Deck 5: Linkage, Recombination, and Eukaryotic Gene Mapping
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Deck 5: Linkage, Recombination, and Eukaryotic Gene Mapping
1
Linked genes:
A)assort randomly.
B)can't crossover and recombine.
C)are allelic.
D)segregate together.
E)will segregate independently.
A)assort randomly.
B)can't crossover and recombine.
C)are allelic.
D)segregate together.
E)will segregate independently.
D
2
Two linked genes, (A)and (B), are separated by 18 cM.A man with genotype Aa Bb marries a woman who is aa bb.The man's father was AA BB.What is the probability that their first child will be Aa bb?
A)0)18
B)0)41
C)0)09
D)0)25
E)0)50
A)0)18
B)0)41
C)0)09
D)0)25
E)0)50
C
3
Linked genes are:
A)allelic.
B)dominant.
C)on different chromosomes.
D)on the same chromosome.
E)recessive lethal.
A)allelic.
B)dominant.
C)on different chromosomes.
D)on the same chromosome.
E)recessive lethal.
D
4
Recombination frequencies can be calculated by:
A)counting the number of recombinant and parental offspring when an individual who is heterozygous for two genes is testcrossed.
B)performing a chi-square test on the F2 progeny when an individual who is homozygous for two genes is initially crossed to an individual who is homozygous recessive for these two genes.
C)counting the number of offspring who are expressing the dominant phenotype when a heterozygous individual for two genes is testcrossed.
D)performing a chi-square test of the progeny of a cross between parents who are both heterozygous for the same two genes.
E)counting the number of offspring that are found in the cross of an individual who is heterozygous for two genes with another parent who is homozygous dominant for one of these genes and homozygous recessive for the other gene.
A)counting the number of recombinant and parental offspring when an individual who is heterozygous for two genes is testcrossed.
B)performing a chi-square test on the F2 progeny when an individual who is homozygous for two genes is initially crossed to an individual who is homozygous recessive for these two genes.
C)counting the number of offspring who are expressing the dominant phenotype when a heterozygous individual for two genes is testcrossed.
D)performing a chi-square test of the progeny of a cross between parents who are both heterozygous for the same two genes.
E)counting the number of offspring that are found in the cross of an individual who is heterozygous for two genes with another parent who is homozygous dominant for one of these genes and homozygous recessive for the other gene.
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5
Recombination occurs through:
A)crossing over and chromosome interference.
B)chromosome interference and independent assortment.
C)somatic-cell hybridization and chromosome interference.
D)complete linkage and chromosome interference.
E)crossing over and independent assortment.
A)crossing over and chromosome interference.
B)chromosome interference and independent assortment.
C)somatic-cell hybridization and chromosome interference.
D)complete linkage and chromosome interference.
E)crossing over and independent assortment.
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6
Genetic distances within a given linkage group:
A)cannot exceed 100 cM.
B)are dependent on crossover frequencies between paired, nonsister chromatids.
C)are measured in centiMorgans.
D)cannot be determined.
E)are both dependent on crossover frequencies between paired, nonsister chromatids and measured in centiMorgans.
A)cannot exceed 100 cM.
B)are dependent on crossover frequencies between paired, nonsister chromatids.
C)are measured in centiMorgans.
D)cannot be determined.
E)are both dependent on crossover frequencies between paired, nonsister chromatids and measured in centiMorgans.
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7
A linkage group:
A)involves on or more genes found on different chromosomes.
B)is when two genes interact with each other to form a phenotype.
C)is formed between one or more genes residing close together on a chromosome.
D)involves genetic recombination of greater than 50% among two genes.
E)is when one gene effects the expression of another gene.
A)involves on or more genes found on different chromosomes.
B)is when two genes interact with each other to form a phenotype.
C)is formed between one or more genes residing close together on a chromosome.
D)involves genetic recombination of greater than 50% among two genes.
E)is when one gene effects the expression of another gene.
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8
You are doing lab work with a new species of beetle.You isolated lines that breed true for either blue shells and long antenna, or green shells and short antenna.Crossing these lines yields F1 progeny with blue shells and long antenna.Crossing F1 progeny with beetles that have green shells and short antenna yield the following progeny:
A chi-square test is done to test for independent assortment.What is the resulting chi-square value and how many degree(s)of freedom should be used in its interpretation?
A)27.1 and one degree of freedom
B)14.9 and three degrees of freedom
C)14.9 and two degrees of freedom
D)27.1 and three degrees of freedom
E)0)42 and two degrees of freedom

