Genetics From Genes To Genomes 6th Edition by Leland Hartwell - Test Bank

Genetics From Genes To Genomes 6th Edition by Leland Hartwell – Test Bank

 

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Sample Questions Are Posted Below

 

Genetics, 6e (Hartwell)

Chapter 5   Linkage, Recombination, and the Mapping of Genes on Chromosomes

 

In Drosophila, the genes y, f, and v are all X-linked. Females who are homozygous for recessive alleles of all three genes (y f v / y f v) are crossed to wild-type males. The resulting trihybrid F1 females are testcrossed. The F2 are distributed as follows:

 

y  f  v	3210
y  f  v+	72
y  f+  v	1024
y  f+  v+	678
y+  f  v	690
y+  f   v+	1044
y+  f+  v	60
y+  f+  v+	3222
	10,000

 

1) Which of the following linkage maps correctly shows the order and distance between the y, f, and v genes?

1. A) f——35 m.u.——y——15 m.u.——v
2. B) f——22 m.u.——y——15 m.u.——v
3. C) y——35 m.u.——f——22 m.u.——v
4. D) y——22 m.u.——v——15 m.u.——f
5. E) y——15 m.u.——v——22 m.u.——f

 

Answer:  E

Section:  05.03

Topic:  Mapping: Locating Genes Along a Chromosome

Learning Objective:  05.03.02 Refine genetic maps based on data from three-point testcrosses.

Bloom’s:  4. Analyze

Accessibility:  Keyboard Navigation

 

 

 

2) What is the coefficient of coincidence in this region?

1. A) 0
2. B) 0.2
3. C) 0.4
4. D) 0.6
5. E) 0.8

 

Answer:  C

Section:  05.03

Topic:  Mapping: Locating Genes Along a Chromosome

Learning Objective:  05.03.02 Refine genetic maps based on data from three-point testcrosses.

Bloom’s:  3. Apply

Accessibility:  Keyboard Navigation

Females heterozygous for the recessive second chromosome mutations px, sp, and cn are mated to a male homozygous for all three mutations. The offspring are as follows:

 

px  sp  cn	1461
px  sp  cn+	3497
px  sp+  cn	1
px  sp+  cn+	11
px+  sp  cn	9
px+  sp  cn+	0
px+  sp+  cn	3482
px+  sp+  cn+	1539
	10,000

 

3) What is the genotype of the females that gave rise to these progeny?

1. A) px+sp cn / px sp+cn+
2. B) px+sp cn+ / px sp+cn
3. C) px+sp+ cn+ / px sp cn
4. D) px sp cn+ / px+ sp+cn
5. E) insufficient data

 

Answer:  D

Section:  05.01; 05.03

Topic:  Gene Linkage and Recombination; Mapping: Locating Genes Along a Chromosome

Learning Objective:  05.01.02 Differentiate between parental and recombinant gametes.; 05.03.02 Refine genetic maps based on data from three-point testcrosses.

Bloom’s:  4. Analyze

Accessibility:  Keyboard Navigation

 

4) Which of the three genes is in the middle?

1. A) px
2. B) sp
3. C) cn
4. D) insufficient data

 

Answer:  A

Section:  05.03

Topic:  Mapping: Locating Genes Along a Chromosome

Learning Objective:  05.03.02 Refine genetic maps based on data from three-point testcrosses.

Bloom’s:  4. Analyze

Accessibility:  Keyboard Navigation

5) Which of the following linkage maps correctly shows the order and distance between the px, sp, and cn genes?

1. A) sp——0.21 m.u.——px——30.01 m.u.——cn
2. B) sp——30.01 m.u.——px——0.21 m.u.——cn
3. C) sp——0.2 m.u.——px——30 m.u.——cn
4. D) px——0.2 m.u.——sp——30.2 m.u.——cn
5. E) px——30.2 m.u.——sp——0.2 m.u.——cn

 

Answer:  A

Section:  05.03

Topic:  Mapping: Locating Genes Along a Chromosome

Learning Objective:  05.03.02 Refine genetic maps based on data from three-point testcrosses.

Bloom’s:  4. Analyze

Accessibility:  Keyboard Navigation

 

6) What is the coefficient of coincidence in this region?

1. A) 0
2. B) 0.16
3. C) 0.33
4. D) 0.5
5. E) 0.66

 

Answer:  B

Section:  05.03

Topic:  Mapping: Locating Genes Along a Chromosome

Learning Objective:  05.03.02 Refine genetic maps based on data from three-point testcrosses.

Bloom’s:  3. Apply

Accessibility:  Keyboard Navigation

 

 

In peas, tall (T) is dominant to short (t), red flowers (R) is dominant to white flowers (r), and wide leaves (W) is dominant to narrow leaves (w). A tall plant that has red flowers and wide leaves is crossed to a short plant that has white flowers and narrow leaves. The resulting progeny are shown in the table.

 

tall, red, wide	381
tall, white, wide	122
short, red, wide	118
short, white, wide	379
	1000

 

7) What is the genotype of the tall plant that has red flowers and wide leaves?

1. A) T R W / t r w
2. B) T R W / t r W​
3. C) T R W / T R W​
4. D) T R W / T r w​
5. E) T r W / t R W​

 

Answer:  B

Section:  05.01

Topic:  Gene Linkage and Recombination

Learning Objective:  05.01.02 Differentiate between parental and recombinant gametes.

Bloom’s:  4. Analyze

Accessibility:  Keyboard Navigation

 

8) This cross is not useful to determine if one of the genes is linked to the others. Which gene?

1. A) gene T
2. B) gene R
3. C) gene W
4. D) This cross shows that all three genes are linked.

 

Answer:  C

Section:  05.03

Topic:  Mapping: Locating Genes Along a Chromosome

Learning Objective:  05.03.01 Establish relative gene positions using two-point cross data.; 05.03.02 Refine genetic maps based on data from three-point testcrosses.

Bloom’s:  2. Understand

Accessibility:  Keyboard Navigation

 

 

9) If two or more of the genes are linked, what map distance separates them?

1. A) 4 m.u.
2. B) 12 m.u.
3. C) 24 m.u.
4. D) 50 m.u.
5. E) None of the genes are linked to each other.

 

Answer:  C

Section:  05.01; 05.03

Topic:  Gene Linkage and Recombination; Mapping: Locating Genes Along a Chromosome

Learning Objective:  05.01.04 Explain how a testcross can provide evidence for or against linkage.; 05.03.01 Establish relative gene positions using two-point cross data.

