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3 years ago

If it is 12:00 noon in California,what time is it in Virginia?

Biology

2 answers:

Anettt [7]3 years ago

6 0

California is 3 hours ahead, so, 3:00 pm

AlekseyPX3 years ago

4 0

This is easy because they are diferent time zones
pacific: (California )
Mts 1 hour ahead of pacific
middle 2 hours ahead ahead of pacific
Eastern:(Virginia )3 hours ahead of pacific
So the time would be 3:00 pm

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According to the given information, the allele for the red-green colorblindness is inherited in an X linked recessive manner. Let's assume that the allele X^c is responsible for red-green colorblindness. The woman is normal but had a colorblind father (X^cY). Fathers give their X chromosomes to the daughters while their Y chromosome is transmitted to their sons. The sons get their X chromosomes from the mother.

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In flies dumpy wings and ebony bodies are each mutations recessive to wild type, and you expect them to behave in a Mendelian fa

Dimas [21]

Answer:

See the answers below

Explanation:

Assuming that D (d) represents the allele for wing type and E (e) represents the allele for body type, crossing DdEe with DdEe will yield the following offspring according to the Punnet's square (see the attached image):

(a) 9/16 D_E_ wild type

3/16 D_ee wild wing, ebony body

3/16 ddE_ dumpy wing, wild body

1/16 ddee dumpy wing, ebony body

(b) Chi square $X^2$ = $\frac{(O - E)^2}{E}$ , where O = observed frequency and E = expected frequency.

Phenotype O E $X^2$

Wild type 473 9/16 x 830 = 466.875 $\frac{(473 - 466.875)^2}{466.875}$ = 0.08

wild w/ebony b 156 3/16 x 830 = 155.625 $\frac{(156 - 155.625)^2}{155.625}$ = 0.0009

Dumpy w/wild b 149 3/16 x 830 = 155.625 $\frac{(149 - 155.625)^2}{155.625}$ = 0.28

dumpy w/ebony b 52 1/16 x 830 = 51.875 $\frac{(52 - 51.875)^2}{51.875}$ = 0.0003

Total $X^2$ = 0.3612

Degree of freedom = 4 - 1 = 3

Tabulated value of $X^2$ (0.05)= 7.815

(c) Since the calculated Chi square value is less than the tabulated value, we conclude that the observed outcome agrees with the expected outcome and that the cross followed the standard Mendelian pattern.

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