Genetics Exam 2 Practice Problems
A chromosome has the following segments, where • represents the centromere: A B • C D E F G What types of chromosome mutations are required to change this chromosome into each of the following chromosomes? (In some cases, more than one chromosome mutation may be required.) a. A B A B • C D E F G
(a) Tandem duplication of AB
A chromosome has the following segments, where • represents the centromere: A B • C D E F G What types of chromosome mutations are required to change this chromosome into each of the following chromosomes? (In some cases, more than one chromosome mutation may be required.) b. A B • C D E A B F G
(b) displaced duplication of AB
A chromosome has the following segments, where • represents the centromere: A B • C D E F G What types of chromosome mutations are required to change this chromosome into each of the following chromosomes? (In some cases, more than one chromosome mutation may be required.) c. A B • C F E D G
(c) paracentric inversion of DEF
A chromosome has the following segments, where • represents the centromere: A B • C D E F G What types of chromosome mutations are required to change this chromosome into each of the following chromosomes? (In some cases, more than one chromosome mutation may be required.) d. A • C D E F G
(d) deletion of B
A chromosome has the following segments, where • represents the centromere: A B • C D E F G What types of chromosome mutations are required to change this chromosome into each of the following chromosomes? (In some cases, more than one chromosome mutation may be required.) e. A B • C D E
(e) deletion of FG
A chromosome has the following segments, where • represents the centromere: A B • C D E F G What types of chromosome mutations are required to change this chromosome into each of the following chromosomes? (In some cases, more than one chromosome mutation may be required.) f. A B • E D C F G
(f) paracentric inversion of CDE
A chromosome has the following segments, where • represents the centromere: A B • C D E F G What types of chromosome mutations are required to change this chromosome into each of the following chromosomes? (In some cases, more than one chromosome mutation may be required.) g. C • B A D E F G
(g) pericentric inversion of ABC
A chromosome has the following segments, where • represents the centromere: A B • C D E F G What types of chromosome mutations are required to change this chromosome into each of the following chromosomes? (In some cases, more than one chromosome mutation may be required.) h. A B • C F E D F E D G
(h) duplication and inversion of DEF
A chromosome has the following segments, where • represents the centromere: A B • C D E F G What types of chromosome mutations are required to change this chromosome into each of the following chromosomes? (In some cases, more than one chromosome mutation may be required.) i. A B • C D E F C D F E G
(i) duplication of CDEF, inversion of EF.
Briefly explain the differences between F+, F-, Hfr, and F' cells
- An F+ cell will contain a circular plasmid separate from the chromosome -The Hfr cell has the f factor integrated into its chromosome. - In F' strains the f factor exists as a separate circular plasmid but the plasmid carries bacterial genes that were originally part of the bacterial chromosome. -The F- strain does not contain the f factor and can receive DNA from cells that contain the F factor. (F+, Hfr, and F' )
How does it help to explain some of the characteristics of mitochondria and chloroplasts?
- Chloroplasts and mitochondria contain genomes that encode for proteins, tRNAs, and rRNAs. - The chloroplast and mitochondrial genome sizes, circular chromosome structure, and other aspects of genome structure are similar to those of eubacterial cells. - Moreover, the chloroplast and mitochondrial ribosomes are similar in size and function to eubacterial ribosomes. DNA sequences in mitochondrial and chloroplast genomes are most similar to those in eubacteria. - also helps explain the double membranes of mitochondria and chloroplasts -where the outer membrane originated as the host cell's endosomal membrane and the inner membrane derived from the endosymbiont's plasma membrane.
What similarities and differences exist in the enzymatic activities of DNA polymerases I and III? What is the function of each DNA polymerase in bacterial cells?
- DNA polymerase I has a 3' to 5' as well as a 5' to 3' exonuclease activity. - DNA polymerase III has only a 3' to 5' exonuclease activity. (1) DNA polymerase I carries out proofreading. It also removes and replaces the RNA primers used to initiate DNA synthesis. (2) DNA polymerase III is the primary replication enzyme and also has a proofreading function in replication.
What types of genomes do viruses have?
- Double stranded DNA - single stranded DNA, - double stranded RNA - single stranded RNA.. depends on the type of virus.
What is semiconservative replication?
