Genetics Test 2

when doing the r scale (0---1--2--3---4) what is the total

+1 of last number shown

F plasmid conjugation

- Transfer of a strand of F plasmid from F+ donor to F- recipient thru mating bridge f- cell becomes f+ cell w/ f plasmid, r plasmids (genes for antibiotic resistance) can be transferred rapidly in this way

For linked genes A, B, and C, the map distance A-B is 5 map units and the map distance B-C is 25 map units. If there are 10 double crossover events out of 1000 offspring, what is the interference?

0.2 the coefficient of coincidence is 0.01-0.0125=0.8 so interference is 1-0.8=0.2

Phenotype: Number Observed: Spineless 321 wild type 38 claret, spineless 130 claret 18 claret, hairless 309 hairless, claret, spineless 32 hairless 140 hairless, spineless 12 In the fruit fly, Drosophila melanogaster, a spineless (no wing bristles) female fly is mated to a male that is claret (dark eyes) and hairless (no thoracic bristles). Phenotypically wild-type F1 female progeny were mated to fully homozygous (mutant) males, and 1000 progeny were obtained what is the coefficient of confidence?

1

Genes X, Y, and Z are linked. Crossover gametes between genes X and Y are observed with a frequency of 25%, and crossover gametes between genes Y and Z are observed with a frequency of 5%. What is the expected frequency of double crossover gametes among these genes?

1.25% 0.25*0.05=0.0125

What is probability of aa;bb;cc;dd;ee

1/256 1/4*1/2*1/4*1/2...

Phenotype: Number Observed: Spineless 321 wild type 38 claret, spineless 130 claret 18 claret, hairless 309 hairless, claret, spineless 32 hairless 140 hairless, spineless 12 In the fruit fly, Drosophila melanogaster, a spineless (no wing bristles) female fly is mated to a male that is claret (dark eyes) and hairless (no thoracic bristles). Phenotypically wild-type F1 female progeny were mated to fully homozygous (mutant) males, and 1000 progeny were obtained what is the map distance between hairless and spineless

10 map units

The genes for mahogany eyes and ebony body are approximately 25 map units apart on chromosome 3 in Drosophila. Assume that a mahogany-eyed female was mated to an ebony-bodied male and that the resulting F1 phenotypically wild-type females were mated to mahogany, ebony males to obtain an F2generation.\ If 1000 F2 offspring are obtained, how many flies of each phenotype are expected? __wild type : __ mahogeny : __ ebony : __ mahogeny, ebony (Remember the two genes are 25 mu apart on the same chromosome. In order to determine how many offspring of each type are produced, you must determine if the F1 females carries both mutations on one chromosome, and wildtype alleles on the other or if she carries one mutation and one wildtype allele on each chromosome.)

125:375:375:125

If two genes are found on different chromosomes, or if they are far enough apart on the same chromosome that the chance of a crossover between them is very high, the genes are considered to be unlinked. Unlinked genes follow Mendel's law of independent assortment. If, however, two genes tend to "travel together" because they are near one another on the same chromosome, they are said to be linked. Linked genes do not follow Mendel's law of independent assortment. In this tutorial, you will compare the inheritance patterns of unlinked and linked genes. A wild-type tomato plant (Plant 1) is homozygous dominant for three traits: solid leaves (MM), normal height (DD), and smooth skin (PP). Another tomato plant (Plant 2) is homozygous recessive for the same three traits: mottled leaves (mm), dwarf height (dd), and peach skin (pp). In a cross between these two plants (MMDDPP x mmddpp), all offspring in the F1 generation are wild type and heterozygous for all three traits (MmDdPp). Now suppose you perform a testcross on one of the F1 plants (MmDdPp x mmddpp). The F2 generation can include plants with these eight possible phenotypes: solid, normal, smooth solid, normal, peach solid, dwarf, smooth solid, dwarf, peach mottled, normal, smooth mottled, normal, peach mottled, dwarf, smooth mottled, dwarf, peach Assuming that the three genes undergo independent assortment, predict the phenotypic ratio of the offspring in the F2 generation

1:1:1:1:1:1:1:1

In a three‑point mapping experiment for the genes y‑w‑ec, the following percentages of events are observed: NCO events: 65%; SCO events between y and w: 15%; SCO events between y and ec: 17%; DCO events: 3% What is the map distance between y and ec?