A)27.1 and one degree of freedom
B)14.9 and three degrees of freedom
C)14.9 and two degrees of freedom
D)27.1 and three degrees of freedom
E)0)42 and two degrees of freedom
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9
Dr.Disney has been raising exotic fruit flies for decades.Recently, he discovered a strain of fruit flies that in the recessive condition have baby blue eyes that he designates as bb.He also has another strain of fruit flies that in the recessive condition have pink wings that are designated as pw.He is able to establish flies that are homozygous for both mutant traits.He mates these two strains with each other.Dr.Disney then takes phenotypically wild-type females from this cross and mates them with double recessive males.In the resulting testcross progeny, he observes 500 flies that are of the following make-up:
Assuming the wild-type alleles for these two genes are b+ and pw+, what is the CORRECT testcross of the F1 flies?
A)bb+ pw+/bb pw × bb pw/bb pw
B)bb+ pw+/bb pw × bb pw+/bb+ pw
C)bb+ pw/bb pw+ × bb pw/bb pw
D)bb+ pw/bb pw+ × bb+ pw+/bb pw
E)bb+ pw+/bb pw × bb+ pw/bb pw+

A)bb+ pw+/bb pw × bb pw/bb pw
B)bb+ pw+/bb pw × bb pw+/bb+ pw
C)bb+ pw/bb pw+ × bb pw/bb pw
D)bb+ pw/bb pw+ × bb+ pw+/bb pw
E)bb+ pw+/bb pw × bb+ pw/bb pw+
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10
Linked genes always exhibit:
A)phenotypes that are similar.
B)recombination frequencies of less than 50%.
C)homozygosity when involved in a testcross.
D)a greater number of recombinant offspring than parental offspring when involved in a testcross.
E)a lack of recombinant offspring when a heterozygous parent is testcrossed.
A)phenotypes that are similar.
B)recombination frequencies of less than 50%.
C)homozygosity when involved in a testcross.
D)a greater number of recombinant offspring than parental offspring when involved in a testcross.
E)a lack of recombinant offspring when a heterozygous parent is testcrossed.
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11
Recombination occurs through:
A)crossing over and chromosome interference.
B)chromosome interference and independent assortment.
C)somatic-cell hybridization and chromosome interference.
D)complete linkage and chromosome interference.
E)crossing over and independent assortment.
A)crossing over and chromosome interference.
B)chromosome interference and independent assortment.
C)somatic-cell hybridization and chromosome interference.
D)complete linkage and chromosome interference.
E)crossing over and independent assortment.
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12
A genetic map shows which of the following?
A)The distance in numbers of nucleotides between two genes
B)The number of genes on each of the chromosomes of a species
C)The linear order of genes on a chromosome
D)The location of chromosomes in the nucleus when they line up at metaphase during mitosis
E)The location of double crossovers that occur between two genes
A)The distance in numbers of nucleotides between two genes
B)The number of genes on each of the chromosomes of a species
C)The linear order of genes on a chromosome
D)The location of chromosomes in the nucleus when they line up at metaphase during mitosis
E)The location of double crossovers that occur between two genes
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13
A testcross includes:
A)one parent who is homozygous recessive for one gene pair and a second parent who is homozygous recessive for a second gene pair.
B)one parent who is homozygous dominant for one or more genes and a second parent who is homozygous recessive for these same genes.
C)two parents who are both heterozygous for two or more genes.
D)one parent who shows the dominant phenotype for one or more genes and a second parent who is homozygous recessive for these genes.
E)one parent who shows the recessive phenotype for one or more genes and a second parent who is homozygous dominant for these genes.
A)one parent who is homozygous recessive for one gene pair and a second parent who is homozygous recessive for a second gene pair.
B)one parent who is homozygous dominant for one or more genes and a second parent who is homozygous recessive for these same genes.
C)two parents who are both heterozygous for two or more genes.
D)one parent who shows the dominant phenotype for one or more genes and a second parent who is homozygous recessive for these genes.
E)one parent who shows the recessive phenotype for one or more genes and a second parent who is homozygous dominant for these genes.
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14
You are doing lab work with a new species of beetle.You isolated lines that breed true for either blue shells and long antenna, or green shells and short antenna.Crossing these lines yields F1 progeny with blue shells and long antenna.Crossing F1 progeny with beetles that have green shells and short antenna yield the following progeny:
Assuming that the genes are linked, what is the map distance in cM between them?
A)33.3 cM
B)25.0 cM
C)49.5 cM
D)8)0 cM
E)The genes are actually assorting independently.

