How did P<0.005 come into being? How do you determine what is due to chance and what's not? (a)Brown body,straight wingsYellow body,curved wingsy*y,cv*cvуу сусу1 A testcross is carriedout betweenCrosscockroaches differingin two characteristics.63 yty cvtcv28 yty cvcvbrown body, straight wingsbrown body, curved wingsyellow body, straight wingsyellow body, curved wings33 yy cvtcv77 уу сусу(b) Contingency tableSegregation of y* and y 3.with genotypesfor one locus2 To test for independentassortment of allelesencoding the two traits,a table is constructed.Rowalong the top...y*yyytotalscv*cv 635 Numbers of each5genotype areplaced in thetable cells, andthe row totals,column totals,3396Segregationof cvt and cvсу су2877105..and genotypesfor the other locusColumntotals91110201along the left side.Grand total and grand totalare computed.(c)Number expectedrow total X column totalNumberGenotypeobservedgrand total96 X91y*y cvtcv63= 43.462016 The expectednumbers of105 X91y*y cvcv28= 47.54201progeny,assumingindependentassortment,are calculated.96 X110yy cy*cv33= 52.54201105X110уу сусу77= 57.46201(d)2 = observed-expected)2expected7 A chi-squarevalue iscalculated.(63 – 43.46)2(28 – 47.54)2 , (33 – 52.54)?+52.54(77 - 57.46)2+43.4647.5457.46= 8.79 + 8.03 + 7.27 + 6.64= 30.738 The probabilityis less than 0.005,df = (number of rows - 1) X (number of columns – 1)| indicating thatthe difference(e)df = (2 – 1) X (2 – 1) = 1 X 1 = 1between thenumbers ofobserved andP< 0.005Conclusion: The genes for body color and type of wingare not assorting independently and must be linked.expected progenyis probably notdue to chance.Pierce, Genetics: A Conceptual Approach, 7e © 2020 W. H. Freeman and Company

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How did P<0.005 come into being? How do you determine what is due to chance and what's not?

Biology: The Unity and Diversity of Life (MindTap Course List)

15th Edition

ISBN:9781337408332

Author:Cecie Starr, Ralph Taggart, Christine Evers, Lisa Starr

Publisher:Cecie Starr, Ralph Taggart, Christine Evers, Lisa Starr

Chapter13: Observing Patterns In Inherited Traits

Section: Chapter Questions

Problem 5GP

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Question

How did P<0.005 come into being? How do you determine what is due to chance and what's not?

(a)
Brown body,
straight wings
Yellow body,
curved wings
y*y,cv*cv
уу сусу
1 A testcross is carried
out between
Cross
cockroaches differing
in two characteristics.
63 yty cvtcv
28 yty cvcv
brown body, straight wings
brown body, curved wings
yellow body, straight wings
yellow body, curved wings
33 yy cvtcv
77 уу сусу
(b) Contingency table
Segregation of y* and y 3.with genotypes
for one locus
2 To test for independent
assortment of alleles
encoding the two traits,
a table is constructed.
Row
along the top...
y*y
yy
totals
cv*cv 63
5 Numbers of each
5
genotype are
placed in the
table cells, and
the row totals,
column totals,
33
96
Segregation
of cvt and cv
су су
28
77
105
..and genotypes
for the other locus
Column
totals
91
110
201
along the left side.
Grand total and grand total
are computed.
(c)
Number expected
row total X column total
Number
Genotype
observed
grand total
96 X91
y*y cvtcv
63
= 43.46
201
6 The expected
numbers of
105 X91
y*y cvcv
28
= 47.54
201
progeny,
assuming
independent
assortment,
are calculated.
96 X110
yy cy*cv
33
= 52.54
201
105X110
уу сусу
77
= 57.46
201
(d)
2 = observed-expected)2
expected
7 A chi-square
value is
calculated.
(63 – 43.46)2
(28 – 47.54)2 , (33 – 52.54)?
+
52.54
(77 - 57.46)2
+
43.46
47.54
57.46
= 8.79 + 8.03 + 7.27 + 6.64
= 30.73
8 The probability
is less than 0.005,
df = (number of rows - 1) X (number of columns – 1)| indicating that
the difference
(e)
df = (2 – 1) X (2 – 1) = 1 X 1 = 1
between the
numbers of
observed and
P< 0.005
Conclusion: The genes for body color and type of wing
are not assorting independently and must be linked.
expected progeny
is probably not
due to chance.
Pierce, Genetics: A Conceptual Approach, 7e © 2020 W. H. Freeman and Company

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