20 map units the map distance between any two genes is the sum of the percentages of all detectable recomb. events so 17+3=20

place the three loci on the map by dragging eah label to the correct bin. if two loci are situated very close together on the map, place them in the same bin

3 min: phe 1+ 2 min: gly3+ lys2+ 1 min:

Phenotype: Number Observed: Spineless 321 wild type 38 claret, spineless 130 claret 18 claret, hairless 309 hairless, claret, spineless 32 hairless 140 hairless, spineless 12 In the fruit fly, Drosophila melanogaster, a spineless (no wing bristles) female fly is mated to a male that is claret (dark eyes) and hairless (no thoracic bristles). Phenotypically wild-type F1 female progeny were mated to fully homozygous (mutant) males, and 1000 progeny were obtained what is the map distance between claret and hairles

30 map units

Assume that the genes from the previous example are located along the chromosome in order X,Y, and Z. What is the probability of recombination between genes x and z?

30% 25+5=30

If two genes are on the same chromosome exhibit complete linkage, what is the expected F2 phenotypic ratio from a selfed heterozygote with the genotype a+b+//ab?

3:1

When two data points are being observed ( ex. shape and color) are being considered separately, F2 individuals exhibit a _______ ratio

3:1

Phenotype: Number Observed: Spineless 321 wild type 38 claret, spineless 130 claret 18 claret, hairless 309 hairless, claret, spineless 32 hairless 140 hairless, spineless 12 In the fruit fly, Drosophila melanogaster, a spineless (no wing bristles) female fly is mated to a male that is claret (dark eyes) and hairless (no thoracic bristles). Phenotypically wild-type F1 female progeny were mated to fully homozygous (mutant) males, and 1000 progeny were obtained what is the map distance between claret and spineless

40 map units

If 1000 F2 offspring are obtained, how many flies of each phenotype are expected? ___ wild-type : ____ tan : ___ bare : ____ tan, bare (Remember the two genes are 15 mu apart on the same chromosome. In order to determine how many offspring of each type are produced, you must determine if the F1 females carries both mutations on one chromosome, and wildtype alleles on the other or if she carries one mutation and one wildtype allele on each chromosome.)

425:75:75:425

When the pea test was completed (green and yellow, wrinkly and smooth) the ratio was

9:3:3:1

for a dihybrid cross the ratio between two unlinked genes will be

9:3:3:1

parental types should represent ____% of total testcross progeny

>50

Conjugation

A temporary union of two organisms for the purpose of DNA transfer.

What happens when we extend Mendel's crossing design to more than one gene?

A/a monohybrid A/a; B/b or AaBb= dihybrid AaBbCc= trihybrid unlinked genes: A/a; B/b or AaBb linked genes: AB/ab

The genes for mahogany eyes and ebony body are approximately 25 map units apart on chromosome 3 in Drosophila. Assume that a mahogany-eyed female was mated to an ebony-bodied male and that the resulting F1 phenotypically wild-type females were mated to mahogany, ebony males to obtain an F2generation. which of following phenotypes would be expected in the F2 offspring? wild type mahogeny-eyed ebony-bodied mahogeny-eyed, ebony-bodied

All of the above

Which of the following statements about conjugation is true?

DNA is transferred from an F+ cell to an F-cell

A plaque is a substance that causes mutation in bacteria

FALSE

In a bacterial cross in which the donor (Hfr) is a+b+ and the recipient strain (F-) is a-b-, it is expected that recombinant bacteria will all be a+b+.

FALSE

The cross GE/ge × ge/ge produces the following progeny: GE/ge 404; ge/ge 396; gE/ge 97; Ge/ge 103. From these data one can conclude that the G and E loci assort independently.

