make a chi square data.
body is normal and wing is mutant
E=not ebony
e=ebony
V=not vestigial
v=vestigial
Homework Answers
The chi-square statistic: The chi-square statistic is 2.2537. The p-value is .133293. This result is not significant at p < .05.
| EE | Ee | Marginal Row Totals | |
| VV | 3 (5.56) [1.18] | 17 (14.44) [0.45] | 20 |
| vv | 17 (14.44) [0.45] | 35 (37.56) [0.17] | 52 |
| Marginal Column Totals | 20 | 52 |
72 (Grand Total) |
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make a chi square data.
body is normal and wing is mutant
E=not ebony
e=ebony
V=not...
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Scalloped (sd) wing is an X-linked recessive and ebony (e) body color is an autosomal recessive mutation in Drosophila. If a true breeding scalloped female that is wild type for ebony, is mated with a true breeding ebony male that is wild type for scolloped, what proportion of scalloped, normal body colored males is expected in the F2? a. 0 b. 1/16 c. 3/16 d. 1/32 e. 1/8
In Drosophila, b+ is the allele for normal body color and at the same gene b...
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In Drosophila (fruit flies) the genes how, dumpy and ebony are located on chromosome 3. LOF...
In Drosophila (fruit flies) the genes how, dumpy and ebony are located on chromosome 3. LOF = loss of function. Flies homozygous for a LOF mutation (no gene product made) in ebony have dark black bodies. Flies homozygous for a LOF mutation (no gene product made) in dumpyhave truncated (short) wings. Flies homozygous for a partial LOF mutation (some gene product made but significantly less than normal) in how have wings that will not fold down (held out wings; that's...
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In Drosophila (fruit flies) the genes how, dumpy and ebony are located on chromosome 3. LOF...
In Drosophila (fruit flies) the genes how, dumpy and ebony are located on chromosome 3. LOF = loss of function. Flies homozygous for a LOF mutation (no gene product made) in ebony have dark black bodies. Flies homozygous for a LOF mutation (no gene product made) in dumpy have truncated (short) wings. Flies homozygous for a partial LOF mutation (some gene product made but significantly less than normal) in how have wings that will not fold down (held out wings;...
can you please help me with number 4? An example of linked genes in Drosophila The...
can you please help me with number 4?
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I know that the sum of square of normal random variables follow a chi-square distribution. But...
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E10 from Brooker: In fruit flies, curved wings are recessive to straight wings, and ebony body...
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recalling that wild type is tan body, red eyes, straight wings and straight antennae, the data...
recalling that wild type is tan body, red eyes, straight wings and
straight antennae, the data is consistent with the logical
hypothesis that mutant trait-pick a mutant-is -autosomal or
sexlinked-, -dominant or recessive-.
using the f2 geb data and assuming we want to fail to reject the
null hypothesis abd support "logical" hypothesis for this mutant
trait, for the chi square goodness of fit test what is the chi
square statistical value? (value should be to the hundreths)
Here is...
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can you please help me with number 4?
An example of linked genes in Drosophila The genes for wing shape and body color are linked (they are on the same chromosome) Drosophila and linked genes In the example shown left, wild type alleles are dominant and are given an upper case symbol of the mutant phenotype (Cu or Eb). This notation used for Drosophila departs from the convention of using the dominant gene to provide the symbol. This is necessary...
E10 from Brooker: In fruit flies, curved wings are recessive to straight wings, and ebony body is recessive to gray body. A cross was made berween true-breeding flies with curved wings and pray bodies and flies with straight wings and ebony bodies. The F1 offspring were then mated to flies with curved wings and ebony bodies to produce an F2 generation. A. Diagram the genotypes of this cross starting with the parental generation and ending with the Frencration B. What...
recalling that wild type is tan body, red eyes, straight wings and
straight antennae, the data is consistent with the logical
hypothesis that mutant trait-pick a mutant-is -autosomal or
sexlinked-, -dominant or recessive-.
using the f2 geb data and assuming we want to fail to reject the
null hypothesis abd support "logical" hypothesis for this mutant
trait, for the chi square goodness of fit test what is the chi
square statistical value? (value should be to the hundreths)
Here is...
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