A)33.3 cM
B)25.0 cM
C)49.5 cM
D)8)0 cM
E)The genes are actually assorting independently.
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15
Crossing over occurs during:
A)late anaphase.
B)prophase.
C)metaphase.
D)early anaphase
E)telophase.
A)late anaphase.
B)prophase.
C)metaphase.
D)early anaphase
E)telophase.
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16
A physical map often measures _____, whereas a genetic map measures _____.
A)distances between chromosomes; distances between genes
B)map units between genes; physical distances along the chromosome
C)centiMorgans; base pairs
D)distances in base pairs along the chromosome; centiMorgans
E)map units between genes; centiMorgans
A)distances between chromosomes; distances between genes
B)map units between genes; physical distances along the chromosome
C)centiMorgans; base pairs
D)distances in base pairs along the chromosome; centiMorgans
E)map units between genes; centiMorgans
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17
Compared to a physical map, a genetic map:
A)is more accurate.
B)is less accurate.
C)is equally accurate.
D)measures recombinant distance between genes.
E)cannot be made for humans.
A)is more accurate.
B)is less accurate.
C)is equally accurate.
D)measures recombinant distance between genes.
E)cannot be made for humans.
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18
Assume that an individual of AB/ab genotype is testcrossed and four classes of testcross progeny are found in equal frequencies.Which of the following statements is TRUE?
A)The genes A and B are on the same chromosome and closely linked.
B)The genes A and B are on the same chromosome and very far apart.
C)The genes A and B are probably between 10 and 20 map units apart on the same chromosome.
D)The genes A and B are likely located on different chromosomes.
E)Either b or d could be correct.
A)The genes A and B are on the same chromosome and closely linked.
B)The genes A and B are on the same chromosome and very far apart.
C)The genes A and B are probably between 10 and 20 map units apart on the same chromosome.
D)The genes A and B are likely located on different chromosomes.
E)Either b or d could be correct.
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19
Is it possible for two different genes located on the same chromosome to assort independently?
A)No, if two genes are on the same chromosome, they will be linked and the recombination frequency will be less than 50%.
B)Yes, if the two genes are close enough to each other, there are a limited number of crossover events between them.
C)No, there will be a very high crossover interference such that the recombination frequency will be reduced significantly.
D)Yes, if the genes are far enough apart on the same chromosome, a crossover occurs between them in just about every meiotic event.
E)Yes, but only if the two genes are both homozygous.
A)No, if two genes are on the same chromosome, they will be linked and the recombination frequency will be less than 50%.
B)Yes, if the two genes are close enough to each other, there are a limited number of crossover events between them.
C)No, there will be a very high crossover interference such that the recombination frequency will be reduced significantly.
D)Yes, if the genes are far enough apart on the same chromosome, a crossover occurs between them in just about every meiotic event.
E)Yes, but only if the two genes are both homozygous.
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20
Dr.Disney has been raising exotic fruit flies for decades.Recently, he discovered a strain of fruit flies that in the recessive condition have baby blue eyes that he designates as bb.He also has another strain of fruit flies that in the recessive condition have pink wings that are designated as pw.He is able to establish flies that are homozygous for both mutant traits.He mates these two strains with each other.Dr.Disney then takes phenotypically wild-type females from this cross and mates them with double recessive males.In the resulting testcross progeny he observes 500 flies that are of the following make-up:
What is the relationship with respect to location between the two genes?
A)They are far apart on the same chromosome and assorting independently.
B)They are linked and the map distance between them is 41.5 cM.
C)They are on different chromosomes and assorting independently.
D)They are linked and 16.6 cM apart.
E)They are linked and 50.0 cM apart.

A)They are far apart on the same chromosome and assorting independently.
B)They are linked and the map distance between them is 41.5 cM.
C)They are on different chromosomes and assorting independently.
D)They are linked and 16.6 cM apart.
E)They are linked and 50.0 cM apart.
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21
In corn, small pollen (sp)is recessive to normal pollen (sp+)and banded necrotic tissue, called zebra necrotic (zn), is recessive to normal tissue (zn+).The genes that produce these phenotypes are closely linked on chromosome 10.If no crossing over occurs between these two loci, give the types of progeny expected from the following cross: 
A)sp+ zn+/sp zn; sp zn/sp zn
B)sp+ zn/sp zn; sp zn+/sp zn
C)sp+ zn+/sp+ zn+; sp+ zn+/sp zn; sp zn/sp zn
D)sp+ zn/sp+ zn; sp+ zn/sp zn+; sp zn+/sp+ zn; sp zn+/sp zn+
E)sp+ zn+/sp zn; sp+ zn/sp zn

A)sp+ zn+/sp zn; sp zn/sp zn
B)sp+ zn/sp zn; sp zn+/sp zn
C)sp+ zn+/sp+ zn+; sp+ zn+/sp zn; sp zn/sp zn
D)sp+ zn/sp+ zn; sp+ zn/sp zn+; sp zn+/sp+ zn; sp zn+/sp zn+
E)sp+ zn+/sp zn; sp+ zn/sp zn
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22
Two genes, A and B, are located 30 map units apart.The dihybrid shown below is mated to a tester aa bb.What proportion of the offspring is expected to be dominant for both traits? 
A)0%
B)15%
C)30%
D)35%
E)70%