- In semiconservative replication, the original two strands of the double helix serve as templates for new strands of DNA. - When replication is complete, two double-stranded DNA molecules will be present. - Each will consist of one original template strand and one newly synthesized strand that is complementary to the template.
Compare and contrast prokaryotic and eukaryotic chromosomes. How are they alike and how do they differ?
- Prokaryotic chromosomes are usually circular - eukaryotic chromosomes are linear - Prokaryotic chromosomes generally contain the entire genome - eukaryotic chromosome has only a portion of the genome: The eukaryotic genome is divided into multiple chromosomes. - Prokaryotic chromosomes are generally smaller and have only a single origin of DNA replication. - Eukaryotic chromosomes are often many times larger than prokaryotic chromosomes and contain multiple origins of DNA replication. - Prokaryotic chromosomes are typically condensed into nucleoids, which have loops of DNA compacted into a dense body. - Eukaryotic chromosomes contain DNA packaged into nucleosomes, which are further coiled and packaged into successively higher-order structures. The condensation state of eukaryotic chromosomes varies with the cell cycle.
What are the four major types of aneuploidy?
- nullisomy - monosomy - trisomy - tetrasomy
What are some advantages of using bacteria and viruses for genetic studies?
- small size and genome - rapid reproduction - easy to culture - Numerous methods available for genetics engineering - Genomes are small - Techniques to isolate/manipulate genes are available - They have medical importance - They can be genetically engineered to be of commercial value
If a double-stranded DNA molecule is 15% thymine, what are the percentages of all the other bases?
15% adenine, 35% guanine, and 35% cytosine
Species I is diploid (2 n = 4) with chromosomes AABB; related species II is diploid (2 n = 6) with chromosomes MMNNOO. Give the chromosomes that would be found in individuals with the following chromosome mutations: g. Nullisomy in species II for chromosome N
2n - 2 6-2= 4 chromosomes MMMN
Species I has 2n=14 and species II has 2n=20. Give all possible chromosomes numbers that may be found in the following individuals: a. An autotriploid of species I
2n = 14 Or 2X = 14 Therefore X = 7 Autotriploidy means 3X Therefore chromosome no.=3X7 = 21
Species I has 2 n = 16 chromosomes. How many chromosomes will be found per cell in each of the following mutants in this species? b. Autotriploid
2n = 16 Auto's derive extra chromosome SET from SAME species n=8 3n= 24 24 chromosomes
Species I has 2 n = 8 chromosomes and species II has 2 n = 14 chromosomes. What would the expected chromosome numbers be in individuals with the following chromosome mutations? Give all possible answers. b. Autotetraploidy in species II
2n =14 n=7 4n = 28 chromosomes
Species I has 2n=14 and species II has 2n=20. Give all possible chromosomes numbers that may be found in the following individuals: b. An autotetraploids of species II.
2n=20 or 2X = 20 X = 10 Therefore autotetraploidy =4X Therefore no. of chromosome = 4X10 =40
One nucleotide strand of a DNA has the base sequence illustrated below. 5'-ATTGCTACGG-3' Give the base sequence and label the 5' and 3' ends of the complementary DNA nucleotide strand.
3′—TAACGATGCC—5′
Which of the following relations or ratios would be true for a double-stranded DNA molecule?
A + C = G + T A + G/C + T=1 A/T = G/C
Which of the following relations will be true for the percentage of bases in double-stranded DNA?
A double-stranded DNA molecule will contain equal percentages of A and T nucleotides and equal percentages of G and C nucleotides. The combined percentage of A and T bases added to the combined percentage of the G and C bases should equal 100
How does a purine differ from a pyrimidine? What purines and pyrimidine are found in DNA and RNA?
A purine consists of a six-sided ring attached to a five-sided ring. A pyrimidine consists of only a six-sided ring. In both DNA and RNA, the purines found are adenine and guanine. DNA and RNA differ in their pyrimidine content. The pyrimidine cytosine is found in both RNA and DNA. However, DNA contains the pyrimidine thymine, whereas RNA contains the pyrimidine uracil but not thymine.
List some of the characteristics that make bacteriophages ideal organisms for many types of genetic studies.
