Solution Manual for Genetics Laboratory Investigations, 14th Edition

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’s Manual
GENETICS
Laboratory Investigations
Fourteenth Edition
Thomas R. Mertens
Ball State University
Robert L. Hammersmith
Ball State University
SOLUTION MANUAL

iii
To The
Many suggestions helpful to users of
Genetics Laboratory Investigations have
been incorporated directly into the fourteenth edition. Sources of genetic stocks
and specific chemicals, supplies, and procedures are often included with the
investigations. Additional information can be found in this
Manual.
We wish to call to the attention of those adopting
Genetics Laboratory
Investigations the need for long-range planning on the part of the course instructor
or coordinator. For example, if you were to do one investigation per week and wait
until the fourteenth week of the semester to begin Investigation 13 on "open-ended"
Drosophila matings, it would be impossible to complete the investigation in the
remaining weeks of the semester. Since the manual has been designed to meet
instructional needs in a variety of academic settings, you can be selective in
choosing which investigations to do and in determining the sequence in which to do
them. We never use all of the 26 investigations in one semester, nor would we expect
others to do so.
While the
Manual provides answers to many of the questions in
the investigations, those questions dependant on data collected in the investigations
are generally not answered here. In at least two instances (Investigations 3 and 24),
we have provided sets of data collected by students in our classes that instructors
may wish to share with their students. Finally, we invite your comments concerning the
usefulness of our manual; please call to our attention errors or unclear writing that
can be corrected in future printings. We hope that
Genetics Laboratory Investigations,
Fourteenth Edition will serve you and your students well.

iv
Table of Contents
Answers to Queries and Problems
Page
To the ..................................................................iii
Investigation 1:
Drosophila and Maize Experiments in Genetics:
Monohybrid and Dihybrid Crosses ................................... 1
Investigation 2: Principles of Probability ......................................... 6
Investigation 3: The Chi-Square Test ............................................... 8
Investigation 4: Cell Reproduction: Mitosis ........................................13
Investigation 5: Meiosis in Animals: Oogenesis and Spermatogenesis .................15
Investigation 6: Meiosis in Angiosperms: Microsporogenesis .........................17
Investigation 7: Polytene Chromosomes from
Drosophila Salivary Glands ..............19
Investigation 8: Sex Chromosomes and Gene Transmission .............................20
Investigation 9: The Sex Check: A Study of Sex Chromatin in Human Cells ............23
Investigation 10: Human Chromosomes .................................................24
Investigation 11: Linkage and Crossing Over .........................................26
Investigation 12: Genetics of Ascospore Color in
Sordaria:
An Investigation of Linkage and Crossing
Over Using Tetrad Analysis ........................................28
Investigation 13: Open-Ended Experiments Using
Drosophila:
Locating a Mutant Gene in Its Chromosome ..........................30
Investigation 14: The Genetic Material: Isolation of DNA ............................33
Investigation 15: Restriction Endonuclease Digestion and
Gel Electrophoresis of DNA ........................................35
Investigation 16: Amplification of DNA Polymorphisms by Polymerase
Chain Reaction (PCR) and DNA Fingerprinting .......................38
Investigation 17: Transformation of
Escherichia coli ................................40
Investigation 18: Gene Action: Synthesis of β-Galactosidase
in
Escherichia coli ...............................................41
Investigation 19: Chromatographic Characterization of
Drosophila melanogaster Mutants ...................................42
Investigation 20: Bacterial Mutagenesis .............................................43
Investigation 21: Gene Recombination in Phage .......................................44
Investigation 22: Polygenic Inheritance: Fingerprint Ridge Count ....................45
Investigation 23: Population Genetics: The Hardy-Weinberg Principle .................46

v
Investigation 24: Population Genetics: The Effects of Selection
and Genetic Drift ................................................ 48
Investigation 25: Applied Human Genetics ........................................... 50
Investigation 26: NCBI and Genomic Data Mining ..................................... 53
SUPPLEMENTAL LABORATORY TOPICS
Supplemental 1 ..................................................................... 56
Supplemental 2 ..................................................................... 59
Supplemental 3 ..................................................................... 61