Bloom’s:  4. Analyze

Accessibility:  Keyboard Navigation

 

 

 

A dihybrid testcross is made to determine if genes C and D are linked. The results are shown in the table.

 

Parent genotypes:	Cc Dd × cc dd
	Cc Dd	222
	Cc dd	280
Progeny genotypes:	cc Dd	280
	cc dd	218
		1000

 

 

10) The chi-square value is the sum for all progeny classes of (observed-expected)2/expected. Using the chi-square test for goodness of fit, calculate the chi-square value to test the null hypothesis that genes C and D are unlinked. What is the chi-square value?

1. A) 0
2. B) 0.0576
3. C) 10.8
4. D) 14.4
5. E) cannot be determined

 

Answer:  D

Section:  05.04

Topic:  The Chi-Square Test and Linkage Analysis

Learning Objective:  05.04.02 Discuss the concept of the null hypothesis and its use in data analysis.

Bloom’s:  4. Analyze

Accessibility:  Keyboard Navigation

 

 

 

11) Using the chi-square test for goodness of fit, how many degrees of freedom are in this data set?

1. A) 1
2. B) 2
3. C) 3
4. D) 4

 

Answer:  C

Section:  05.04

Topic:  The Chi-Square Test and Linkage Analysis

Learning Objective:  05.04.03 Evaluate the significance of experimental data based on the chi-square test.

Bloom’s:  2. Understand

Accessibility:  Keyboard Navigation

 

12) Given this data, use Table 5.2 to find the most accurate range within which the p value falls.

1. A) 0.001 < p < 0.01
2. B) 0.01 < p < 0.05
3. C) 0.05 < p < 0.10
4. D) 0.10 < p < 0.50

 

Answer:  A

Section:  05.04

Topic:  The Chi-Square Test and Linkage Analysis

Learning Objective:  05.04.03 Evaluate the significance of experimental data based on the chi-square test.

Bloom’s:  2. Understand

Accessibility:  Keyboard Navigation

13) What is a reasonable conclusion based on the chi-square analysis?

1. A) There is a high probability that the deviation from the expected results is due chance.
2. B) One can say with a high degree of confidence that genes C and D are linked.
3. C) The analysis supports the null hypothesis.
4. D) Genes C and D are most likely unlinked.

 

Answer:  B

Section:  05.04

Topic:  The Chi-Square Test and Linkage Analysis

Learning Objective:  05.04.03 Evaluate the significance of experimental data based on the chi-square test.

Bloom’s:  2. Understand

Accessibility:  Keyboard Navigation

 

 

 

14) If only 100 progeny had been counted and the same proportions of progeny genotypes observed, how would the p value and the conclusion drawn about linkage change?

1. A) The p value would increase, and the likelihood of linkage decreases.
2. B) The p value would decrease, and the likelihood of linkage increases.
3. C) Neither the p value nor the likelihood of linkage would change.
4. D) The p value would decrease, and the likelihood of linkage decreases.

 

Answer:  A

Section:  05.04

Topic:  The Chi-Square Test and Linkage Analysis

Learning Objective:  05.04.03 Evaluate the significance of experimental data based on the chi-square test.

Bloom’s:  4. Analyze

Accessibility:  Keyboard Navigation

 

15) The R and S genes are linked and 10 map units apart. In the cross R s / r S × r s / r s what fraction of the progeny will be R S / r s?

1. A) 5%
2. B) 10%
3. C) 25%
4. D) 40%
5. E) 45%

 

Answer:  A

Section:  05.01; 05.02

Topic:  Gene Linkage and Recombination; Recombination – A Result of Crossing-Over During Meiosis

Learning Objective:  05.01.02 Differentiate between parental and recombinant gametes.; 05.02.03 Discuss the relationship between the recombination frequency and the map distance separating two loci on a chromosome.

Bloom’s:  4. Analyze

Accessibility:  Keyboard Navigation

16) If the map distance between genes A and B is 20 map units and the map distance between genes B and C is 35 map units, what is the map distance between genes A and C?

1. A) 15 map units
2. B) 55 map units
3. C) More information is needed to distinguish between 15 and 55 map units.
4. D) Gene C must be located on a different nonhomologous chromosome.

 

Answer:  C

Section:  05.03

Topic:  Mapping: Locating Genes Along a Chromosome

Learning Objective:  05.03.01 Establish relative gene positions using two-point cross data.

Bloom’s:  4. Analyze

Accessibility:  Keyboard Navigation

 

17) In Drosophila, singed bristles (sn) and cut wings (ct) are both caused by recessive, X-linked alleles. The wild-type alleles (sn+ and ct+) are responsible for straight bristles and intact wings, respectively. A female homozygous for sn and ct+ is crossed to a sn+ct male. The F1 flies are interbred. The F2 males are distributed as follows:

 

sn ct	13
sn ct+	36
sn+ ct	39
sn+ ct+	12

 

What is the map distance between sn and ct?

1. A) 12 m.u.
2. B) 13 m.u.
3. C) 25 m.u.
4. D) 50 m.u.
5. E) 75 m.u.

 

Answer:  C

Section:  05.01; 05.03

Topic:  Gene Linkage and Recombination; Mapping: Locating Genes Along a Chromosome

Learning Objective:  05.01.04 Explain how a testcross can provide evidence for or against linkage.; 05.03.01 Establish relative gene positions using two-point cross data.

Bloom’s:  4. Analyze

Accessibility:  Keyboard Navigation

 

 

18) In Drosophila, singed bristles (sn) and cut wings (ct) are both caused by recessive, X-linked alleles. The wild type alleles (sn+ and ct+) are responsible for straight bristles and intact wings, respectively. A female homozygous for sn and ct+ is crossed to a sn+ ct male. The F1 flies are interbred. The F2 males are distributed as follows:

 

sn ct	13
sn ct+	36
sn+ ct	39
sn+ ct+	12

 

Of these 4 genotypic classes of offspring, which arose from a parental gamete produced by the F1 females? (Select all that apply.)

1. A) sn+ ct+
2. B) sn ct
3. C) sn+ ct
4. D) sn ct+

 

Answer:  C, D

Section:  05.01

Topic:  Gene Linkage and Recombination

Learning Objective:  05.01.02 Differentiate between parental and recombinant gametes.