A simple structure consisting of just two contrasting elements
Species I is diploid (2 n = 4) with chromosomes AABB; related species II is diploid (2 n = 6) with chromosomes MMNNOO. Give the chromosomes that would be found in individuals with the following chromosome mutations: e. Tetrasomy in species I for chromosome A.
AAAABB or AABBBB 2n+2 6 chromosomes
Species I is diploid (2 n = 4) with chromosomes AABB; related species II is diploid (2 n = 6) with chromosomes MMNNOO. Give the chromosomes that would be found in individuals with the following chromosome mutations: a. Autotriploidy in species I.
AAABBB gives an extra set of chromosomes 6 chromosomes
Species I is diploid (2 n = 4) with chromosomes AABB; related species II is diploid (2 n = 6) with chromosomes MMNNOO. Give the chromosomes that would be found in individuals with the following chromosome mutations: c. Monosomy in species I.
AAB or ABB 2n-1 3 chromosomes
Which bases are capable of forming hydrogen bonds with each another?
Adenine is capable of forming TWO hydrogen bonds with thymine. Guanine is capable of forming THREE hydrogen bonds with cytosine.
Species I has 2 n = 8 chromosomes and species II has 2 n = 14 chromosomes. What would the expected chromosome numbers be in individuals with the following chromosome mutations? Give all possible answers. f. Allotetraploidy including species I and II
Allotetraploid can give 3 cases 2n from s1 and 2n from s2 8 + 14 =22 chromosomes 1n from s1 and 3n from s2 4 + 21 = 25 3n from s1 and 1n from s2 12 + 7 = 19 chromosomes
Species I has 2n=14 and species II has 2n=20. Give all possible chromosomes numbers that may be found in the following individuals: d. An allotetraploid formed from species I and species II.
Allotetraploid from species I and II 2n of I and n of II 14+20=34 No. of chromosome present will be in 2n, 2n = 68.
Species I is diploid (2 n = 4) with chromosomes AABB; related species II is diploid (2 n = 6) with chromosomes MMNNOO. Give the chromosomes that would be found in individuals with the following chromosome mutations: f. Allotriploidy including species I and II.
Allotriploid formed from species I and II There will be n from sp I and 2n from species II Or 2n from species 1 ad n from species 2 First case n from sp. 1 and 2n from sp.2 n = 2+3=5 Therefore chromosome no. will be 2n of this = 5 X 2= 10 Second case 2n of 1 and n of II n= 4+6=10 no. of chromosome = 10 X 2 =20
Species I has 2n=14 and species II has 2n=20. Give all possible chromosomes numbers that may be found in the following individuals: c. An allotriploid formed from species I and species II.
Allotriploid formed from species I and II There will be n from sp I and 2n from species II Or 2n from species 1 and n from species 2 First case n from sp. 1 and 2n from sp.2 n = 7+20=27 Therefore chromosome no. will be 2n of this = 27X2=54 Second case 2n of 1 and n of II n= 14+10=24 no. of chromosome = 24X2 =48.
John Smith is a pig farmer. For the past 5 years, Smith has been adding vitamins and low doses of antibiotics to his pig food; he says that these supplements enhance the growth of the pigs. Within the past year, however, several of his pigs died from infections of common bacteria., which failed to respond to large doses of antibiotics. b) What advice might you offer Smith to prevent this problem in the future?
As a clinician's advise, John is suggested to replace his pigs with a new breeding stock with normal immune response. He is also advised to immediately discontinue the practice of antibiotic supplementation, however, occasional supplementation can be performed. Providing a balanced diet will also improve the quality of pigs and their health in the farm.
The percentage of cytosine in a double-stranded DNA molecule is 40%. What is the percentage of thymine?