1
ANSWERS TO QUERIES AND PROBLEMS
Investigation 1
I. MEDIUM
Demerec and Kaufmann’s
Drosophila guide includes a number of recipes for
different media. 1 The ingredients of another useful medium, which has been attributed
to C.B. Bridges of the California Institute of Technology is as follows:
A. Ingredients of C. B. Bridge’s
Drosophila Medium
20 g of agar
200 g of cornmeal
145 ml of Karo Syrup
145 ml of Br’er Rabbit Molasses
2400 ml of distilled water
4.5 g of Dowicil-200 dissolved in 15 ml of distilled water 2
B. Preparing The Medium
First, dissolve the agar by heating it with 1000 ml of the distilled
water, and then add an additional 1000 ml of water. Carefully bring the
mixture to a boil. Before adding the cornmeal to this mixture, mix the
cornmeal with the remaining 400 ml of unheated distilled water. (Mixing
the cornmeal with the unheated water prevents it from forming lumps when
you add it to the hot agar suspension.) Add the cornmeal to the dissolved
agar mixture, stirring continuously to prevent the medium from sticking.
Now add the molasses and syrup, and boil the whole mixture for about 15
minutes. Dissolve the Dowicil-200 (mold inhibitor) in 15 ml of distilled
water; add this to the main mixture and mix thoroughly. The cornmeal
should not settle out in the final product.
Pour the medium into chemically clean bottles. These bottles may be of
any size, but 4-oz. wide-mouthed bottles or half-pint milk bottles are
satisfactory. Transfer the medium to a beaker for pouring and fill the
bottles to a depth of about 1 in. Take care to prevent the medium from
coming into contact with the neck of the bottle. Place a piece of
nonabsorbent paper, such as brown wrapping paper, in each bottle so that
it extends down into the medium (a double piece of paper is more
satisfactory). Use paper that is about 1 in. wide and be certain that it
extends upward to a point about 0.5 in. below the neck of the bottle. This
paper provides a dry place on which Drosophila larvae can pupate. Cotton
plugs may be used to stopper the bottles, but foam diSPo plugs are more
convenient. 3
____________________________
1
Drosophila Guide may be downloaded from Carnegie Institution for Science,
Books Online, http://carnegiescience. Edu/publications/books_online.
2 Dowicil-200 is available from Sigma-Aldrich, P.O. Box 14508, St Louis,
MO 63178
3 Foam diSPo plugs are available from Baxter Scientific Products Division
of VWR Scientific Products (Sargent-Welch), P.O. Box 5229, Buffalo Grove,
IL 60089, and Carolina Biological Supply Company: www.carolina.com.
____________________________

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2
Once the bottles of medium are plugged, sterilize them in an autoclave
at 20 lbs. pressure for 20 minutes. Other items to be used in handling the
flies may be sterilized at the same time.
Water will condense on the insides of the bottles as they cool after
sterilization. Allow 48 hours in a well-ventilated room for this water to
evaporate. Immediately before you place flies into the bottles of medium,
shake a small amount of dry yeast onto the medium. The yeast will grow and
serve as food for the developing fly larvae.
V. EXPERIMENTAL MATINGS
C. Making Crosses
1. F 1 females will be mated to F 1 males to produce the F 2 . This mating can
occur prior to or after placing flies in a fresh bottle.
2. A test cross is a controlled mating to a recessive homozygote; therefore,
the F 1 female must not have mated before she is placed with the recessive
male.
3. Any cross designed to determine the genotype of the individual tested; a
cross to an individual homozygous recessive for genes in question.
VI. SUGGESTED MONOHYBRID CROSSES
A.
Drosophila Crosses
3 and 4. Sample answer: red eyes. The F 1 flies all had red (wild type) eye
color, and the F 2 consisted of about 3 red to 1 sepia (brown).
Genes that behave as dominants may be symbolized with capital
letters or with a + subscripted above the gene symbol.
VII. INDEPENDENT ASSORTMENT
A. Dihybrid Cross With
Drosophila
2. F 1 phenotype: Wild type for both traits.
F 2 phenotypic frequencies: 9 wild type (long wing, gray body) : 3 long,
ebony : 3 vestigial, gray : 1 vestigial, ebony.
5. Table 1.5
Phenotype
Observed
Number
Expected
Number
Deviation
O-E
long, gray ________________ ________________ ________________
vestigial, gray ________________ ________________ ________________
long, ebony ________________ ________________ ________________
vestigial, ebony ________________ ________________ ________________
6. P 1
e/e vg+/vg+ (ebony) ×
e+/e+ vg/vg (vestigial)
F 1
e+/e vg+/vg (wild type)
F 2 1
e+/e+ vg+/vg+, 2
e+/e vg+/vg+, 2
e+/e+ vg+/vg, 4
e+/e vg+/vg (all wild
type); 1
e+/e+ vg/vg, 2
e+/e vg/vg (both vestigial); 1
e/e vg+/vg+, 2
e/e
vg+/vg (both ebony); 1
e/e vg/vg (ebony, vestigial)