FALSE

To construct a mapping cross of linked genes, it is important that the genotypes of some of the gametes produced by the heterozygote can be deduced by examining the phenotypes of the progeny.

FALSE To construct a mapping cross of linked genes, it is important that the genotypes of all the gametes produced by the heterozygote can be deduced by examining the phenotypes of the progeny, taking into consideration that the homozygote produced only recessive gametes. Gametes and their genotypes can never be observed directly.

True or false? A bacterial strain that is pro + thi + leu − met − will grow on minimal media plus leucine and thiamine.

FALSE the strain is auxotrophic for leucine and methionine meaning that it cannot synthesize those nutrients, thus, it requires those two nutrients to be added to minimal media for growth.

Crossing over during orophase 1 of meiosis occurs between alleles on sister chromatids

FALSE, Crossing over during meiosis occurs between alleles on nonsister chromatids.

The "interrupted mating technique" provides a genetic map in Drosphilia

False

Use the figure to predict the results of Lederberg and Tatum's experiment.(Figure 1) Bacterial cultures from which flask(s) will grow on minimal medium plus pro and thi? Select all that apply.

Flask 1 and 3 Minimal medium can only support the growth of bacteria that are wild type for all growth requirements, and if supplemented with proline and thiamine, can support growth of bacteria that are pro - and thi -, if they are wild type for other genes.

Use the figure to predict the results of Lederberg and Tatum's experiment. (Figure 1) Bacterial cultures from which flask(s) will grow on complete medium? Select all that apply.

Flask 1,2. and 3 complete medium is extensively supplemented and can support the growth of auxotrophs and prototrophs

Use the figure to predict the results of Lederberg and Tatum's experiment. (Figure 1) Bacterial cultures from which flask(s) will grow on minimal medium plus val and arg? Select all that apply.

Flask 2 and 3 Minimal medium can only support the growth of bacteria that are wild type for all growth requirements. If valine and arginine are added to this medium, it can support growth of bacteria that are val - and arg -, if they are wild type for other genes.

Use the figure to predict the results of Lederberg and Tatum's experiment. (Figure 1) Bacterial cultures from which flask(s) will grow on minimal medium plus val? Select all that apply.

Flask 3 Minimal medium only supports the growth of bacteria that are wild type for all growth requirements. When minimal medium is supplemented with proline, it can support the growth of pro - bacteria, but these bacteria must be wild type for other genes. Because strains 1 and 2 have additional mutations, and are auxotrophs, they will not grow under this condition. However, genetic exchange can occur between these two auxotrophic strains when combined together in flask 3, which can produce prototrophs that are wild type for all growth requirements and capable of growing in this condition.

Use the figure to predict the results of Lederberg and Tatum's experiment. (Figure 1) Bacterial cultures from which flask(s) will grow on minimal medium? Select all that apply.

Flask 3 The bacteria in strains 1 and 2 are auxotrophs and cannot grow in minimal medium. However, because genetic exchange can occur between these strains when combined together in flask 3, prototrophs which are wild type for all growth requirements can be produced. They are capable of growing in this simple medium.

Use the figure to predict the results of Lederberg and Tatum's experiment. (Figure 1) Bacterial cultures from which flask(s) will grow on minimal medium plus pro? Select all that apply.

Flask 3 Minimal medium only supports the growth of bacteria that are wild type for all growth requirements. When minimal medium is supplemented with proline, it can support the growth of pro - bacteria, but these bacteria must be wild type for other genes. Because strains 1 and 2 have additional mutations, and are auxotrophs, they will not grow under this condition. However, genetic exchange can occur between these two auxotrophic strains when combined together in flask 3, which can produce prototrophs that are wild type for all growth requirements and capable of growing in this condition.