As its a double stranded DNA G is always complimentary with C A is complementary with T. Hence %G= %C, and % A= %T. if 40% are C then DNA strand should have 40% of G. 100 -(40 C) - (40 G) - 2X = 20 - 2x = x= 10 10% thymine
Species I has 2 n = 16 chromosomes. How many chromosomes will be found per cell in each of the following mutants in this species? c. Autotetraploid
Auto's derive extra chromosome SET from SAME species n = 8 4n = 32 32 Chromosomes
What is the difference between a paracentric and a pericentric inversion? A.the number of genes inverted B.the number of centromeres inverted C.the placement of centromeres in the inversions D.if the inversion is on autosomal of sex chromosomes
B.the number of centromeres inverted Paracentric inversions does not occur in the centromere. Crossing over only occurs in the out The loop occur in one arm of the chromosome. The result is a large loop in where recombination occurs in only part of the loop. Dicentric bridges (two loops) can form in the center of the chromosome where recombination does not occur. An acentric loop also occurs in the centromere region. The result is recombination in one loop. Pericentric inversion does occur in the centromere (in some cases). The most important part is that the chromosome can be elongated or shorted after the inversions have taken place. Deletions and duplications can occur during crossing over during the inversion. The result is recombination in two loops.
Fraenkel-Conrat
Conducted experiments showing that RNA can serve as the genetic material in some viruses
Why is DNA gyrase necessary for replication?
DNA gyrase (also referred to as topoisomerase) reduces supercoiling (relaxes tension) which builds up during DNA unwinding, preventing DNA breakage.
How does an RNA nucleotide differ from a DNA nucleotide?
DNA nucleotides, or deoxyribonucleotides, have a deoxyribose sugar that lacks an oxygen molecule at the 2' carbon of the sugar molecule. Ribonucleotides, or RNA nucleotides, have a ribose sugar with an oxygen linked to the 2' carbon of the sugar molecule. Ribonucleotides may contain the nitrogenous base uracil, but not thymine. DNA nucleotides contain thymine, but not uracil.
Why is primase required for replication?
DNA polymerases cannot begin without a primer containing a free 3 '-OH.
Fred Griffith
Demonstrated that heat-killed material from bacteria could genetically transform live bacteria
Kossel
Determined that DNA contains nitrogenous bases
Miescher
Determined that DNA is acidic and high in phosphorous
Oswald Avery
Determined that DNA is responsible for transformation in bacteria
Barbara McClintock
Discovered Transposable Elements (jumping genes)
Erwin Chargaff
Discovered regularity in the ratios of different bases in DNA
Phoebus Aaron Levene
Discovered that DNA consists of repeating nucleotides
A diploid plant cell contains 2 billion (2 x 109) base pairs of DNA b. Give the number of molecules of each type of histone protein associated with the genomic DNA
Each nucleosome has two molecules of H2A, H2B, H3, abd H4 so for the four of them each has 2x10^7 For H1 there is only one molecule per nucleosome so it has 1X10^7 H1
A bacterium synthesizes DNA at each replication fork at a rate of 1000 nucleotides per second. If this bacterium completely replicates its circular chromosome by theta replication in 30 minutes, how many base pairs of DNA will its chromosome contain?
Each replication complex is synthesizing DNA at each fork at a rate of 1000 nucleotides per second. So for each second, 2000 nucleotides are being synthesized by both forks (1000 nucleotides per second × 2 forks = 2000 nucleotides per second) or 120,000 nucleotides per minute. If the bacterium requires 30 minutes to replicate its chromosome, then the size of the chromosome is 3,600,000 nucleotides (120,000 nucleotides per minute × 30 minutes = 3,600,000)
What is the difference between euchromatin and heterochromatin?
Euchromatin undergoes regular cycles of condensation during mitosis and decondensation during interphase Heterochromatin remains highly condensed throughout the cell cycle, except transiently during replication. Nearly all transcription takes place in euchromatic regions with little or no transcription within heterochromatin.
What are hairpins and how do they form?
Hairpins are a type of secondary structure found in single strands of nucleotides. The formation of hairpins occurs when sequences of nucleotides on the single strand are inverted complementary repeats of one another.
Rarely, the conjugation from Hfr and F- cells produces two Hfr cells. Explain how this event takes place.
Hfr strains contain an F factor integrated into the bacterial chromosome. The F factor mediates transfer of the bacterial chromosome. During conjugation of an Hfr strain with an F-strain, the transfer process begins within the F factor. So literally, part of the F factor is the first to arrive in F-cell. However, the remaining part of the F factor is transferred last. Because nearly 100 minutes are required to completely transfer the donor chromosome, the two cells must remain in contact for the entire 100 minutes. So if the donor and recipient cell are not disturbed and the transfer process is not interrupted, then the entire Hfr strain's chromosome, including the F factor, can be donated to the F-cell. Following recombination between the donor and recipient chromosomes results in the production of a second
A diploid plant cell contains 2 billion (2 x 10^9) base pairs of DNA a. How many nucleosomes are present in the cell?