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3
B. Dihybrid Crosses in Maize
A number of alternative corn experiments are available commercially and
may be used to demonstrate the classical dihybrid F 2 ratio of 9:3:3:1.
Either endosperm traits alone, seedling traits alone, or a combination of
one endosperm and one seedling trait may be involved in such crosses.
Consult the current catalogs of such firms as the Carolina Biological
Supply Company to determine what materials are available. Among the
possibilities are the following:
o
Y/Y wx/wx ×
y/y Wx/Wx F 1
Y/y Wx/wx F 2 9:3:3:1
yellow waxy white starchy yellow starchy
o
R/R Su/Su ×
r/r su/su F 1
R/r Su/su F 2 9:3:3:1
purple starchy yellow sweet purple starchy
o An F 2 segregating for two seedlings traits such as a gene for dwarf (e.g.,
d 5) and one for albinism (e.g.,
lw 3 ) would produce an F 2 ratio of 9 tall
green : 3 tall albino : 3 dwarf green : 1 dwarf albino.
o An F 2 in which one trait is an endosperm trait (e.g.,
Su, starchy vs.
su,
sugary) and the other trait a seedling condition (e.g.,
Gl, normal vs.
gl,
glossy) will also assort into a 9:3:3:1 ratio.
2. Gene Symbol Phenotype
Su starchy (smooth)
su sugary (wrinkled)
R colored (purple)
r colorless (yellow -- no purple)
(a) Dominant: starchy (smooth) and colored (purple).
(b) Recessive: sugary (wrinkled) and colorless (yellow).
3. Genotypes of the two possibilities: (1)
R/R Su/Su ×
r/r su/su or (2)
R/R
su/su ×
r/r Su/Su.
(a) Phenotypes: (1) purple starchy × yellow sugary or (2) purple sugary ×
yellow starchy.
(b) More than one original cross could have been used as shown above.
4.
R/r Su/su purple, starchy.
Table 1.6
Phenotype
Observed
Number
Expected
Number
Deviation
O-E
purple, starchy _________________ _________________ _________________
purple, sugary _________________ _________________ _________________
yellow, starchy _________________ _________________ _________________
yellow, sugary _________________ _________________ _________________

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4
VIII. GENE INTERACTIONS
A. Gene Interactions in
Drosophila (This particular mating can be part of an
open-ended experiment in which different crosses involving eye color mutants
can be made. See Investigation 13.)
Parental (scarlet)
sc/sc Bw/Bw × (brown)
Sc/Sc bw/bw
F1
Sc/sc Bw/bw Wild-type (brick-red)
F2 Symbol Phenotype Expected Ratio
Sc/- Bw/- Wild-type 9/16
Sc/- bw/bw brown 3/16
sc/sc Bw/- scarlet 3/16
sc/sc bw/bw white 1/16
B. Gene Interactions in Maize
Among the possible kinds of ears of corn that show epistasis or gene
interaction are the following:
Phenotypes of the
True-Breeding Parents
Phenotype of
the F 1 Kernels
Phenotypes and Ratios
of the F 2 Kernels
1. purple × white purple 9 purple : 3 red : 4 white
2. yellow × yellow yellow 13 yellow : 3 purple
3. white × white purple 9 purple : 7 white
4. purple × yellow purple 12 purple : 3 yellow : 1 white
Given below are the genotypes for the four crosses cited above. Note
that official maize gene symbols are used.
1.
Pr/Pr R/R x
pr/pr r/r F 1
Pr/pr R/r F 2 9
Pr/- R/- (purple)
purple white purple
3
pr/pr R/- (red)
4 --
r/r (white)
2.
C I
/C I R/R ×
C/C r/r F 1
CI / C R/r F 2 9
CI /- R/- (yellow)
yellow yellow yellow
13 3
CI /-
r/r (yellow)
1
C/C r/r
3
C/C R/- (purple)
3.
C/C r/r x
c/c R/R F 1
C/c R/r F 2 9
C/- R/- (purple)
white white purple
3
C/- r/r
7 3
c/c R/- (white)
1
c/c r/r

