Transcript Diapositive 1

Lecture 10
Recapitulation, Review ,Tutorial
Chi-Square Calculations and the Null Hypothesis
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When we assume that data will fit a given ratio such as 1:1, 3:1, or 9:3:3:1, we
establish what is called the null hypothesis (H0). It is so named because the hypothesis
assumes that there is no real difference between the measured values (or ratio) and
the predicted values (or ratio).
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Any difference can be attributed purely to chance. The validity of the null hypothesis
for a given set of data is measured using statistical analysis.
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Depending on the results of this analysis, the null hypothesis may either
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(1) be rejected
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or (2) fail to be rejected.
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If it is rejected, the observed deviation from the expected result is judged not to be
attributable to chance alone. In this case, the null hypothesis and the underlying
assumptions leading to it must be reexamined. If the null hypothesis fails to be
rejected, any observed deviations are attributed to chance.
Quiz
1)- In one of Mendel’s dihybrid crosses, he observed 315
round yellow, 108 round green, 101 wrinkled yellow, and 32
wrinkled green F2 plants.
Analyze these data using the Chi-square test to see if:
• (a) they fit a 9:3:3:1 ratio.
• (b) the round : wrinkled data fit a 3:1 ratio.
• (c) the yellow : green data fit a 3:1 ratio.
• Consider two hypothetical recessive autosomal genes a and b, where a
heterozygote is testcrossed to a double homozygous mutant. Predict the
phenotypic ratios under the following conditions:
• (a) a and b are located on separate autosomes.
• (b) a and b are linked on the same autosome but are so far apart that a
crossover always occurs between them.
• (c) a and b are linked on the same autosome but are so close together that
a crossover almost never occurs.
• With two pairs of genes involved (P/p and Z/z), a testcross
(ppzz) with an organism of unknown genotype indicated that
the gametes produced were in the following
• proportions:
• PZ, 42.4%; Pz, 6.9%; pZ, 7.1%; pz, 43.6%
• Draw all possible conclusions from these data.
• In rabbits, black color (B) is dominant to brown (b), while full color (C) is
dominant to chinchilla (cch). The genes controlling these traits are linked.
Rabbits that are heterozygous for both traits and express black, full color
were crossed with rabbits that express brown, chinchilla, with the
following results:
• 31 brown, chinchilla
• 34 black, full color
• 16 brown, full color
• 19 black, chinchilla
• Determine the arrangement of alleles in the heterozygous parents and the
map distance between the two genes.
• The genes dumpy (dp), clot (cl), and apterous (ap)
are linked on chromosome II of Drosophila. In a
series of two-point crosses, the following genetic
distances were determined. What is the sequence of
the three genes?
• dp–ap= 42
• dp–cl =3
• ap–cl= 39
• In a testcross between a plant of unknown genotype and
phenotype and a plant that is homozygous recessive for both traits,
the following progeny were obtained:
• Colored, green 88
• Colored, yellow 12
• Colorless, green 8
• Colorless, yellow 92
• Explain how these results were obtained by determining the exact
genotype and phenotype of the unknown plant, including the
precise arrangement of the alleles on the homologs.