IN SHORT 200 nucleotides per nucleosomes base pairs of dna is another way of saying how many nucleotides it has 2 X 10^9 / 2 X 10^2 = 1 x 10 ^7 nucleosomes
Alfred Hershey
Identified DNA as the genetic material in bacteriophage
Species I has 2 n = 16 chromosomes. How many chromosomes will be found per cell in each of the following mutants in this species? d. Trisomic
If it doesn't have auto or allol account it for one Trisomic = 2n+1 16+1 = 17 chromosomes
What is the difference between autopolyploidy and allopolyploidy? How does each arise?
In autopolyploidy, all sets of chromosomes are derived from a single species. In allopolyploidy, the sets of chromosomes are derived from two or more different species. Autopolyploidy may arise through nondisjunction in an early 2n embryo or through meiotic nondisjunction that produces a gamete with extra sets of chromosomes. Allopolyploidy is usually preceded by hybridization between two different species, followed by chromosome doubling.
Briefly describe the differences between the lytic cycle of virulent phages and the lysogenic cycle of temperate phages.
In the lytic cycle, the viral genes immediately turn the host cell into a producing factory and then the bacterial cell wall is ruptured and the progeny phages are released. Thus the host is destroyed.In the lysogenic cycle, the host remains intact and through excision the progeny is released.
How can inversions in which no genetic information is lost or gained cause phenotypic effects?
Inversions alter the wild-type phenotype by causing breaks in genes or moving them to new locations thus disrupting normal gene expression
Species I is diploid (2 n = 4) with chromosomes AABB; related species II is diploid (2 n = 6) with chromosomes MMNNOO. Give the chromosomes that would be found in individuals with the following chromosome mutations: d. Trisomy in species II for chromosome M.
MMMNNOO or MMNNNOO or MMNNOOO 2n+1 7 chromosomes
Write a sequence of bases in an RNA molecule that will produce a hairpin structure.
Many possibilities, including: TGCGATACTCATCGCA because they are inverted complementary except for few in the middle which can't link and thus form that "hair pin"
Species I has 2 n = 16 chromosomes. How many chromosomes will be found per cell in each of the following mutants in this species? e. Double monosomic
Monosomic = 2n-1 double monosomic = 2n - 1 -1 16-1-1 = 14 chromosomes
Species I has 2 n = 16 chromosomes. How many chromosomes will be found per cell in each of the following mutants in this species? a. Monosomic
Monosomic = 2n-1 15 chromosomes
The following diagram represents two nonhomologous chromosomes: A B • C D E F G R S • T U V W X What type of chromosome mutation would produce each of the following groups of chromosomes? b. A U V B • C D E F G R S • T W X
Non-reciprocal translocation of UV
Species I has 2 n = 16 chromosomes. How many chromosomes will be found per cell in each of the following mutants in this species? f. Nullisomic
Nullisomic = 2n - 2 16-2 = 14 chromosomes
A chromosome has the following segments, where • represents the centromere: A B C D E • F G What types of chromosome mutations are required to change this chromosome into each of the following chromosomes? (In some cases, more than one chromosome mutation may be required.) d. A F • E D C B G
PERICENTRIC INVERSION, as the gene order and centromere position have changed
What are the three parts of a DNA nucleotide? Learn to draw it !!
Phosphate, Nitrogen containing base, and Deoxyribose sugar
The following diagram represents two nonhomologous chromosomes: A B • C D E F G R S • T U V W X What type of chromosome mutation would produce each of the following groups of chromosomes? c. A B • T U V F G R S • C D E W X
Reciprocal translocation of TUV and CDE
The following diagram represents two nonhomologous chromosomes: A B • C D E F G R S • T U V W X What type of chromosome mutation would produce each of the following groups of chromosomes? d. A B • C W G R S • T U V D E F X
Reciprocal translocation of W and DEF
How does supercoiling arise? What is the difference between positive and negative supercoiling?