Bloom’s:  4. Analyze

Accessibility:  Keyboard Navigation

 

19) Suppose the L and M genes are on the same chromosome but separated by 100 map units. What fraction of the progeny from the cross  L M / l m  ×  l m / l m would be L m / l m?

1. A) 10%
2. B) 25%
3. C) 50%
4. D) 75%
5. E) 100%

 

Answer:  B

Section:  05.02

Topic:  Recombination – A Result of Crossing-Over During Meiosis

Learning Objective:  05.02.04 Explain why the value of the recombination frequency between any two genes is limited to 50%.

Bloom’s:  4. Analyze

Accessibility:  Keyboard Navigation

 

 

20) The pairwise map distances for four linked genes are as follows: A-B = 22 m.u., B-C = 7 m.u., C-D = 9 m.u., B-D = 2 m.u., A-D = 20 m.u., A-C = 29 m.u. What is the order of these four genes?

1. A) ABCD
2. B) ADBC
3. C) ABDC
4. D) BADC
5. E) CADB

 

Answer:  B

Section:  05.03

Topic:  Mapping: Locating Genes Along a Chromosome

Learning Objective:  05.03.01 Establish relative gene positions using two-point cross data.

Bloom’s:  4. Analyze

Accessibility:  Keyboard Navigation

 

21) The zipper-like connection between paired homologs in early prophase is known as a

1. A) spindle fiber.
2. B) synaptic junction.
3. C) synaptonemal complex.
4. D) chiasma.
5. E) None of the choices is correct.

 

Answer:  C

Section:  05.02

Topic:  Recombination – A Result of Crossing-Over During Meiosis

Learning Objective:  05.02.02 Describe the role of chiasmata in chromosome segregation during meiosis.

Bloom’s:  1. Remember

Accessibility:  Keyboard Navigation

 

22) The measured distance between genes D and E in a two-point testcross is 50 map units. Where are genes D and E in relation to each other? (Select all that apply.)

1. A) D and E are on different homologous chromosomes.
2. B) D and E are on the same chromosome, at least 50 map units apart.
3. C) D and E are on the same chromosome, exactly 50 map units apart.
4. D) D and E are on the same chromosome, less than 50 map units apart.

 

Answer:  A, B

Section:  05.02

Topic:  Recombination – A Result of Crossing-Over During Meiosis

Learning Objective:  05.02.03 Discuss the relationship between the recombination frequency and the map distance separating two loci on a chromosome.; 05.02.04 Explain why the value of the recombination frequency between any two genes is limited to 50%.

Bloom’s:  3. Apply

Accessibility:  Keyboard Navigation

23) The R and S genes are linked and 10 map units apart. In the cross R s / r S × r s / r s what percentage of the progeny will be R s / r s?

1. A) 5%
2. B) 10%
3. C) 25%
4. D) 40%
5. E) 45%

 

Answer:  E

Section:  05.01; 05.02

Topic:  Gene Linkage and Recombination; Recombination – A Result of Crossing-Over During Meiosis

Learning Objective:  05.01.02 Differentiate between parental and recombinant gametes.; 05.02.03 Discuss the relationship between the recombination frequency and the map distance separating two loci on a chromosome.

Bloom’s:  4. Analyze

Accessibility:  Keyboard Navigation

 

24) If the recombination frequency between two genes is close to 50%, what could be true about the location of the two genes? (Select all that apply.)

1. A) They are on nonhomologous chromosomes.
2. B) They are far apart on the same chromosome.
3. C) They are very close together on the same chromosome.
4. D) None of the choices could be true.

 

Answer:  A, B

Section:  05.02

Topic:  Recombination – A Result of Crossing-Over During Meiosis

Learning Objective:  05.02.04 Explain why the value of the recombination frequency between any two genes is limited to 50%.

Bloom’s:  2. Understand

Accessibility:  Keyboard Navigation

 

 

 

25) A dihybrid testcross is made between genes H and I. Four categories of offspring are produced: H I, H i, h I, and h i. How many degrees of freedom would there be in a chi-square test for goodness of fit of the null hypothesis that the H and I genes are unlinked?

1. A) 1
2. B) 2
3. C) 3
4. D) 4
5. E) 0

 

Answer:  C

Section:  05.04

Topic:  The Chi-Square Test and Linkage Analysis

Learning Objective:  05.04.03 Evaluate the significance of experimental data based on the chi-square test.

Bloom’s:  2. Understand

Accessibility:  Keyboard Navigation

26) Which process(es) can generate recombinant gametes? (Select all that apply.)

1. A) crossing-over between two linked heterozygous loci
2. B) independent assortment of two unlinked heterozygous loci
3. C) segregation of alleles in a homozygote
4. D) crossing-over between two linked homozygous loci

 

Answer:  A, B

Section:  05.01

Topic:  Gene Linkage and Recombination

Learning Objective:  05.01.02 Differentiate between parental and recombinant gametes.

Bloom’s:  2. Understand

Accessibility:  Keyboard Navigation

 

27) Crossing-over takes place in bivalents (tetrads) consisting of ________ chromatids, and one crossover involves ________ chromatids.

1. A) 2; 2
2. B) 2; 4
3. C) 4; 2
4. D) 4; 4
5. E) 8; 4

 

Answer:  C

Section:  05.02

Topic:  Recombination – A Result of Crossing-Over During Meiosis

Learning Objective:  05.02.01 Explain the physical process by which recombination takes place.

Bloom’s:  1. Remember

Accessibility:  Keyboard Navigation

 

 

 

28) In Drosophila, the genes y (yellow body) and car (carnation eyes) are located at opposite ends of the X chromosome. In doubly heterozygous females (y+ car+ / y car), a single chiasma is observed somewhere along the X chromosome in 90% of the examined oocytes. No X chromosomes with multiple chiasmata are observed. What percentage of the male progeny from such a female would be recombinant for y and car?

1. A) 5%
2. B) 10%
3. C) 45%
4. D) 55%
5. E) 90%

 

Answer:  C

Section:  05.02; 05.03

Topic:  Recombination – A Result of Crossing-Over During Meiosis; Mapping: Locating Genes Along a Chromosome

Learning Objective:  05.02.04 Explain why the value of the recombination frequency between any two genes is limited to 50%.; 05.03.01 Establish relative gene positions using two-point cross data.

Bloom’s:  4. Analyze

Accessibility:  Keyboard Navigation

29) The Q gene locus is 10 map units from the R gene locus which is 40 map units from the S gene locus:

 

Q——10 m.u.——R——40 m.u.——S

 

Which interval would likely show the higher ratio of double to single chiasmata?

