plants. d. COB1 is not transcribed in the mut1 plant.arrow_forwardA paper hypothesizes that white flowers are unable to produce anthocyanins (purple pigments) because they lack a functional “A” protein. However, it is also possible that an unknown gene is responsible for the lack of anthocyanins. Now that they have isolated DNA sequences of the “A” allele, design an experiment to use these DNA sequences to distinguish between these two hypotheses.arrow_forward
- This is a schematic diagram of an agarose gel used to analyze the DNA fragments generated by analyzing individual female flies with primers P1, P2, and P3 in a single PCR reaction. Lane M represents DNA fragments that are size markers (standards). In lanes 1, 2, and 3, sketch in the DNA fragments expected to be produced by PCR for female Drosophila homozygous for wild-type, heterozygous, and homozygous for the white-one mutation, respectively:arrow_forwardHow can you assess dominance and mapping with only one mutant?arrow_forwardUsing PCR, you want to amplify an approximately 1 kbexon of the human autosomal gene encoding theenzyme phenylalanine hydroxylase from the genomicDNA of a patient suffering from the autosomal recessive condition phenylketonuria (PKU).a. Why might you wish to perform this PCR amplification in the first place, given that the sequence ofthe human genome has already been determined?b. Calculate the number of template molecules thatare present if you set up a PCR reaction using1 nanogram (1 × 10−9 grams) of chromosomalDNA from blood cells as the template. Assume thateach haploid genome contains only a single genefor phenylalanine hydroxylase and that the molecular weight of a base pair is 660 grams per mole. Thehaploid human genome contains 3 × 109base pairs.c. Calculate the number of PCR product moleculesyou will obtain if you perform 25 PCR cycles andthe yield from each cycle is exactly twice that ofthe previous cycle. What would be the mass ofthese PCR products taken together?arrow_forward
- In our simple model of DNA amplification we assumed that a successful amplification doubled the DNA each cycle. Reality isn't so kind! Two related problems encountered in DNA profiling are the issues of asymmetric peaks and "drop-out" where an allele produces a small to very-small peak compared to the other allele. Assume you have a sample with a heterozygous set of alleles. Let one of the alleles be successfully doubled each cycle but the second allele is (on average) only increased by 1.8. If the sample is subjected to 30 PCR cycles, how much larger will the "successful" peak be than the "less successful" peak? Set your answer up as a ratio of the "successful" /"unsuccessful" and round your number to 1 decimal place. (for example, if your calculations show that after 30 cycles the successful allele has 55340 copies and the unsuccessful has 10000 copies the ratio is 5.5)arrow_forwardAliens with orang eye color allele (o) is recessive to the dominant black eye color allele (O). The locus of the orange gene from 10 pure breed orange eyes aliens and 10 pure breed black eyes aliens. You notice a difference in the DNA sequence linked to each allele of the orange gene and you decide to use it as a physical marker to follow recombination between this sequence linked to the orange gene. These sequences consist of short tandem repeat (STR) with two different number of repeats, each associated with one of the two orange gene alleles. a PCR test distinguishes the 10 repeat STR and the 6 repeat STR associated with the O and o alleles respectively. Using primers on each side of the STRs you can amplify by PCR this sequence and visualizing the size of the 10 repeat and 6 repeat STRs in an electrophoresis gel. You can follow the two STR sequences (10 and 6 repeats) linkage and recombination frequencies with the orange gene locus. The black eye aliens yield a PCR fragment that is…arrow_forwardSelect all that would be true if I had a nonsense mutation in an exon of a gene: The nonsense mutant allele would be the same size as wildtype by PCR-electrophoresis The nonsense mutant protein would be the same size by Western as the wildtype protein The nonsense mutant allele would be a different size compared to wildtype by PCR- electrophoresis The nonsense mutant protein would be a different size by Western compared to the wildtype proteinarrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
- Biology: The Dynamic Science (MindTap Course List)BiologyISBN:9781305389892Author:Peter J. Russell, Paul E. Hertz, Beverly McMillanPublisher:Cengage Learning
Biology: The Dynamic Science (MindTap Course List)
Biology
ISBN:9781305389892
Author:Peter J. Russell, Paul E. Hertz, Beverly McMillan
Publisher:Cengage Learning
An Introduction to the Human Genome | HMX Genetics; Author: Harvard University;https://www.youtube.com/watch?v=jEJp7B6u_dY;License: Standard Youtube License