A)0%
B)15%
C)30%
D)35%
E)70%
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23
In flower beetles, pygmy (py)is recessive to normal size (py+), and red color (r)is recessive to brown (r+).A beetle heterozygous for these characteristics was test crossed to a beetle homozygous for pygmy and red.The following are progeny phenotypes from this testcross:
Carry out a series of chi-square tests to determine if there is equal segregation of alleles at the py locus.What is the CORRECT chi-square value and how many degree(s)of freedom should be used in its interpretation?
A)0)16 with one degree of freedom
B)0)16 with three degrees of freedom
C)0)48 with one degree of freedom
D)0)48 with two degrees of freedom
E)4)56 with one degree of freedom

A)0)16 with one degree of freedom
B)0)16 with three degrees of freedom
C)0)48 with one degree of freedom
D)0)48 with two degrees of freedom
E)4)56 with one degree of freedom
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24
Interference occurs when:
A)two genes are assorting independently.
B)two genes are far apart on a genetic map.
C)one crossover inhibits another.
D)the number recombinant progeny classes in the testcross of a heterozygote exceeds the number of parental progeny.
E)a crossover causes the termination of the meiosis event in which the crossover is occurring.
A)two genes are assorting independently.
B)two genes are far apart on a genetic map.
C)one crossover inhibits another.
D)the number recombinant progeny classes in the testcross of a heterozygote exceeds the number of parental progeny.
E)a crossover causes the termination of the meiosis event in which the crossover is occurring.
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25
The map distances for genes that are close to each other are more accurate than map distances for genes that are quite far apart because:
A)with genes that are far apart, double crossover and other multiple crossover events often lead to lethal recombinants that reduce the number of recombinant progeny.
B)with genes that are far apart, double crossover and other multiple crossover events often lead to nonrecombinant or parental offspring and thus reduce the true map distance.
C)crossover interference will cause more double crossover and other multiple crossover events to occur than would be expected and thus result in a higher number of recombinant progeny than expected to occur with genes that are far apart.
D)double crossover and other multiple crossover events occur more often when genes are close to each other and can be readily detected, so these map distances are more accurate than those for genes that are far apart.
E)when genes are far apart, single crossover recombinant classes are more difficult to detect than when genes are close together.
A)with genes that are far apart, double crossover and other multiple crossover events often lead to lethal recombinants that reduce the number of recombinant progeny.
B)with genes that are far apart, double crossover and other multiple crossover events often lead to nonrecombinant or parental offspring and thus reduce the true map distance.
C)crossover interference will cause more double crossover and other multiple crossover events to occur than would be expected and thus result in a higher number of recombinant progeny than expected to occur with genes that are far apart.
D)double crossover and other multiple crossover events occur more often when genes are close to each other and can be readily detected, so these map distances are more accurate than those for genes that are far apart.
E)when genes are far apart, single crossover recombinant classes are more difficult to detect than when genes are close together.
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26
In soreflies (a hypothetical insect), the dominant allele, L, is responsible for resistance to a common insecticide called Loritol.Another dominant allele, M, is responsible for the ability of soreflies to sing like birds.A true-breeding mute, Loritol-resistant sorefly was mated to a true-breeding singing, Loritol-sensitive sorefly, and the singing, Loritol-resistant female progeny were test-crossed with true-breeding wild-type (i.e., mute, Loritol-sensitive)males.Of the 400 total progeny produced, 117 were mute and Loritol-resistant, 114 could sing and were Loritol-sensitive, 83 could sing and were Loritol-resistant, and 86 were mute and Loritol-sensitive.What will be the results of a chi-square test for independent assortment?
A)9)70 with three degrees of freedom
B)4)63 with three degrees of freedom
C)6)48 with four degrees of freedom
D)2)54 with one degree of freedom
E)Because there are four classes of offspring, the genes must be assorting independently.
A)9)70 with three degrees of freedom
B)4)63 with three degrees of freedom
C)6)48 with four degrees of freedom
D)2)54 with one degree of freedom
E)Because there are four classes of offspring, the genes must be assorting independently.
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27
You are studying two linked genes in lizards.You have two females and you know that both are the same genotype, heterozygous for both genes (A/a and B/b).You test-cross each female to a male that is fully homozygous recessive for both genes (a/a and b/b)and get the following progeny with the following phenotypes:
How can you explain the drastic difference between these two crosses?
A)The two genes are assorting independently in female 1 and are linked in female 2.
B)The two genes are linked in female 1 and are assorting independently in female 2.
C)The two alleles are in the coupling configuration in female 1 but in the repulsion configuration in female 2.
D)The two alleles are in the repulsion configuration in female 1 but in the coupling configuration in female 2.
E)The two genes are likely located on a sex chromosome in female 1 and are likely located on an autosome in female 2.