1. A) Q-R
2. B) R-S
3. C) The ratios would be the same in the two intervals.
4. D) Two chiasmata never occur in the same interval.

 

Answer:  B

Section:  05.03

Topic:  Mapping: Locating Genes Along a Chromosome

Learning Objective:  05.03.02 Refine genetic maps based on data from three-point testcrosses.

Bloom’s:  2. Understand

Accessibility:  Keyboard Navigation

 

 

 

30) The map of a chromosome interval is:

 

A——10 m.u.——B——40 m.u.——C

 

From the cross A b c / a B C × a b c / a b c, how many double crossovers would be expected out of 1000 progeny?

1. A) 5
2. B) 10
3. C) 20
4. D) 40
5. E) 80

 

Answer:  D

Section:  05.03

Topic:  Mapping: Locating Genes Along a Chromosome

Learning Objective:  05.03.02 Refine genetic maps based on data from three-point testcrosses.

Bloom’s:  4. Analyze

Accessibility:  Keyboard Navigation

31) The cross L p q / l P Q × l p q / l p q is carried out. If the L gene is in the middle, between genes P and Q, what would be the genotypes of the double crossover gametes in this cross?

1. A) L P Q and l p q
2. B) L p Q and l P q
3. C) l p Q and L P q
4. D) L p q and l P Q
5. E) cannot be determined

 

Answer:  A

Section:  05.03

Topic:  Mapping: Locating Genes Along a Chromosome

Learning Objective:  05.03.02 Refine genetic maps based on data from three-point testcrosses.

Bloom’s:  4. Analyze

Accessibility:  Keyboard Navigation

 

 

 

32) Suppose a three-point testcross was conducted involving genes X, Y, and Z. If the most abundant classes of progeny are X Y z and x y Z and the rarest classes are x Y Z and X y z, which gene is in the middle?

1. A) X
2. B) Y
3. C) Z
4. D) cannot be determined

 

Answer:  B

Section:  05.03

Topic:  Mapping: Locating Genes Along a Chromosome

Learning Objective:  05.03.02 Refine genetic maps based on data from three-point testcrosses.

Bloom’s:  4. Analyze

Accessibility:  Keyboard Navigation

 

33) In Drosophila, the genes b, c, and sp are linked and arranged as shown below:

 

b——30 m.u.——c——20 m.u.——sp

 

This region exhibits 90% interference. How many double crossovers would be recovered in a three-point cross involving b, c, and sp out of 1000 progeny?

1. A) 3
2. B) 6
3. C) 54
4. D) 60
5. E) 600

 

Answer:  B

Section:  05.03

Topic:  Mapping: Locating Genes Along a Chromosome

Learning Objective:  05.03.02 Refine genetic maps based on data from three-point testcrosses.

Bloom’s:  4. Analyze

Accessibility:  Keyboard Navigation

 

 

34) In tetrad analysis, which result would indicate that two genes are linked?

1. A) NPD = T.
2. B) PD = T.
3. C) PD = NPD.
4. D) PD > NPD.
5. E) PD > T.

 

Answer:  D

Section:  05.05

Topic:  Tetrad Analysis in Fungi

Learning Objective:  05.05.03 Describe how the relative numbers of PDs and NPDs can be used to establish linkage.

Bloom’s:  2. Understand

Accessibility:  Keyboard Navigation

 

35) In tetrad analysis, NPD asci can result from

(Select all that apply.)

1. A) independent assortment of unlinked genes.
2. B) double crossovers between linked genes.
3. C) single crossovers between linked genes.
4. D) single crossovers between a gene and a centromere.

 

Answer:  A, B

Section:  05.05

Topic:  Tetrad Analysis in Fungi

Learning Objective:  05.05.02 Differentiate between parental ditype (PD), nonparental ditype (NPD), and tetratype (T).

Bloom’s:  2. Understand

Accessibility:  Keyboard Navigation

 

36) When analyzing octads (ordered tetrads), second-division (MII) segregations result from

1. A) single crossovers between linked genes.
2. B) double crossovers between linked genes.
3. C) single crossovers between a gene and a centromere.
4. D) independent assortment of unlinked genes.
5. E) nondisjunction of homologs.

 

Answer:  C

Section:  05.05

Topic:  Tetrad Analysis in Fungi

Learning Objective:  05.05.01 Explain the meaning of the term tetrad as applied to the asci produced by certain fungi.

Bloom’s:  2. Understand

Accessibility:  Keyboard Navigation

 

 

37) Tetrad analysis shows that crossing-over occurs at the four-strand stage (i.e., after replication) because, when two genes are linked,

1. A) NPD > T.
2. B) T > NPD.
3. C) T > PD.
4. D) PD > NPD.
5. E) PD > T.

 

Answer:  B

Section:  05.05

Topic:  Tetrad Analysis in Fungi

Learning Objective:  05.05.02 Differentiate between parental ditype (PD), nonparental ditype (NPD), and tetratype (T).

Bloom’s:  3. Apply

Accessibility:  Keyboard Navigation

 

38) Sturtevant’s detailed mapping studies of the X chromosome of Drosophila supported what genetic principle?

1. A) That genes are arranged in a linear order on the chromosomes.
2. B) That genes are carried on chromosomes.
3. C) That sex determination is controlled by the X and Y chromosomes.
4. D) That segregation of an allelic gene pair is accompanied by disjunction of homologous chromosomes.
5. E) That different pairs of chromosomes assort independently.

 

Answer:  A

Section:  05.02

Topic:  Recombination – A Result of Crossing-Over During Meiosis

Learning Objective:  05.02.03 Discuss the relationship between the recombination frequency and the map distance separating two loci on a chromosome.

Bloom’s:  2. Understand

Accessibility:  Keyboard Navigation

 

 

39) Suppose an individual is heterozygous for alternate alleles of gene A (Aa). Under what conditions would a crossover in a somatic cell of this individual lead to a clone of cells that are homozygous for a? (Pick the most precise answer.)

1. A) The crossover would have to occur between the A locus and the centromere and involve two homologous (nonsister) chromatids.
2. B) The crossover would have to occur between the A locus and the end of the chromosome and involve two homologous (nonsister) chromatids.
3. C) The crossover would have to occur between the A locus and the end of the chromosome and involve two nonhomologous chromosomes.
4. D) The crossover would have to occur between the A locus and the centromere and involve two sister chromatids.

