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

Human Anatomy & Physiology (11th Edition)
11th Edition
ISBN:9780134580999
Author:Elaine N. Marieb, Katja N. Hoehn
Publisher:Elaine N. Marieb, Katja N. Hoehn
Chapter1: The Human Body: An Orientation
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Problem 1RQ: The correct sequence of levels forming the structural hierarchy is A. (a) organ, organ system,...
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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?

 

(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
Transcribed Image Text:(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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