A)The two genes are assorting independently in female 1 and are linked in female 2.
B)The two genes are linked in female 1 and are assorting independently in female 2.
C)The two alleles are in the coupling configuration in female 1 but in the repulsion configuration in female 2.
D)The two alleles are in the repulsion configuration in female 1 but in the coupling configuration in female 2.
E)The two genes are likely located on a sex chromosome in female 1 and are likely located on an autosome in female 2.
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28
A low coefficient of coincidence indicates that:
A)far fewer double crossover recombinant progeny were recovered from a testcross than would be expected from the map distances of the genes involved.
B)crossing over has been enhanced for genes that are located near the centromere of chromosomes because there is less interference of one crossover on the occurrence of a second crossover event.
C)single crossover recombinant classes in the progeny have been increased because the genes involved produce lethal phenotypes when in parental gene combinations.
D)there is a large map distance between one of the outside genes in the heterozygous parent and the middle gene, while there is a short map distance between the middle gene and the other outside gene.
E)the physical distance between two genes is very short compared with the genetic map distance between these two genes.
A)far fewer double crossover recombinant progeny were recovered from a testcross than would be expected from the map distances of the genes involved.
B)crossing over has been enhanced for genes that are located near the centromere of chromosomes because there is less interference of one crossover on the occurrence of a second crossover event.
C)single crossover recombinant classes in the progeny have been increased because the genes involved produce lethal phenotypes when in parental gene combinations.
D)there is a large map distance between one of the outside genes in the heterozygous parent and the middle gene, while there is a short map distance between the middle gene and the other outside gene.
E)the physical distance between two genes is very short compared with the genetic map distance between these two genes.
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29
Three-factor testcrosses are only informative in gene mapping when which of the following occurs?
A)One parent is homozygous recessive for the three genes, and the other parent is homozygous dominant.
B)All three genes are located on separate chromosomes, and one parent is homozygous dominant for at least two of these genes.
C)Both parents are homozygous for the three genes.
D)One parent is heterozygous for the three genes, and the other parent is homozygous recessive.
E)One of the genes must be located on a sex chromosome and be heterozygous, and the other two genes must be located on an autosome and be homozygous.
A)One parent is homozygous recessive for the three genes, and the other parent is homozygous dominant.
B)All three genes are located on separate chromosomes, and one parent is homozygous dominant for at least two of these genes.
C)Both parents are homozygous for the three genes.
D)One parent is heterozygous for the three genes, and the other parent is homozygous recessive.
E)One of the genes must be located on a sex chromosome and be heterozygous, and the other two genes must be located on an autosome and be homozygous.
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30
In soreflies (a hypothetical insect), the dominant allele, L, is responsible for resistance to a common insecticide called Loritol.Another dominant allele, M, is responsible for the ability of soreflies to sing like birds.A true-breeding mute, Loritol-resistant sorefly was mated to a true-breeding singing, Loritol-sensitive sorefly, and the singing, Loritol-resistant female progeny were test-crossed with true-breeding wild type (i.e., mute, Loritol-sensitive)males.Of the 400 total progeny produced, 117 were mute and Loritol-resistant, 114 could sing and were Loritol-sensitive, 83 could sing and were Loritol-resistant, and 86 were mute and Loritol-sensitive.A chi-square test is done to determine if there is equal segregation of alleles at the L locus.What will be the chi-square value obtained and how many degrees of freedom would be used to interpret this value?
A)0)09 and one degree of freedom
B)0)56 and two degrees of freedom
C)0 and one degree of freedom
D)9)72 and four degrees of freedom
E)A chi-square test is not the appropriate statistical test to answer this question.
A)0)09 and one degree of freedom
B)0)56 and two degrees of freedom
C)0 and one degree of freedom
D)9)72 and four degrees of freedom
E)A chi-square test is not the appropriate statistical test to answer this question.
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31
Assume that A and B are two linked genes on an autosome in Drosophila.A testcross is made where AB/ab flies are crossed to ab/ab flies and the progeny are counted and shown below.However, it is known that the Aa/bb genotype is lethal before the flies hatch and does not appear among the testcross progeny counted.What is the most precise map distance that can be calculated from these data? Aa Bb = 235
Aa bb = 225
Aa Bb = 20
A)4)2 cM
B)4)0 cM
C)16.4 cM
D)8)0 cM
E)50 cM
Aa bb = 225
Aa Bb = 20
A)4)2 cM
B)4)0 cM
C)16.4 cM
D)8)0 cM
E)50 cM
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32
You are examining the following human pedigree and want to determine if the rare dominant disease allele (D)is linked to a specific DNA sequence location, referred to as a "molecular marker." You are using two molecular markers, R1 and R2.Parental and progeny genotypes and phenotypes are indicated.Note that the father is a dihybrid at both loci, but the mother is homozygous recessive at both loci and linked to R1 (d-R1/d-R1).There is complete penetrance of the trait and a linkage phase of D-R1/d-R2 in the father.Assuming that the marker and the gene are linked, what is the best estimate of the map distance between the two loci? 
A)12 cM
B)50 cM
C)16 cM
D)5 cM
E)25 cM