 

Answer:  A

Section:  05.06

Topic:  Mitotic Recombination and Genetic Mosaics

Learning Objective:  05.06.01 Explain how mitotic recombination leads to the mosaic condition termed twin spots.

Bloom’s:  2. Understand

Accessibility:  Keyboard Navigation

 

40) If an individual is heterozygous at two loci (A b / a B) that are located on the same chromosome arm with A closer to the centromere than B, under what conditions would a crossover in a somatic cell generate a twin spot?

1. A) The crossover would have to occur between the A locus and the centromere and involve two homologous, nonsister chromatids.
2. B) The crossover would have to occur between the A locus and the B locus and involve two homologous, nonsister chromatids.
3. C) The crossover would have to occur between the B locus and the end of the chromosome and involve two homologous, nonsister chromatids.
4. D) A double crossover would have to occur, with one crossover between the A locus and the centromere and a second crossover between the A and B loci, and both crossovers would have to involve two homologous, nonsister chromatids.
5. E) No crossover in a somatic cell could generate a twin spot.

 

Answer:  A

Section:  05.06

Topic:  Mitotic Recombination and Genetic Mosaics

Learning Objective:  05.06.01 Explain how mitotic recombination leads to the mosaic condition termed twin spots.

Bloom’s:  2. Understand

Accessibility:  Keyboard Navigation

 

 

41) Individuals heterozygous for the RB+ and RB− alleles can develop tumors as a result of (Select all that apply.)

1. A) a mitotic crossover that leads to homozygosity for RB+in some cells and RB−in other cells.
2. B) a somatic mutation in the RB+allele that leads to homozygosity for RB−.
3. C) a somatic mutation in the RB−allele that leads to homozygosity for RB+.
4. D) the fact that RB−is dominant to RB+.

 

Answer:  A, B

Section:  05.06

Topic:  Mitotic Recombination and Genetic Mosaics

Learning Objective:  05.06.01 Explain how mitotic recombination leads to the mosaic condition termed twin spots.

Bloom’s:  2. Understand

Accessibility:  Keyboard Navigation

 

42) Which type of tetrad contains two recombinant and two parental spores?

1. A) PD
2. B) NPD
3. C) T
4. D) ordered tetrads
5. E) None of these types contain two recombinant and two parental spores.

 

Answer:  C

Section:  05.05

Topic:  Tetrad Analysis in Fungi

Learning Objective:  05.05.02 Differentiate between parental ditype (PD), nonparental ditype (NPD), and tetratype (T).

Bloom’s:  1. Remember

Accessibility:  Keyboard Navigation

 

 

 

In Drosophila, singed bristles (sn) and carnation eyes (car) are both caused by recessive X-linked alleles. The wild-type alleles (sn+ and car+) are responsible for straight bristles and red eyes, respectively. A sn car female is mated to a sn+ car+ male and the F1 progeny are interbred. The F2 are distributed as follows:

 

sn car	55
sn car+	45
sn+ car	45
sn+ car+	55
	200

 

 

43) If you want to analyze this data for evidence of linkage between sn and car, what is the null hypothesis?

1. A) Genes sn and car are linked.
2. B) Genes sn and car are not linked.
3. C) Genes sn and car are located close together on the same chromosome.
4. D) Crossing-over sometimes occurs between sn and car.

 

Answer:  B

Section:  05.04

Topic:  The Chi-Square Test and Linkage Analysis

Learning Objective:  05.04.02 Discuss the concept of the null hypothesis and its use in data analysis.

Bloom’s:  2. Understand

Accessibility:  Keyboard Navigation

 

 

 

44) What is the χ2 value for a chi-square test for goodness of fit of the null hypothesis?

1. A) 0.5
2. B) 1.0
3. C) 2.0
4. D) 0.4
5. E) 20

 

Answer:  C

Section:  05.04

Topic:  The Chi-Square Test and Linkage Analysis

Learning Objective:  05.04.03 Evaluate the significance of experimental data based on the chi-square test.

Bloom’s:  3. Apply

Accessibility:  Keyboard Navigation

45) What is the p value from this test? (Pick the most accurate choice.)

1. A) p > 0.5
2. B) 0.1 < p < 0.5
3. C) p < 0.1
4. D) p < 0.05
5. E) p < 0.01

 

Answer:  A

Section:  05.04

Topic:  The Chi-Square Test and Linkage Analysis

Learning Objective:  05.04.03 Evaluate the significance of experimental data based on the chi-square test.

Bloom’s:  3. Apply

Accessibility:  Keyboard Navigation

 

46) What does the data analysis allow you to conclude about linkage between sn and car?

(Select all that apply.)

1. A) There is a high probability that the deviations from the expected number of F2in each genotype class are due to chance.
2. B) The data do not allow rejection of the null hypothesis.
3. C) The p value is high meaning that the data is significant.
4. D) There is good evidence that cn and car are linked.

 

Answer:  A, B

Section:  05.04

Topic:  The Chi-Square Test and Linkage Analysis

Learning Objective:  05.04.03 Evaluate the significance of experimental data based on the chi-square test.

Bloom’s:  3. Apply

Accessibility:  Keyboard Navigation

 

In humans, the genes for red-green color blindness (R = normal, r = color blind) and hemophilia A (H = normal, h = hemophilia) are both X-linked and only 3 map units apart.

 

47) Suppose a woman has four sons, and two are color blind but have normal blood clotting and two have hemophilia but normal color vision. What is the probable genotype of the woman?

1. A) H R / h r
2. B) H r / h r
3. C) h r / h R
4. D) H r / h R
5. E) H R / H r

 

Answer:  D

Section:  05.01; 05.02

Topic:  Gene Linkage and Recombination; Recombination – A Result of Crossing-Over During Meiosis

Learning Objective:  05.01.02 Differentiate between parental and recombinant gametes.; 05.02.03 Discuss the relationship between the recombination frequency and the map distance separating two loci on a chromosome.

Bloom’s:  3. Apply

Accessibility:  Keyboard Navigation

48) A woman whose mother is color blind and whose father has hemophilia A is pregnant with a boy. If the alleles for color blindness and hemophilia A are rare in the population, what is the probability that the baby will have normal vision and normal blood clotting?