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5
4.
y/y R/R ×
Y/Y r/r F 1
Y/y R/r F 2 9
Y/- R/-
purple yellow purple 12 (purple)
3
y/y R/-
3
Y/- r/r (yellow)
1
y/y r/r (white)
Other F 2 ears of “genetic corn” showing epistasis can be obtained.
All of the ears cited in this investigation involve two segregating gene
loci.
The epistasis ratios (9:3:4, 13:3, 9:7, and 12:3:1) are all modifications of the
classical 9:3:3:1 F 2 ratio. Ears of genetic corn in which three gene loci are
segregating may also be obtained. For example, the F 2 trihybrid epistasis ratio
of 27:37 (a variation of the classical 27:9:9:9:3:3:3:1) could be used for this
purpose.
The epistatic F 2 maize ears are becoming more difficult to obtain from vendors.
For this reason, we have printed FIGURE 1.11 in a larger format so that students
can count each of these different maize ears. Careful counting of kernels will
provide reasonable data for analysis. For the ear in (B), hue of the kernel is
more important than intensity of the color. For example, light red and dark red
kernels are all counted as red, and the same is true for purple.

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6
Investigation 2
II. INDEPENDENT EVENTS OCCURRING SIMULTANEOUSLY
1. Both heads: 1/2 × 1/2 = 1/4; one head, one tail: 1/2 × 1/2 = 1/4; head on
one coin and tail on the other: 1/4 + 1/4 = 1/2; both coins tails: 1/2 ×
1/2 = 1/4. Two coins fall heads, heads about 1/4 of the time; heads, tails
(and vice versa) about 1/2 of the time; and tails, tails, about 1/4 of the
time. Stated as a ratio instead of a fraction, the expected result is
1:2:1.
Table 2.3
Classes Combinations Class Occurring Observed Expected (O-E)
3 heads HHH 1/2 × 1/2 × 1/2 = 1/8 7
2 heads,
1 Tail
HHT, HTH,
THH
3(1/2 × 1/2 × 1/2) = 3/8 21
1 head,
2 Tails
HTT, THT,
TTH
3(1/2 × 1/2 × 1/2) = 3/8 21
3 tails TTT 1/2 × 1/2 × 1/2 = 1/8 7
Total 8 possible 8/8 = 1 56 56
4. Table 2.4
Classes Combinations Probability of Each Class
Occurring
4 heads HHHH 1/2 × 1/2 × 1/2 × 1/2 = 1/16
3 heads : 1 tail HHHT, HHTH,
HTHH, THHH
4(1/2 × 1/2 × 1/2 × 1/2) = 4/16
2 heads : 2 tails HHTT, HTTH, THHT,
TTHH, HTHT, THTH
6(1/2 × 1/2 × 1/2 × 1/2) = 6/16
3 tails : 1 head HTTT, THTT
TTHT, TTTH
4(1/2 × 1/2 × 1/2 × 1/2) = 4/16
4 tails TTTT 1/2 × 1/2 × 1/2 × 1/2 = 1/16
5. a. (1/2) 4 = 1/16
b. 4(1/2) 3
(1/2) = 4/16 = 1/4
c. 6(1/2) 2
(1/2) 2 = 6/16 = 3/8
d. Two boys and two girls. There are more ways (6) in which a family can
consist of 2 boys and 2 girls.
e. A boy 1/2, a girl 1/2.