A)12 cM
B)50 cM
C)16 cM
D)5 cM
E)25 cM
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33
In flower beetles, pygmy (py)is recessive to normal size (py+), and red color (r)is recessive to brown (r+).A beetle heterozygous for these characteristics was test crossed to a beetle homozygous for pygmy and red.The following are progeny phenotypes from this testcross:
Carry out a series of chi-square tests to determine if the two loci are assorting independently.What is the CORRECT chi-square value and how many degree(s)of freedom should be used in its interpretation?
A)112 with one degree of freedom
B)265 with three degrees of freedom
C)367 with four degree of freedom
D)16.5 with three degrees of freedom
E)367 with three degrees of freedom

A)112 with one degree of freedom
B)265 with three degrees of freedom
C)367 with four degree of freedom
D)16.5 with three degrees of freedom
E)367 with three degrees of freedom
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34
An individual has the following genotype.Gene loci (A)and (B)are 15 cM apart.What are the CORRECT frequencies of some of the gametes that can be made by this individual? 
A)Ab = 7.5%; AB = 42.5%
B)ab = 25%; aB = 50%
C)AB = 7.5%; aB = 42.5%
D)aB = 15%; Ab = 70%
E)aB = 70%; Ab = 15%

A)Ab = 7.5%; AB = 42.5%
B)ab = 25%; aB = 50%
C)AB = 7.5%; aB = 42.5%
D)aB = 15%; Ab = 70%
E)aB = 70%; Ab = 15%
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35
Why are the progeny of a testcross generally used to map loci? Why not the F2 progeny from an F1 × F1 cross?
A)Only recombinant offspring would be found in the progeny of an F1 × F1 cross.
B)The progeny of an F1 × F1 cross would be found in a 9:3:3:1 ratio when two genes are involved, whereas the progeny of a testcross would result in a 1:1:1:1 ratio.
C)It is easier to classify recombinant and parental offspring of a testcross than with the progeny of an F1 × F1 cross.
D)In a testcross more of the progeny would be expected to display the dominant phenotype than in the progeny of an F1 × F1 cross.
E)A testcross is more useful for mapping genes that are located near each other but when genes are quite far apart on the same chromosome, an F1 × F1 cross actually is more useful.
A)Only recombinant offspring would be found in the progeny of an F1 × F1 cross.
B)The progeny of an F1 × F1 cross would be found in a 9:3:3:1 ratio when two genes are involved, whereas the progeny of a testcross would result in a 1:1:1:1 ratio.
C)It is easier to classify recombinant and parental offspring of a testcross than with the progeny of an F1 × F1 cross.
D)In a testcross more of the progeny would be expected to display the dominant phenotype than in the progeny of an F1 × F1 cross.
E)A testcross is more useful for mapping genes that are located near each other but when genes are quite far apart on the same chromosome, an F1 × F1 cross actually is more useful.
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36
If the recombination frequency between genes (A)and (B)is 5.3%, what is the distance between the genes in map units on the linkage map?
A)53 cM
B)5)3 cM
C)0)53 cM
D)10.6 cM
E)25 cM
A)53 cM
B)5)3 cM
C)0)53 cM
D)10.6 cM
E)25 cM
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37
You just bought two black guinea pigs, one male and one female, of the same genotype from the pet store that are known to be heterozygous (Bb).You also know that black fur (BB)is dominant over white fur (bb)and that a lethal recessive allele is located only one cM away from the recessive b allele, and your animals are both heterozygous for this gene also.What is the probability of finding a white individual among the progeny if you cross these two animals?
A)0)25
B)0)002475
C)0)000025
D)0)004975
E)0)495
A)0)25
B)0)002475
C)0)000025
D)0)004975
E)0)495
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38
A series of two-point crosses among fruit flies is carried out between genes for brown eyes (bw), arc wings (a), vestigial wings (vg), ebony body color (e), and curved wings (cv).The following number of nonrecombinant and recombinant progeny were obtained from each cross:
Using these data from two-point crosses, what it the BEST genetic map (in cM)that can be developed?
A)cv 5 bw 13 a 34 vg with e assorting independently
B)bw 5 cv 24 vg 32 a with e assorting independently
C)a 5 bw 13 vg 24 e with vg assorting independently
D)cv 13 bw 5 a 27 vg with e assorting independently
E)bw 5 a 24 cv 13 vg with e assorting independently