1. A) 0
2. B) 0.03
3. C) 0.485
4. D) 0.47
5. E) 0.015

 

Answer:  E

Section:  05.01; 05.03

Topic:  Gene Linkage and Recombination; Mapping: Locating Genes Along a Chromosome

Learning Objective:  05.01.02 Differentiate between parental and recombinant gametes.; 05.03.01 Establish relative gene positions using two-point cross data.

Bloom’s:  3. Apply

Accessibility:  Keyboard Navigation

 

 

 

In Drosophila, the autosomal recessive pr and cn mutations cause brown and bright-red eyes, respectively (wild-type flies have brick-red eyes). Flies who are homozygous recessive at both pr and cn have orange eyes. A female who has wild-type eyes is crossed to an orange-eyed male. Their progeny have the following distribution of eye colors:

 

wild-type	8
brown	241
bright-red	239
orange	12
	500

 

49) Which phenotypes are parental?

1. A) wild-type and orange
2. B) brown and bright-red
3. C) wild-type and brown
4. D) bright-red and orange
5. E) There is no way to determine this.

 

Answer:  B

Section:  05.01

Topic:  Gene Linkage and Recombination

Learning Objective:  05.01.02 Differentiate between parental and recombinant gametes.; 05.01.04 Explain how a testcross can provide evidence for or against linkage.

Bloom’s:  4. Analyze

Accessibility:  Keyboard Navigation

50) What is the genotype of the wild-type mother of these progeny?

1. A) pr cn / pr+cn+
2. B) pr+cn / pr+cn
3. C) pr+cn / pr cn+
4. D) pr cn+ / pr cn+
5. E) pr cn / pr cn

 

Answer:  C

Section:  05.01

Topic:  Gene Linkage and Recombination

Learning Objective:  05.01.02 Differentiate between parental and recombinant gametes.; 05.01.04 Explain how a testcross can provide evidence for or against linkage.

Bloom’s:  4. Analyze

Accessibility:  Keyboard Navigation

 

 

 

51) The mother of these progeny resulted from a cross between two flies from true-breeding lines. What are the genotypes of these two lines?

1. A) pr cn+/ pr cn+and pr+ cn / pr+ cn
2. B) pr+cn+/ pr+ cn+ and pr cn / pr cn
3. C) pr+cn+/ pr cn and pr cn / pr cn
4. D) pr+cn / pr cn and pr cn+/ pr cn
5. E) More than one of these could be true.

 

Answer:  A

Section:  05.01

Topic:  Gene Linkage and Recombination

Learning Objective:  05.01.02 Differentiate between parental and recombinant gametes.; 05.01.04 Explain how a testcross can provide evidence for or against linkage.

Bloom’s:  4. Analyze

Accessibility:  Keyboard Navigation

 

52) What is the map distance between the pr and cn genes?

1. A) 20 m.u.
2. B) 2 m.u.
3. C) 4 m.u.
4. D) 46 m.u.
5. E) 8 m.u.

 

Answer:  C

Section:  05.01; 05.03

Topic:  Gene Linkage and Recombination; Mapping: Locating Genes Along a Chromosome

Learning Objective:  05.01.04 Explain how a testcross can provide evidence for or against linkage.; 05.03.01 Establish relative gene positions using two-point cross data.

Bloom’s:  4. Analyze

Accessibility:  Keyboard Navigation

 

 

Consider a pair of homologous chromosomes heterozygous for three genes (e.g. A B C / a b c) during prophase I of meiosis. Let the sister chromatids of one homolog be numbered 1 and 2; and the sister chromatids of the other homolog be numbered 3 and 4.

 

53) A crossover that would result in genetic recombination (e.g., A b c or a B C) could involve which pair of chromatids? (Select all that apply.)

1. A) 1 & 2
2. B) 3 & 4
3. C) 1 & 3
4. D) 2 & 4
5. E) 1 & 4
6. F) 2 & 3

 

Answer:  C, D, E, F

Section:  05.02

Topic:  Recombination – A Result of Crossing-Over During Meiosis

Learning Objective:  05.02.01 Explain the physical process by which recombination takes place.

Bloom’s:  2. Understand

Accessibility:  Keyboard Navigation

 

54) Assume a double crossover occurs in this pair of chromosomes that results in chromatids of the genotypes A b C and a B c. If the first crossover (the one between A and B) involves chromatids 1 (A B C) & 4 (a b c), which chromatids could be involved in the second crossover?

1. A) 1 & 2
2. B) 3 & 4
3. C) 1 & 3
4. D) 2 & 4
5. E) 1 & 4
6. F) 2 & 3

 

Answer:  E

Section:  05.02

Topic:  Recombination – A Result of Crossing-Over During Meiosis

Learning Objective:  05.02.01 Explain the physical process by which recombination takes place.

Bloom’s:  2. Understand

Accessibility:  Keyboard Navigation

 

 

A haploid yeast strain has mating type a and genotype his4 TRP1. A different haploid yeast strain has mating type α and genotype HIS4 trp1. The two strains mate and produce diploid offspring with genotype his4 TRP1 / HIS4 trp1. The diploid cells undergo meiosis to produce a variety of tetrad types. In the following questions, the genotypes of the spores in a tetrad are shown. For each tetrad, tell what type it is: parental (PD), non-parental (NPD), or tetratype (T).

 

55) his4 TRP1; his4 trp1; HIS4 trp1; HIS4 TRP1

1. A) PD
2. B) NPD
3. C) T
4. D) cannot be determined

 

Answer:  C

Section:  05.05

Topic:  Tetrad Analysis in Fungi

Learning Objective:  05.05.02 Differentiate between parental ditype (PD), nonparental ditype (NPD), and tetratype (T).

Bloom’s:  4. Analyze

Accessibility:  Keyboard Navigation

 

56) his4 TRP1; his4 TRP1; HIS4 trp1; HIS4 trp1

1. A) PD
2. B) NPD
3. C) T
4. D) cannot be determined

 

Answer:  A

Section:  05.05

Topic:  Tetrad Analysis in Fungi

Learning Objective:  05.05.02 Differentiate between parental ditype (PD), nonparental ditype (NPD), and tetratype (T).

Bloom’s:  4. Analyze

Accessibility:  Keyboard Navigation

 

57) his4 trp1; his4 trp1; HIS4 TRP1; HIS4 TRP​

1. A) PD
2. B) NPD
3. C) T
4. D) cannot be determined

 

Answer:  B

Section:  05.05

Topic:  Tetrad Analysis in Fungi

Learning Objective:  05.05.02 Differentiate between parental ditype (PD), nonparental ditype (NPD), and tetratype (T).

