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VikaD [51]
3 years ago
7

Please i need help with thi my life DEPENDS ON IT I WILL GIVE BRAINLY TO

Biology
2 answers:
Stells [14]3 years ago
6 0

Answer:

1 is metaphase 1

2 is cytokineses

3 is prophase 1

4 is anaphase 1

Explanation:

4vir4ik [10]3 years ago
5 0
Is this the hub? Because my friend found out a way to cheat
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1.<br> Meiosis in Brennan's Mom's ovaries
Y_Kistochka [10]

Answer:

Ok... thanks for the points

Explanation:

4 0
3 years ago
Please help at soon as possible!
Butoxors [25]
For starters, the left image is a plant cell and the right image is an animal cell.

The black dots in both images would be drawn to the nucleus. All cells have nuclei. The black oval in the right image would be drawn to the cell membrane. Only animal cells have these. The rectangle with a gap would be drawn to the cell wall. Only plant cells have cell walls. The space surrounding the nucleus in both cells would be drawn to cytoplasm. All cells have cytoplasm. The dotted circles would be drawn to chloroplast. Only plant cells have these.

Hope I could help!
7 0
3 years ago
You are a botanist looking at the plant color in a new breed of Snapdragon. The two parent plants are Blue and Green and when yo
irakobra [83]

Answer:

Incomplete dominance is the inheritance pattern where the dominant allele did not mask the recessive allele completely and form a mix of both alleles. Here the inheritance is the incomplete inheritance. The ratio of F2 generation is 1:2:1.

Given:

R1R1 = 42

R2R2 = 39

R1R2 = 86

Total R1 alleles = 2*42+86 = 170

Total R2 alleles = 2*39+86 = 164

Total alleles = 334

Frequency of allele R1 = 170/334 = 0.51

Frequency of allele R2 = 164 / 334 = 0.49

Expected number of each phenotype:

Total population = 167

Blue = R1R1 = 0.51 * 0.51 * 167 = 43.44

Green = R2R2 = 0.49 * 0.49 * 167 = 40.10

Cyan = 2*R1*R2 = 2*0.51*0.49*167 = 83.46

Phenotype     Observed(O)    Expected (E)    O-E      (O-E)2     (O-E)2/E

Blue                 42                   43.44               -1.44       2.0736     0.0477

cyan                86                   83.46                 2.54     6.4516     0.0773

green              39                    40.1                    -1.1       1.2100      0.0302

Total              167                     167                                               0.1552

Chi-square value = 0.155

Degrees of freedom = no. of phenotypes – 1

Df = 3-1 = 2

Critical value = 5.99

Chi-square value of 0.155 is less than the critical value of 5.99. So we accept the null hypothesis.

4 0
3 years ago
That's my question plz help me
galina1969 [7]
What’s your question sir
4 0
3 years ago
Can i have some help plz
Veseljchak [2.6K]

Answer:

ok tell me

Explanation:

i can edit this so once you tell me i can change it

3 0
3 years ago
Read 2 more answers
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