You have screened for several new recessive mutations in a species of wasp. Wasps homozygous for apricot (aa) have pale orange eyes. (Wild-type eyes are brown.) Wasps homozygous for blunt (bb) have short wings. (Wild-type wings are long.) You make a pure-breeding double-mutant (apricot, blunt) line and cross it with wild-type wasps. The F1 is wild-type in appearance. You testcross the F1 dihybrids with the double-mutant line (the "tester" genotype) and obtain four phenotypes in the testcross progeny:wild typeapricot, bluntapricotblunt You have also identified a third recessive mutation that you call crimson. Wasps homozygous for crimson have a red abdomen. (Wild-type abdomens are yellow.)You make pure-breeding crimson (cc) and triple-mutant (apricot, blunt, crimson) lines. To map the three loci, you perform a trihybrid (three-point) testcross: You first cross crimson wasps with (apricot, blunt) wasps to make an F1, and then you testcross the F1 with the triple-mutant line.The testcross progeny have eight phenotypes. For each phenotype, determine the genotype of the gamete it received from the F1 trihybrid.

GAMETE FROM TRIHYBRID wild type: ABC apricot, blunt, crimson: abc apricot, blunt: abC crimson: ABc blunt: AbC apricot, crimson: aBc blunt, crimson: Abc apricot: aBC

Hfr and time of entry maps

Hfr is similar to F+ plasmid, however Hfr is integrated in the chromosome of the strain, therefore is called high frequency recombination factor. Time of entry mapping can be used to order genes for a particular Hfr strain

Assume that a cross is made between AaBb and aabb plants and that the offspring fall into approximately equal numbers of the following groups: AaBb, Aabb, aaBb, aabb. These results are consistent with the following circumstance:

Independent assortment

What term is applied when two genes fail to assort independently, that is they tend to segregate together during gamete formation?

Linkage

Assume that a cross is made between AaBb and aabb plants and that the offspring occur in the following numbers: 106 AaBb, 48 Aabb, 52 aaBb, 94 aabb. These results are consistent with the following circumstance:

Linkage with approximately 33 map units between the two gene loci

How can the order of three linked genes (A, B, and C) on the same chromosome be determined?

Look for double-crossover phenotypes involving the wild-type and mutant alleles of genes A,B, and C offspring with a double-crossover phenotype will occur less frequently and will therefore be easy to identify. The three possible gene orders are then compared to the double crossover genotypes to determine which gene order fits the observed pattern

To make a rough map of the lys2, gly3, and phe1 loci, you perform an interrupted conjugation experiment between the following two strains: Hfr lys2+ gly3+ phe1+ AmpS × F- lys2- gly3- phe1- AmpR (Note: AmpS indicates sensitivity to ampicillin. AmpR indicates resistance to ampicillin.) You interrupt conjugation after 1 min, 2 min, and 3 min and plate exconjugants on minimal medium containing ampicillin and various combinations of lysine, glycine, and phenylalanine. You then score the resulting plates for the presence of colonies.

Minimal medium + lysine+ glycine: 1 min : no colonies 2 min: no colonies 3 min: colonies colony genotypes: lys2+ or lys2- phe1+ gly3+or gly3- minimal medium + phenylalanine + glycine 1 min : no colonies 2 min: colonies 3 min: colonies colony genotypes: lys2+ phe1+ or phe 1- gly3+or gly3- Minimal medium + phenylaline+ lycine: 1 min : no colonies 2 min: colonies 3 min: colonies colony genotypes: lys2+ or lys2- phe1+ or phe 1- gly3+

Which of the following statements about gamete formation during meiosis is false? * Parental genes can be formed only if there is no crossing over during meiosis * Complete linkage results in formation of only parental gametes *Recombinant gametes contain combinations of alleles not found in parent cell. *Parental gametes contain the same combinations of linked genes as found in parent cell.

Parental gametes can be formed only if there is no crossing over during meiosis is false.

Trihybrid cross

Segregation and independent assortment applied to three pairs of constraining traits (64 boxes)

Mitochondria (mtDNA) and chloroplasts (cpDNA)

Single circular model very limited array of genes (critical for specialized function) (basic metabolic functions)

A symbiotic relationship between a phage and a bacterium apparently occurs in the process of lysogeny

TRUE

An Hfr cell can initiate chromosome transfer from one E. Coli to another.