Bloom’s:  4. Analyze

Accessibility:  Keyboard Navigation

58) HIS4 trp1; HIS4 TRP1; his4 TRP1; his4 trp1​

1. A) PD
2. B) NPD
3. C) T
4. D) cannot be determined

 

Answer:  C

Section:  05.05

Topic:  Tetrad Analysis in Fungi

Learning Objective:  05.05.02 Differentiate between parental ditype (PD), nonparental ditype (NPD), and tetratype (T).

Bloom’s:  4. Analyze

Accessibility:  Keyboard Navigation

 

59) HIS4 trp1; HIS4 trp1; his4 TRP1; his4 TRP1

1. A) PD
2. B) NPD
3. C) T
4. D) cannot be determined

 

Answer:  A

Section:  05.05

Topic:  Tetrad Analysis in Fungi

Learning Objective:  05.05.02 Differentiate between parental ditype (PD), nonparental ditype (NPD), and tetratype (T).

Bloom’s:  4. Analyze

Accessibility:  Keyboard Navigation

 

In Neurospora, the white-spore allele (ws) results in white spores rather than the wild-type black spores. Diploid ws / ws+ cells undergo meiosis followed by one round of mitosis to form ordered tetrads. Indicate if the pattern of spores seen in the tetrads are consistent with first-division or second-division segregation.

 

60) ws; ws; ws; ws; ws+; ws+; ws+; ws+

1. A) first-division segregation pattern
2. B) second-division segregation pattern

 

Answer:  A

Section:  05.05

Topic:  Tetrad Analysis in Fungi

Learning Objective:  05.05.04 Explain how ordered and unordered tetrad analysis can map the positions of genes and (for ordered tetrads) centromeres.

Bloom’s:  1. Remember

Accessibility:  Keyboard Navigation

 

 

61) ws; ws; ws+; ws+; ws; ws; ws+; ws+

1. A) first-division segregation pattern
2. B) second-division segregation pattern

 

Answer:  B

Section:  05.05

Topic:  Tetrad Analysis in Fungi

Learning Objective:  05.05.04 Explain how ordered and unordered tetrad analysis can map the positions of genes and (for ordered tetrads) centromeres.

Bloom’s:  1. Remember

Accessibility:  Keyboard Navigation

 

62) ws; ws; ws+; ws+; ws+; ws+; ws; ws

1. A) first-division segregation pattern
2. B) second-division segregation pattern

 

Answer:  B

Section:  05.05

Topic:  Tetrad Analysis in Fungi

Learning Objective:  05.05.04 Explain how ordered and unordered tetrad analysis can map the positions of genes and (for ordered tetrads) centromeres.

Bloom’s:  1. Remember

Accessibility:  Keyboard Navigation

 

63) ws+; ws+; ws+; ws+; ws; ws; ws; ws

1. A) first-division segregation pattern
2. B) second-division segregation pattern

 

Answer:  A

Section:  05.05

Topic:  Tetrad Analysis in Fungi

Learning Objective:  05.05.04 Explain how ordered and unordered tetrad analysis can map the positions of genes and (for ordered tetrads) centromeres.

Bloom’s:  1. Remember

Accessibility:  Keyboard Navigation

 

 

 

64) ws+; ws+; ws; ws; ws+; ws+; ws; ws

1. A) first-division segregation pattern
2. B) second-division segregation pattern

 

Answer:  B

Section:  05.05

Topic:  Tetrad Analysis in Fungi

Learning Objective:  05.05.04 Explain how ordered and unordered tetrad analysis can map the positions of genes and (for ordered tetrads) centromeres.

Bloom’s:  1. Remember

Accessibility:  Keyboard Navigation

65) Two genes are considered linked when there are more F2 progeny with recombinant genotypes than parental genotypes in the offspring of a dihybrid testcross.

 

Answer:  FALSE

Section:  05.01

Topic:  Gene Linkage and Recombination

Learning Objective:  05.01.04 Explain how a testcross can provide evidence for or against linkage.

Bloom’s:  2. Understand

Accessibility:  Keyboard Navigation

 

66) Chiasmata are structures that show where recombination occurred between sister chromatids.

 

Answer:  FALSE

Section:  05.02

Topic:  Recombination – A Result of Crossing-Over During Meiosis

Learning Objective:  05.02.02 Describe the role of chiasmata in chromosome segregation during meiosis.

Bloom’s:  1. Remember

Accessibility:  Keyboard Navigation

 

67) Chiasmata can be seen through a light microscope and are sites of recombination.

 

Answer:  TRUE

Section:  05.02

Topic:  Recombination – A Result of Crossing-Over During Meiosis

Learning Objective:  05.02.02 Describe the role of chiasmata in chromosome segregation during meiosis.

Bloom’s:  1. Remember

Accessibility:  Keyboard Navigation

 

 

 

68) A linkage group includes all of the genes on a chromosome, including genes that are so far apart from each other on a chromosome that they assort independently during meiosis.

 

Answer:  TRUE

Section:  05.03

Topic:  Mapping: Locating Genes Along a Chromosome

Learning Objective:  05.03.04 Describe the relationship between linkage groups and chromosomes.

Bloom’s:  2. Understand

Accessibility:  Keyboard Navigation

 

69) The hypothesis that predicts no linkage between genes is known as the null hypothesis.

 

Answer:  TRUE

Section:  05.04

Topic:  The Chi-Square Test and Linkage Analysis

Learning Objective:  05.04.02 Discuss the concept of the null hypothesis and its use in data analysis.

Bloom’s:  2. Understand

Accessibility:  Keyboard Navigation

70) If the p value corresponding to a given χ2 value and number of degrees of freedom is lower than 0.05, then the null hypothesis is rejected and it can be said with some confidence that the two genes being evaluated are linked.

 

Answer:  TRUE

Section:  05.04

Topic:  The Chi-Square Test and Linkage Analysis

Learning Objective:  05.04.02 Discuss the concept of the null hypothesis and its use in data analysis.

Bloom’s:  2. Understand

Accessibility:  Keyboard Navigation

 

71) Genes that are not syntenic are not linked.

 

Answer:  TRUE

Section:  05.03

Topic:  Mapping: Locating Genes Along a Chromosome

Learning Objective:  05.03.04 Describe the relationship between linkage groups and chromosomes.