A)cv 5 bw 13 a 34 vg with e assorting independently
B)bw 5 cv 24 vg 32 a with e assorting independently
C)a 5 bw 13 vg 24 e with vg assorting independently
D)cv 13 bw 5 a 27 vg with e assorting independently
E)bw 5 a 24 cv 13 vg with e assorting independently
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39
Two linked genes, (A)and (B), are separated by 18 cM.A man with genotype Aa Bb marries a woman who is aa bb.The man's father was AA BB.What is the probability that their first two children will both be ab/ab?
A)0)168
B)0)0081
C)0)032
D)0)062
E)0)13
A)0)168
B)0)0081
C)0)032
D)0)062
E)0)13
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40
You just bought two black guinea pigs, one male and one female, of the same genotype from the pet store that are known to be heterozygous (Bb).You also know that black fur (BB)is dominant over white fur (bb), and that a lethal recessive allele is located only one cM away from the recessive b allele and your animals are both heterozygous for this gene also.You decide to start raising your own guinea pigs, but after mating these animals several times, you discover they produce only black offspring among the first 12 progeny.How would you BEST explain this result?
A)The B locus is on the X chromosome, so it can never produce a white phenotype.
B)The B allele is actually codominant with the b allele, so a white phenotype cannot be produced.
C)The recessive lethal allele is in tight coupling linkage with the b allele, so almost all of the bb offspring will also be ll and thus die before they can be observed.
D)The dominant L allele is in tight repulsion linkage with the B allele, so it will be impossible to produce the Bb genotype that would express the white phenotype.
E)Normally, it would be expected that 25% of the offspring would be white, but in this case, random deviations resulted in no white offspring.
A)The B locus is on the X chromosome, so it can never produce a white phenotype.
B)The B allele is actually codominant with the b allele, so a white phenotype cannot be produced.
C)The recessive lethal allele is in tight coupling linkage with the b allele, so almost all of the bb offspring will also be ll and thus die before they can be observed.
D)The dominant L allele is in tight repulsion linkage with the B allele, so it will be impossible to produce the Bb genotype that would express the white phenotype.
E)Normally, it would be expected that 25% of the offspring would be white, but in this case, random deviations resulted in no white offspring.
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41
Discuss the differences, and at least one similarity, between recombination and independent assortment.
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42
In maize (corn), assume that the genes A and B are linked and 30 map units apart.If a plant of Ab/aB is selfed, what proportion of the progeny would be expected to be of ab/ab genotype?
A)2)25%
B)15%
C)9%
D)30%
E)4)5%
A)2)25%
B)15%
C)9%
D)30%
E)4)5%
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43
What is the parental genotype? 
A)a B d / A b D
B)a B D / A b d
C)A B d / a b D
D)A B D / a b d
E)None of the answers is correct.

A)a B d / A b D
B)a B D / A b d
C)A B d / a b D
D)A B D / a b d
E)None of the answers is correct.
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44
Geneticists often assume that map distances less than 7 to 8 map units or cM are quite accurate.Map distances that exceed this threshold significantly are assumed to be less accurate and the level of accuracy decline increases as map distances increase.Briefly explain this observation.
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45
Consider the following three-point (trihybrid)testcross:
Calculate the number of double crossovers you would expect to observe if 1000 progeny result from this testcross assuming a coefficient of coincidence of 0.25.
A)About 14
B)About 26
C)About 10
D)About 4
E)None, because there is crossover interference

A)About 14
B)About 26
C)About 10
D)About 4
E)None, because there is crossover interference
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46
Assume that you discover a new human gene that you believe is located on the Y chromosome although not in the region (pseudoautosomal)of the Y that is homologous with part of the X chromosome.How would you map this gene with respect to the other genes on the Y chromosome?
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47
The results of linkage analysis for DNA marker A and the p53 gene are shown below.What is the best estimate for the approximate genetic distance between marker A and the p53 gene in humans? 
A)1 cM
B)5 cM
C)10 cM
D)20 cM
E)30 cM

A)1 cM
B)5 cM
C)10 cM
D)20 cM
E)30 cM
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48
Forensic biologist use well defined Y chromosomal markers to analyze case evidence.If there were five markers on the Y chromosome, each with five different alleles, how do you predict these would be inherited across men involved in criminal investigations?
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49
Consider the following three-factor (trihybrid)testcross:
Calculate the number of expected individuals of a+a bb c+c genotype if 1000 progeny result from this testcross.
A)About 102
B)About 46
C)About 130
D)About 65
E)About 250