TRUE

If two gene loci are on nonhomologous chromosomes, genes at these loci are expected to assort independently.

TRUE

Lysogeny is most likely associated with transduction. T/ F

TRUE

The cross GE/ge × ge/ge produces the following progeny: GE/ge 404; ge/ge 396; gE/ge 97; Ge/ge 103. From these data one can conclude that the recombinant progeny are gE/ge and Ge/ge.

TRUE

The cross GE/ge × ge/ge produces the following progeny: GE/ge 404; ge/ge 396; gE/ge 97; Ge/ge 103. From these data one can conclude that there are 20 map units between the G and E loci.

TRUE

Viral mutations and variants are often categorized by changes in host range and/or plaque morphology

TRUE

How is a merozygote formed?

The F factor and several adjacent genes are excised from the chromosome of an F+ cell and transferred to an F− strain.

Which of the following statements about mapping bacterial genes by conjugation is NOT true?

The closer a gene to the Hfr origin the more like it will be transferred to the recipient during conjugation

F' plasmid

The plasmid form of the F factor. Fetility factor allows genes to be transferred from 1 bacterium carrying the factor to another that is lacking the factor by conjugation

Two genes that are separated by 10 map units show a recombination percentage of 10%

True

Bacteriophage T4 can absorb to its host because of binding sites in its tail fibers that recognize areas of the E coli cell wall

True The tips of the tail fibers of bacteriophage T4 contain binding sites that recognize specific areas of its host's cell wall.

You have screened for several new recessive mutations in a species of wasp. Wasps homozygous for apricot (aa) have pale orange eyes. (Wild-type eyes are brown.) Wasps homozygous for blunt (bb) have short wings. (Wild-type wings are long.) You make a pure-breeding double-mutant (apricot, blunt) line and cross it with wild-type wasps. The F1 is wild-type in appearance. You testcross the F1 dihybrids with the double-mutant line (the "tester" genotype) and obtain four phenotypes in the testcross progeny:wild typeapricot, bluntapricotblunt Determine the following for each phenotype in the testcross progeny: the haploid genotype of the gamete it received from the F1 dihybrid its full diploid genotype whether the gamete it received from the F1 dihybrid was recombinant (REC) or nonrecombinant (nonREC).

Wild Type: gamete from dihybrid: AB gamete from tester: ab full genotype: AaBb Recombinant or nonrecombinant?: Non Apricot, blunt: gamete from dihybrid: ab gamete from tester: ab full genotype: aabb recombinant or nonrecombinant?: non apricot: gamete from dihybrid: aB gamete from tester: ab full genotype: aaBb recombinant or nonrecombinant? REC blunt: gamete from dihybrid: Ab gamete from tester: ab full genotype: Aabb recombinant or non recombinant? REC

You have screened for several new recessive mutations in a species of wasp. Wasps homozygous for apricot (aa) have pale orange eyes. (Wild-type eyes are brown.) Wasps homozygous for blunt (bb) have short wings. (Wild-type wings are long.) You make a pure-breeding double-mutant (apricot, blunt) line and cross it with wild-type wasps. The F1 is wild-type in appearance. You testcross the F1 dihybrids with the double-mutant line (the "tester" genotype) and obtain four phenotypes in the testcross progeny:wild typeapricot, bluntapricotblunt The eight gamete classes for the testcross in Part B are shown in the table below. For each gamete, determine if it represents a recombinant (REC) or nonrecombinant (nonREC) gamete for the three pairs of loci: apricot and blunt (the A-B column), blunt and crimson (the B-C column), and apricot and crimson (the A-C column). Be sure to determine which alleles are on each chromosome in the F1 trihybrid by analyzing her parents before assigning REC or nonREC to specific gametes.