Supercoiling arises from topoisomerases catalyzing the overwinding (positive supercoiling) or underwinding (negative supercoiling) of the DNA double helix. Supercoiling may occur: (1)when the DNA molecule does not have free ends, as in circular DNA molecules, or (2)when the DNA molecule is folded into loops that are bound to proteins that prevent the ends of the DNA strands from rotating about each other, as in eukaryotic chromosomes.
.A chromosome has the following segments, where • represents the centromere: A B C D E • F G What types of chromosome mutations are required to change this chromosome into each of the following chromosomes? (In some cases, more than one chromosome mutation may be required.) c. A B A B C D E • F G
Tandem duplication of segment AB
Species I has 2 n = 16 chromosomes. How many chromosomes will be found per cell in each of the following mutants in this species? h. Tetrasomic
Tetrasomic = 2n + 2 16 + 2 = 18
What different types of chemical bonds are found in DNA and where are they found?
The deoxyribonucleotides in a single chain or strand of DNA are held by covalent bond called phosphodiester linkages between the 3' end of the deoxyribose sugar of a nucleotide and the 5' end of the deoxyribose sugar of the next nucleotide in the chain. Two chains of deoxyribonucleotides are held together by hydrogen bonds between the complementary nitrogenous bases of the nucleotides in each chain.
What is the endosymbiosis theory?
The endosymbiotic theory proposes that mitochondria and chloroplasts evolved from formerly free-living bacteria that became endosymbionts within a larger eukaryotic cell.
What is a Robertsonian translocation?
The fusion of two non-homologous chromosomes, often with the deletion of a small amount of nonessential genetic material. - RT - fuse long arms of two acrocentric chromosome to from new metacentric chromosome - short arms lost - reduces chromosome number - most common rearrangement in humans 1/900
John Smith is a pig farmer. For the past 5 years, Smith has been adding vitamins and low doses of antibiotics to his pig food; he says that these supplements enhance the growth of the pigs. Within the past year, however, several of his pigs died from infections of common bacteria., which failed to respond to large doses of antibiotics. a) Can you offer an explanation for the increased rate of mortality due to infection in the pigs?
The information deciphers that John has been providing low doses of antibiotics to his pig for last 5 years. This is a huge time period for sustained delivery of small doses of antibiotics. Initially this dose had provided beneficial effects on the health and growth of pigs but gradually, the microflora of pig intestine became resistant to these antibiotics. This made the future generations of the pig by 4-5 years resistant to multiple antibiotics. Thus, the pigs became gradually immuno-compromised and developed resistance against numerous antibiotics. When an infection evaded the group of these pigs, they could not be treated with any of these antibiotics because the pigs were already resistant to them. Thus, this lead to sudden mass mortality of pigs in John's farm. This is why long term antibiotic treatments, even at exceedingly low doses, is never advised.
Describe the composition and structure of the nucleosome.
The nucleosome core particle contains two molecules each of histones H2A, H2B, H3, and H4, which form a protein core with 145-147 bp of DNA wound around the core. For each nucleosome, one molecule of a fifth histone, H1, binds to DNA entering and exiting the nucleosome core to clamp the DNA around the nucleosome.
What substrates are used in DNA synthesis?
The substrates for DNA synthesis are the four types of deoxyribonucleoside triphosphates: - deoxyadenosine triphosphate, - deoxyguanosine triphosphate, - deoxycytosine triphosphate - deoxythymidine triphosphate.
Explain how interrupted conjugation is used to map bacterial genes.
To map genes by conjugation, an interrupted mating procedure is used. During the conjugation process, an Hfr strain is mixed with an F- strain. The two strains must have different genotypes and must remain in physical contact for the transfer to occur. At regular intervals, the conjugation process is interrupted. The chromosomal transfer from the Hfr strain always begins with a portion of the integrated F factor and proceeds in a linear fashion. To transfer the entire chromosome would require approximately 100 minutes. The time required for individual genes to be transferred is relative to their position on the chromosome and the direction of transfer initiated by the F factor. Gene distances are typically mapped in minutes of conjugation. The genes that are transferred by conjugation to the recipient must be incorporated into the recipient's chromosome by recombination to be expressed
Rosalind Franklin
Took x-ray diffraction used in constructing the structure of DNA
How do translocations in which no genetic information is lost or gained produce phenotypic effects?