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7
III. BINOMIAL EXPANSION
1. a. (1/2) 5 = 1/32 =
a5 d. 10(1/2) 2 (1 / 2) 3 = 10/32 = 5/16 = 10
a 2
b 3
b. 5(1/2) 4
(1/2) = 5/32 = 5
a4 b e. 5(1/2)(1/2) 4 = 5/32 = 5
ab 4
c. 10(1/2) 3 (1/2) 2 = 10/32 = 5/16 = 10
a3 b2 f. (1/2) 5 = 1/32 =
b 5
2. a. 1 boy and 5 girls: 6!/5!1! (1/2)(1/2) 5 = 6/64 = 3/32
b. 3 boys and 3 girls: 6!/3!3! (1/2) 3 (1/2) 3 = 20/64 = 5/16
c. All 6 girls: 6!/0!6! (1/2) 0 (1/2) 6 = (1/2) 6 = 1/64
3. A normal child: 3/4; an albino: 1/4.
a. All 4 normal: (3/4) 4 = 81/256
b. 3 normal and 1 albino: 4(3/4) 3 (1/4) = 108/256 = 27/64
c. 2 normal and 2 albino: 6(3/4) 2 (1/4) 2 = 54/256
d. 1 normal and 3 albinos: 4(3/4)(1/4) 3 = 12/256
e. All 4 albinos: (1/4) 4 = 1/256
IV. EITHER-OR SITUATIONS (MUTUALLY EXCLUSIVE EVENTS)
1. Either
C or
c gametes; 1/2 + 1/2 = 1 or 100%
2. Either the genotype
AA or the genotype Aa: 1/4 + 2/4 = 3/4
a. Either
aaB- or
aabb: 3/16 + 1/16 = 4/16 = 1/4
b. Either
aabb or
AaBb: 1/16 + 4/16 = 5/16
c. Either
A-bb or
AAbb: 3/16 + 1/16 = 5/16
d. Either
A-B- or
aabb: 9/16 + 1/16 = 10/16 = 5/8
V. PROBABILITY AND GENETIC COUNSELING
a. 4 × 7: 1(
aa) × 1(
Aa) × 1/2 = 1/2
b. 5 × 1: 1(
Aa) × 2/3(
Aa) × 1/4 = 2/12 = 1/6
c. 6 × 13: 1(
Aa) × 1/2(
Aa) × 1/4 = 1/8
d. 10 × 14: 2/3(
Aa) × 1/2(
Aa) × 1/4 = 2/24 = 1/12
e. 3 × 17: 2/3(
Aa) × 1/3(
Aa) × 1/4 = 2/36 = 1/18
Note: #17 has a 1/3 probability because his overall is his mother’s
probability of being heterozygous (2/3) times his probability (1/2) if his
mother was heterozygous.
f. 3 × 15: 2/3(
Aa) × 1/2(
Aa) × 1/4 = 2/24 = 1/12
g. 16 × 17: 1/2(
Aa) × (2/3 × 1/2)(
Aa) x 1/4 = 2/48 = 1/24

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8
Investigation 3
I. A. APPLICATION OF CHI-SQUARE. COLORED AND WHITE BEANS
Purchase four four-pound bags of white beans from a local supermarket. Place the
beans from two of the bags into a large container, cover them with an aqueous solution
of Basic Fuchsin (approximately 0.05%), and allow the beans to stain for approximately
30 minutes. Pour off the stain and rinse the beans several times with water; drain and
spread the beans out on a sheet of plastic to dry. After the stained beans have dried,
place them in a large container. Add the remaining two bags of white beans and mix
thoroughly. Sixteen pounds of beans easily provide enough beans for a class of 30.
As an alternative method of getting a one-to-one ratio of beans, we have used
equal volumes (not weight) of white navy beans to black beans. Make sure when you buy
the beans that the navy beans are of the same size as the black beans. Be sure to
count multiple samples of the bean mix to see that they are indeed a 1:1 ratio.
Sample Data from a Class of 24
Sample # Red White Total Sample # Red White Total
1 324 354 678 13 322 306 628
2 327 326 653 14 329 352 681
3 290 341 631 15 340 332 672
4 319 321 640 16 331 318 649
5 337 298 635 17 334 302 636
6 338 307 645 18 312 311 623
7 339 317 656 19 328 298 626
8 329 330 659 20 294 291 585
9 308 325 633 21 294 346 640
10 317 346 663 22 284 290 574
11 295 316 611 23 309 302 611
12 340 304 644 24 316 282 598
Use these or similar data for chi-square tests as described in Part I A of this
investigation. Complete Tables 3.3 and 3.4 as instructed.
II. CALCULATION OF CHI-SQUARE
A. Practice Problems
1.
O E O–E (O–E) 2 (O–E) 2 /E
Normal 272 249 23 529 2.1245
Dumpy 60 83 –23 559 6.3735
Totals 332 332 2
χ = 8.4980
’s Calculation of 2
χ = 8.4980