Wild type: gamete from trihybrid: ABC A-B: nonREC B-C: REC A-C: REC apricot, blunt, crimson: gamete from trihybrid: abc A-B: nonREC B-C: REC A-C: REC apricot, blunt: gamete from trihybrid: abC A-B: nonREC B-C: nonREC A-C: nonREC crimson: gamete from trihybrid: ABc A-B: nonREC B-C: nonREC A-C: nonREC blunt: gamete from trihybrid: AbC A-B:REC B-C: nonREC A-C:REC apricot, crimson: gamete from trihybrid: aBc A-B REC B-C nonREC A-C REC blunt, crimson: gamete from trihybrid: Abc A-B REC B-C REC A-C nonREC apricot: gamete from trihybrid: aBC A-B REC B-C REC A-C nonREC

You have screened for several new recessive mutations in a species of wasp. Wasps homozygous for apricot (aa) have pale orange eyes. (Wild-type eyes are brown.) Wasps homozygous for blunt (bb) have short wings. (Wild-type wings are long.) You make a pure-breeding double-mutant (apricot, blunt) line and cross it with wild-type wasps. The F1 is wild-type in appearance. You testcross the F1 dihybrids with the double-mutant line (the "tester" genotype) and obtain four phenotypes in the testcross progeny:wild typeapricot, bluntapricotblunt You examine 450 progeny from the cross in Part C and classify them according to phenotype. The results are shown in the table below.

Wild type: gamete from trihybrid: ABC number observed: 17 A-B: nonREC B-C: REC A-C: REC apricot, blunt, crimson: gamete from trihybrid: abc number observed: 19 A-B: nonREC B-C: REC A-C: REC apricot, blunt: gamete from trihybrid: abC number observed: 161 A-B: nonREC B-C: nonREC A-C: nonREC crimson: gamete from trihybrid: ABc number observed: 163 A-B: nonREC B-C: nonREC A-C: nonREC blunt: gamete from trihybrid: AbC Number observed: 39 A-B:REC B-C: nonREC A-C:REC apricot, crimson: gamete from trihybrid: aBc Number observed: 42 A-B REC B-C nonREC A-C REC blunt, crimson: gamete from trihybrid: Abc number observed: 5 A-B REC B-C REC A-C nonREC apricot: gamete from trihybrid: aBC number observed: 4 A-B REC B-C REC A-C nonREC total progeny=450

Chiasmata

X-shaped regions where crossing over occurred on non -sister chromatids

If two genes are found on different chromosomes, or if they are far enough apart on the same chromosome that the chance of a crossover between them is very high, the genes are considered to be unlinked. Unlinked genes follow Mendel's law of independent assortment. If, however, two genes tend to "travel together" because they are near one another on the same chromosome, they are said to be linked. Linked genes do not follow Mendel's law of independent assortment. In this tutorial, you will compare the inheritance patterns of unlinked and linked genes. A wild-type tomato plant (Plant 1) is homozygous dominant for three traits: solid leaves (MM), normal height (DD), and smooth skin (PP). Another tomato plant (Plant 2) is homozygous recessive for the same three traits: mottled leaves (mm), dwarf height (dd), and peach skin (pp). In a cross between these two plants (MMDDPP x mmddpp), all offspring in the F1 generation are wild type and heterozygous for all three traits (MmDdPp). Now suppose you perform a testcross on one of the F1 plants (MmDdPp x mmddpp). The F2 generation can include plants with these eight possible phenotypes: solid, normal, smooth solid, normal, peach solid, dwarf, smooth solid, dwarf, peach mottled, normal, smooth mottled, normal, peach mottled, dwarf, smooth mottled, dwarf, peach now suppose that the three tomato genes from Part A did not assort independently but instead were linked to one another on the same chromosome. Would you expect the phenotypic ratio in the offspring to change? If so how?

all 8 possible phenotypes could occur, but a greater proportion of the offspring would have the parental phenotypes