Translocations can produce phenotypic effects if the translocation breakpoint disrupts a gene or if a gene near the breakpoint is altered in its expression because of relocation to a different chromosomal environment.
Watson and Crick
Worked out the helical structure of DNA by building models
The following diagram represents two nonhomologous chromosomes: A B • C D E F G R S • T U V W X What type of chromosome mutation would produce each of the following groups of chromosomes? a. A B • C D R S • T U V W X E F G b. A U V B • C D E F G R S • T W X c. A B • T U V F G R S • C D E W X d. A B • C W G R S • T U V D E F X
a. nonreciprocal translocation of EFG
Species I is diploid (2 n = 4) with chromosomes AABB; related species II is diploid (2 n = 6) with chromosomes MMNNOO. Give the chromosomes that would be found in individuals with the following chromosome mutations: b. Allotetraploidy including species I and II.
allotetraploidy of species I and II = for (species I, n=2, and for species II, n=3. cross b/w them then 2n = 5 so allotertaploidy of sp I and II is = 4n = 10 chromosome.
Species I has 2 n = 8 chromosomes and species II has 2 n = 14 chromosomes. What would the expected chromosome numbers be in individuals with the following chromosome mutations? Give all possible answers. a. Allotriploidy including species I and II
allotrioploidy can have two cases 2n from s1 and n from s2 8 + 7 = 15 chromosomes or n from s1 and 2n from s2 4 + 14 = 18 chromosomes
How does specialized transduction differ from transduction?
both have genes transferred via homologous recombination but specialized transduction also transfers genes via helper phages ( allows integration of lambda gal into newly infected cells); frequency of specialized transduction is higher than that of generalized transduction
A chromosome has the following segments, where • represents the centromere: A B C D E • F G What types of chromosome mutations are required to change this chromosome into each of the following chromosomes? (In some cases, more than one chromosome mutation may be required.) a. A B E • F G
deletion
A chromosome has the following segments, where • represents the centromere: A B C D E • F G What types of chromosome mutations are required to change this chromosome into each of the following chromosomes? (In some cases, more than one chromosome mutation may be required.) e. A B C D E E D C • F G
duplication and paracentric inversion
Species I has 2 n = 8 chromosomes and species II has 2 n = 14 chromosomes. What would the expected chromosome numbers be in individuals with the following chromosome mutations? Give all possible answers. d. Monosomy in species II
monosomy = 2n -1 14 - 1 = 13 chromosomes
Species I has 2 n = 16 chromosomes. How many chromosomes will be found per cell in each of the following mutants in this species? g. Autopentaploid
n = 8 penta = 5 5n = 40 chromosomes
Explain how transduction and transformation can be used to map genes. How these methods are similar and how they are different?
n transformation, the relative frequency at which genes are transferred or co-transformed indicates the distance between the two genes. Closer gene pairs are co-transformed more frequently. Physical contact between two cells is not necessary. Recipient cells take up DNA from the environment directly. In transduction, a viral vector is needed for the transfer of DNA. DNA from the donor cell is packaged into viral protein coat. The virus containing the bacterial DNA when infects other bacteria, it transfers the bacterial DNA to them. Only genes that are close together are cotransduced. Therefore, the rate of cotransduction gives an indication of the physical distance between the two genes.
A chromosome has the following segments, where • represents the centromere: A B C D E • F G What types of chromosome mutations are required to change this chromosome into each of the following chromosomes? (In some cases, more than one chromosome mutation may be required.) b. A E D C B • F G
segment BCDE is inverted 180 degrees; but as the centromere has not changed position it is PARACENTRIC INVERSION.
Species I has 2 n = 8 chromosomes and species II has 2 n = 14 chromosomes. What would the expected chromosome numbers be in individuals with the following chromosome mutations? Give all possible answers. e. Tetrasomy in species I
tetrasomy = 2n+2 8 + 2 = 10 Chromosomes
Species I has 2 n = 8 chromosomes and species II has 2 n = 14 chromosomes. What would the expected chromosome numbers be in individuals with the following chromosome mutations? Give all possible answers. c. Trisomy in species I
trisomy = 2n + 1 8 + 1 = 9 chromosomes