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9
a. 1 degree of freedom. c. Less than 1%.
b. Reject.
2.
Phenotype O E O–E (O–E) 2 (O–E) 2 /E
R-Su- 134 149.5 -15.5 240.25 1.6070
R-susu 150 149.5 0.5 .25 0.0017
rrSu- 146 149.5 -3.5 12.25 0.0819
rrsusu 168 149.5 18.5 342.25 2.2893
Totals 598 598 2
χ = 3.9799
’s Calculation of 2
χ = 3.9799
a. 3 degrees of freedom. c. 10% to 30%.
b. Accept.
B. Assigned Problems
1.
Phenotype O E O-E (O-E) 2 (O-E) 2 /E
R-Su- 348 364.5 -16.5 272.25 0.7469
R-susu 119 121.5 -2.5 6.25 0.0514
rrSu- 139 121.5 17.5 306.25 2.5206
rrsusu 42 40.5 1.5 2.25 0.5556
Total 648 648 2
χ = 3.8745
a. 3 degrees of freedom.
b. Accept.
c. 10% to 30%.
d.
Hypothesis 2
χ value P-value Accept or Reject
Hypothesis
3 R- : 1 rr
467 181
2.976 5-10% Accept
3 Su- : 1 susu
487 161
0.0082 90-95% Accept

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10
2.
Phenotype O E O-E (O-E) 2 (O-E) 2 /E
A-B- 104 84.25 19.75 390.0625 4.6298
A-bb 63 84.25 -21.25 451.5625 5.3598
aaB- 69 84.25 -15.25 232.5625 2.7604
aabb 101 84.25 16.75 280.5625 3.3301
Total 337 337.00 2
χ = 16.0801
a. 3 degrees of freedom.
b. Reject.
c. Less than 1%.
d.
Hypothesis 2
χ value P-value Accept or Reject
Hypothesis
1 A- : 1 aa
167 170
0.02671 80-90% Accept
1 B- : 1 bb
173 164
0.24036 50-70% Accept
e. Genes A and B are linked -- lack of independent assortment prevents
obtaining the expected 1:1:1:1 dihybrid test cross ratio. Each gene,
however, behaves according to Mendel’s first law, yielding a 1:1
ratio.
3.
Phenotype O E O-E (O-E) 2 (O-E) 2
/E
A 601 613.77 -12.77 163.0729 0.2657
B 170 134.73 35.27 1243.9729 9.2331
AB 36 44.91 -8.91 79.3881 1.7677
O 690 703.59 -13.59 184.688 0.2625
Totals 1497 1497.00 2
χ = 11.529
a. 3 degrees of freedom.
b. Reject.
c. Less than .01%.
d. A, 40.15%; B, 11.36%; AB, 2.40%; O, 46.09%.

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11
4.
Family Sex
Ratios
O E O-E (O-E) 2 (O-E) 2 /E
All boys 170 151.4375 18.5625 344.5664 2.2753
3 B : 1 G 642 605.75 36.25 1314.0625 2.1693
2 B : 2 G 912 908.625 3.375 11.3906 0.0125
1 B : 3 G 572 605.75 -33.75 1139.06 1.8804
All girls 127 151.4375 -24.4375 597.1914 3.9435
Totals 2423 2423 2
χ = 10.281
a. 4 degrees of freedom.
b. Reject.
c. 1% to 5%.
d.
Sex O E O-E (O-E) 2 (O-E) 2 /E
Male 5002 4846 156 24336 5.0219
Female 4690 4846 –156 24336 5.0219
Totals 9692 9692 2
χ = 10.0438
df = 1
P = < 1%
e. Reject.
f. 5002/4690 = 1.0665/1.0
g. Selective advantage of Y-bearing spermatozoa in fertilizing eggs;
possible production or survival of more Y-bearing spermatozoa so that
more male children are conceived.
5.
Family
Ratios
O E O-E (O-E) 2 (O-E) 2 /E
All A- 1440 1458 –18 121 0.2222
3 A- : 1 aa 1986 1944 42 1764 0.9074
2 A- : 2 aa 970 972 –2 4 0.0041
1 A- : 3 aa 197 216 –19 361 1.6713
All aa 15 18 3 9 0.5000
Totals 4608 4608 2
χ = 3.3050

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