Assume that the genes for tan body and bare wings are 15 map units apart on chromosome 2 in Drosophila. A tan-bodied, bare-winged female was mated to a wild-type male and that the resulting F1 phenotypically wild-type females were mated to tan-bodied, bare-winged males to obtain an F2 generation. What phenotypes would be expected in the F2 offspring? wild type tan body bare wings tan body, bare wings

all of the above

You have screened for several new recessive mutations in a species of wasp. Wasps homozygous for apricot (aa) have pale orange eyes. (Wild-type eyes are brown.) Wasps homozygous for blunt (bb) have short wings. (Wild-type wings are long.) You make a pure-breeding double-mutant (apricot, blunt) line and cross it with wild-type wasps. The F1 is wild-type in appearance. You testcross the F1 dihybrids with the double-mutant line (the "tester" genotype) and obtain four phenotypes in the testcross progeny:wild typeapricot, bluntapricotblunt complete a linkage map

apricot-blunt= 20 mu blunt-crimson= 10 mu

Phenotype: Number Observed: Spineless 321 wild type 38 claret, spineless 130 claret 18 claret, hairless 309 hairless, claret, spineless 32 hairless 140 hairless, spineless 12 In the fruit fly, Drosophila melanogaster, a spineless (no wing bristles) female fly is mated to a male that is claret (dark eyes) and hairless (no thoracic bristles). Phenotypically wild-type F1 female progeny were mated to fully homozygous (mutant) males, and 1000 progeny were obtained with respect to the three genes mentioned in the problem, what are the genotypes of the homozygous parents used in making the phenotypically wild F1 heterozygote

cl+h+sp/cl+h+sp cl h sp+/cl h sp+

Assume that a cross is made between AaBb and aabb plants and that all the offspring are either AaBb or aabb. These results are consistent with the following circumstance:

complete linkage

recombinants for linked genes are produced by

crossing over (recombination)

For polygenic inheritance, __________ and _______________ involved decrease the distinction between "dosage" levels

enviornmental contributions number of genes

independent assortment in diploids

explained by chromosomal behavior

Independent assortment in haploids

explained by chromosomal behavior can observe independent assortment directly

Organelle genes have

extranuclear inheritance

Mapping bacterial genes by conjugation is based on which of the following assumptions?

genes are transferred from donor to recipient in a linear fashion

If 50% recomb. freq. observed in testcross=

genes likely carried on seperate chromosomes

Phenotype: Number Observed: Spineless 321 wild type 38 claret, spineless 130 claret 18 claret, hairless 309 hairless, claret, spineless 32 hairless 140 hairless, spineless 12 In the fruit fly, Drosophila melanogaster, a spineless (no wing bristles) female fly is mated to a male that is claret (dark eyes) and hairless (no thoracic bristles). Phenotypically wild-type F1 female progeny were mated to fully homozygous (mutant) males, and 1000 progeny were obtained which gene is in the middle

hairless

`Dice example: 3 Die 1 green 1 red 1 blue

if you roll 6 red 6 Greem 6 blue what is the probability 1/16 0 sixes? 125/216 all different numbers? 5/9

two mechanisms of recombination

independent assortment (1:1:1:1 ratio = 50% recomb.) crossing over during meiosis (?:?:?:?)

Mendel's Second Law

law of independent assortment unlined genes are inherited independently during gamete formation. 1:1:1:1 ratio of gametes in dihybrid cross (equal segregation)

what if genes are linked

linked genes are on the same chromosome. more likely to be inherited *and expressed* together

You decide to refine your map by doing a recombination experiment. You cross the two strains as in Part A: Hfr lys2+ gly3+ phe1+ AmpS × F- lys2- gly3- phe1- AmpR This time you select for exconjugants on minimal medium containing ampicillin, lysine, and glycine. You count 5,000 phe1+ colonies that grow on this medium. You replica plate these 5,000 colonies onto the following media: minimal medium + ampicillin = 4257 minimal medium + glycine+ampicillin= 4,264 minimal medium + lysine + ampicillin = 4,300 minimal medium + lysine+glycine+ amp= 5,000 Replica plating is a technique that allows the same bacterial colonies to be plated onto different media. A sterile velvet pad is pressed onto a "master" plate to pick up bacteria from each colony. The pad is then pressed onto new plates to reproduce the same colonies on other media. For help understanding which genotypes grow on each medium type, use the Hint. Use these data to deduce the number of exconjugants with each genotype.

