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Licemer1 [7]
3 years ago
5

A squirrel has 40 chromosomes in a somatic cell.

Biology
1 answer:
zhenek [66]3 years ago
4 0

Answer:

How many homologous pairs are found in a diploid cell? 22 pairs

b. How many single chromosomes (unreplicated) are found after the Sphase? 46 chromosomes  

c. How many chromosomes are present in one cell at the end of meiosis II?  15 chromosomes

d. How many chromatids will be found at the beginning of meiosis I? 92 chromatids  

e. How many cells will be produced at the end of meiosis II? four

f. How many tetrads will form in mitosis?  23 different tetrads

Explanation:

hope this helps! make me brainest please

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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
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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) <em>9/16 D_E_  wild type </em>

<em>     3/16 D_ee  wild wing, ebony body</em>

<em>     3/16 ddE_ dumpy wing, wild body</em>

<em>     1/16 ddee dumpy wing, ebony body</em>

(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) <em>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.</em>

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