A)About 102
B)About 46
C)About 130
D)About 65
E)About 250
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50
In a two-point linkage analysis, genes a and b have been found to be 26 cM apart on the same chromosome.A third gene, c, has just been discovered and found to be located between a and b.A three-point linkage analysis with a, b, and c indicates that a and b are actually 33 cM apart, rather than 26 cM.Why does the three-point analysis give a different map distance for a and b than does the two-point linkage analysis, and which is more accurate?
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51
A situation where the coefficient of coincidence is greater than 1.0 would indicate that:
A)the interference is high and one crossover suppresses the occurrence of a second one.
B)no double crossovers were found in the progeny of a testcross, even though some were expected based on probability.
C)double crossovers were found in the progeny of a testcross, but there were fewer of them than would be expected based on probability.
D)there were more double crossovers in the progeny than would be expected based on probability.
E)the genes involved were actually assorting independently.
A)the interference is high and one crossover suppresses the occurrence of a second one.
B)no double crossovers were found in the progeny of a testcross, even though some were expected based on probability.
C)double crossovers were found in the progeny of a testcross, but there were fewer of them than would be expected based on probability.
D)there were more double crossovers in the progeny than would be expected based on probability.
E)the genes involved were actually assorting independently.
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52
How is chi-square test used to test genetic recombination, and what does it tell you?
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53
In addition to determining genotypes, two- and three-factor testcrosses can be used to:
A)map gene loci.
B)screen recessive mutants.
C)measure heritability.
D)determine parental origin.
E)determine the physical location of genes.
A)map gene loci.
B)screen recessive mutants.
C)measure heritability.
D)determine parental origin.
E)determine the physical location of genes.
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54
In Drosophila melanogaster, cut wings (ct)is recessive to normal wings (ct+), sable body (s)is recessive to gray body (s+), and vermilion eyes (v)is recessive to red eyes (v+).All three recessive mutations are X-linked.A female fly with cut wings, sable body, and vermilion eyes is crossed to a male with normal wings, gray body, and red eyes.The F1 females produced by this cross were mated with cut, sable, vermilion males in a testcross.The following are the progeny resulting from the testcross.
What is the CORRECT genetic map with respect to gene order and distances (in cM)for these three genes?
A)
B)
C)
D)
E)

A)

B)

C)

D)

E)

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55
What is the CORRECT gene order? 
A)A b D
B)a B D
C)A D B
D)a d b
E)B a d

A)A b D
B)a B D
C)A D B
D)a d b
E)B a d
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56
What is the map distance between the A and B genes? 
A)1 cM
B)5 cM
C)10 cM
D)12 cM
E)20 cM

A)1 cM
B)5 cM
C)10 cM
D)12 cM
E)20 cM
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57
A single nucleotide polymorphism can be associated with a disease trait based on its genetic _____ to that trait.
A)linkage
B)lod
C)uniqueness
D)recombination distance
E)phenotype
A)linkage
B)lod
C)uniqueness
D)recombination distance
E)phenotype
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58
A genome-wide association study helps identify genetic traits by:
A)independent assortment association.
B)linkage of genetic traits to haplotypes within a population.
C)analysis of single nucleotide polymorphisms.
D)likelihood of linkage between genes on different chromosomes.
E)linkage disequilibrium among other genes.
A)independent assortment association.
B)linkage of genetic traits to haplotypes within a population.
C)analysis of single nucleotide polymorphisms.
D)likelihood of linkage between genes on different chromosomes.
E)linkage disequilibrium among other genes.
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59
A testcross is performed on an individual to examine three linked genes.The most frequent phenotypes of the progeny were Abc and aBC, and the least frequent phenotypes were abc and ABC.What was the genotype of the heterozygous individual that is testcrossed with the correct order of the three genes?
A)Abc/aBC
B)BAC/bac
C)bcA/BCa
D)aBc/AbC
E)bAc/BaC
A)Abc/aBC
B)BAC/bac
C)bcA/BCa
D)aBc/AbC
E)bAc/BaC
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60
In Drosophila melanogaster, cut wings (ct)is recessive to normal wings (ct+), sable body (s)is recessive to gray body (s+), and vermilion eyes (v)is recessive to red eyes (v+).All three recessive mutations are X-linked.A female fly with cut wings, sable body, and vermilion eyes is crossed to a male with normal wings, gray body, and red eyes.The F1 females produced by this cross were mated with cut, sable, vermilion males in a testcross.The following are the progeny resulting from the testcross:
What the interference value shown by this cross?
A)0)42
B)0)25
C)0)58
D)-0.42
E)0)13

A)0)42
B)0)25
C)0)58
D)-0.42
E)0)13
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