lys2+ gly3+ phe1+ = 4,257 lys2+ gly3- phe1+ = 7 lys2- gly3+ phe1+ = 48 lys 2- gly3- phe1+ 693 The four media in this experiment had the same 5,000 colonies plated on them. Minimal medium + ampicillin allowed only the lys2+ gly3+ phe1+ genotype to grow, resulting in 4,257 colonies. The other three media contained supplements and allowed other genotypes to grow in addition to lys2+ gly3+ phe1+ . You can deduce the number of singly mutant exconjugants by subtracting the number of lys2+ gly3+ phe1+ colonies (4,257) from the number of colonies on the singly supplemented plates. The number of lys2+ gly3- phe1+ exconjugants = 4,264 - 4,257 = 7. The number of lys2- gly3+ phe1+ exconjugants = 4,300 - 4,257 = 43. You can deduce the number of doubly mutant exconjugants (lys2- gly3- phe1+ ) by subtracting all the other colonies from the total: 5,000 - 4,257 - 7 - 43 = 693.

temperate phages are those that can enter either the ________ or __________ cycle

lytic ; lysogenic

Bacteriophages engage in two interactive cycles with bacteria. what are these cycles

lytic and lysogenic

Suppose that you perform the cross discussed in Part B: MmDdPp x mmddpp. You plant 1000 tomato seeds resulting from the cross, and get the following results: 420:21:2:52:62:4:23:216 Drag the labels onto the chromosome diagram to identify the locations of and distances between the genes. Use the blue labels and blue targets for the genes; use the white labels and white targets for the distances. Gene m has already been placed on the linkage map.

m-d = 12 cM d-p = 5 cM m-p= 17 cM

Recombination

meiosis produces new allelic combinations (independent assortment or crossing over) new combinations provide raw material for natural selection we can identify that recombination has occured by comparing parental meiotic inputs with gametic outputs

for combined probabilities you __________ the two probabilities together

multiply

what limits prototrophic growth within a culture of a medium

nutrients and oxygen

A clearing made by bacteriophages in a "lawn" of bacteria on an agar plate is called a

plaque

The clearing made by bacteriophages in a "lawn" of bacteria on an agar plate is called a

plaque

Name the general category into which double-stranded circular extrachromosomal DNA elements such as F factors, ColE1, and R would fall.

plasmid

The phenomenon in which one crossover decreases the likelihood of crossovers n nearby regions is called

positive interference

Polygenes are also called

quantitative trait loci (QTL)

a bacteriophage that is capable of entering either a lytic or lysogenic cycle is called a

temperate bacteriophage

linked genes do not follow law of independent assortment but...

the chromosomal basis of segregation remains!

What results would be expected from a plaque assay if the dillution factor of the phage infected culture were too low.

the entire lawn of bacteria would be lysed. The concentration of phage‑infected bacteria increases as the dilution factor decreases. If the dilution factor were too low, there would be so many plaques that all the bacteria would be lysed.

Which of the folllowing statements about the t4 lytic life cycle is false?

the phage DNA is injected into the host and integrates into the bacterial chromosome This statement is false. The phage DNA is injected into the host but is then replicated during the lytic cycle, producing a pool of viral DNA particles. Integration into the bacterial chromosome occurs only during the lysogenic cycle.

In an interrupted mating experiment, the purpose of plating cells on a selective medium is _______.

to ensure that only recombinant genotypes are recovered

Polygenic are phenotypes affected by alleles at

two or more genes, not just multiple alleles at one gene

The inheritance of organelles is

uniparental (maternal) revealed by reciprocal crosses *mutant female xwildtype male yields all mutant *mutant male x wildtype female yields all wildtype *does NOT follow mendelian rules

Why are we interested in recombinants

useful in determining if genes are unlinked (on separate chromosomes) if the frequency of recombinants differs from expected value (independent assortment) they are linked