Bloom’s:  2. Understand

Accessibility:  Keyboard Navigation

 

 

 

72) The genetic distance from one end of a linkage group and the other may exceed 50 m.u. because the distances between many gene pairs are added together to make the map.

 

Answer:  TRUE

Section:  05.02; 05.03

Topic:  Recombination – A Result of Crossing-Over During Meiosis; Mapping: Locating Genes Along a Chromosome

Learning Objective:  05.02.04 Explain why the value of the recombination frequency between any two genes is limited to 50%.; 05.03.04 Describe the relationship between linkage groups and chromosomes.

Bloom’s:  2. Understand

Accessibility:  Keyboard Navigation

 

73) Sectors with a different phenotype in an otherwise uniform yeast colony may be evidence of mitotic recombination.

 

Answer:  TRUE

Section:  05.06

Topic:  Mitotic Recombination and Genetic Mosaics

Learning Objective:  05.06.02 Describe sectored colonies in yeast and their significance in evaluating mitotic recombination.

Bloom’s:  2. Understand

Accessibility:  Keyboard Navigation

 

74) Large sectors suggest a mitotic recombination event late in the growth of a yeast colony.

 

Answer:  FALSE

Section:  05.06

Topic:  Mitotic Recombination and Genetic Mosaics

Learning Objective:  05.06.02 Describe sectored colonies in yeast and their significance in evaluating mitotic recombination.

Bloom’s:  2. Understand

Accessibility:  Keyboard Navigation

 

 

75) A female mouse from a true-breeding wild-type strain was crossed to a male mouse with apricot eyes (ap) and grey body (gy). The F1 mice were wild-type for both traits. When the F1 were interbred, the F2 were distributed as follows:

 

Females

all wild type

200

 

Males

wild type	91
apricot	11
grey	9
apricot, grey	89

 

Which of the following statements is correct?

1. A) ap and gy are unlinked
2. B) ap and gy are linked on an autosome and 10 map units apart
3. C) ap and gy are linked on an autosome and 20 map units apart
4. D) ap and gy are X-linked and 10 map units apart
5. E) ap and gy are X-linked and 20 map units apart

 

Answer:  D

Section:  05.01; 05.03

Topic:  Gene Linkage and Recombination; Mapping: Locating Genes Along a Chromosome

Learning Objective:  05.01.03 Conclude from ratios of progeny in a dihybrid cross whether two genes are linked.; 05.03.01 Establish relative gene positions using two-point cross data.

Bloom’s:  4. Analyze

Accessibility:  Keyboard Navigation

 

76) Suppose the map for a particular human chromosome interval is:

 

a——1 m.u.——b——1 m.u.——c——1 m.u.——d——1 m.u.——e——1 m.u.——f

 

In a man heterozygous for all six genes, what fraction of his sperm would be recombinant in the a-f interval?

1. A) 0%
2. B) 1%
3. C) 2.5%
4. D) 5%
5. E) cannot be determined

 

Answer:  D

Section:  05.01

Topic:  Gene Linkage and Recombination

Learning Objective:  05.01.02 Differentiate between parental and recombinant gametes.

Bloom’s:  3. Apply

Accessibility:  Keyboard Navigation

77) A coefficient of coincidence of 0.5 in a region of three genes means that half as many double crossovers were observed as would have been expected if crossovers in the two intervals were independent.

 

Answer:  TRUE

Section:  05.03

Topic:  Mapping: Locating Genes Along a Chromosome

Learning Objective:  05.03.02 Refine genetic maps based on data from three-point testcrosses.

Bloom’s:  2. Understand

Accessibility:  Keyboard Navigation

 

78) Mitotic recombination occurs between homologous chromosomes. In which of the following would you not expect to encounter mitotic recombination?

1. A) coli
2. B) tobacco plants
3. C) homo sapiens
4. D) drosophila melanogaster

 

Answer:  A

Section:  05.06

Topic:  Mitotic Recombination and Genetic Mosaics

Learning Objective:  05.06.01 Explain how mitotic recombination leads to the mosaic condition termed twin spots.

Bloom’s:  3. Apply

Accessibility:  Keyboard Navigation

 

79) In some fungi, such as the bread mold Neurospora crassa, the arrangement of spores in the ascus directly reflects the order in which they were produced during meiosis. What is the collection of spores produced by meiosis in Neurospora crassa called?

1. A) unordered tetrad
2. B) unordered octad
3. C) ordered tetrad or ordered octad
4. D) ordered pentad

 

Answer:  C

Section:  05.05

Topic:  Tetrad Analysis in Fungi

Learning Objective:  05.05.01 Explain the meaning of the term tetrad as applied to the asci produced by certain fungi.

Bloom’s:  1. Remember

Accessibility:  Keyboard Navigation

 

 

80) Twin spotting provides evidence of what genetic event?

1. A) meiotic recombination
2. B) mitotic recombination
3. C) linkage
4. D) mutation
5. E) biological evolution

 

Answer:  B

Section:  05.06

Topic:  Mitotic Recombination and Genetic Mosaics

Learning Objective:  05.06.01 Explain how mitotic recombination leads to the mosaic condition termed twin spots.

Bloom’s:  1. Remember

Accessibility:  Keyboard Navigation

 

81) Another name for a chromosome is a ________, because it contains alleles that are often inherited together.

1. A) linkage group
2. B) crossing over group
3. C) genetic recombinant
4. D) bivalent

 

Answer:  A

Section:  05.03

Topic:  Mapping: Locating Genes Along a Chromosome

Learning Objective:  05.03.04 Describe the relationship between linkage groups and chromosomes.

Bloom’s:  2. Understand

Accessibility:  Keyboard Navigation

 

82) The diploid garden pea plant has 14 chromosomes. The haploid fungus Neurospora crassa has 7 chromosomes. Neither organism has separate male and female individuals. Therefore, the number of linkage groups in these two organisms is

7. A) the garden pea has 14 linkage groups, and Neurospora has 7.
8. B) the garden pea has 7 linkage groups, and Neurospora has 7.
9. C) the garden pea has 8 linkage groups, and Neurospora has 8.
10. D) the garden pea has 15 linkage groups, and Neurospora has 8.

 

Answer:  B

Section:  05.03

Topic:  Mapping: Locating Genes Along a Chromosome

Learning Objective:  05.03.04 Describe the relationship between linkage groups and chromosomes.

Bloom’s:  2. Understand

Accessibility:  Keyboard Navigation