Python/Pandas. For loop on multiple dataFrames not working correctly

I am trying to process a list of dataframes (example shows 2, reality has much more) in multiple ways using a for loop.
Droping columns in the dataframe referenced in the loop works fine, however, concat doesn't do anything inside the loop. I expect to update the original dataframe referenced in dfs.

UPDATED PROBLEM STATEMENT

Previous examples do not cover this case/ seem to not work.
Example adapted from here: https://stackoverflow.com/questions/50306898/pandas-dataframe-concat-using-for-loop-not-working

Minifying the example leads to the following (code partially borrowed from another question)

import numpy as np
import pandas as pd


data = [['Alex',10],['Bob',12],['Clarke',13]]
data2 = ['m','m','x']
A = pd.DataFrame(data, columns=['Name','Age'])
B = pd.DataFrame(data, columns=['Name','Age'])
C = pd.DataFrame(data2, columns=['Gender'])

#expected result for A:
Anew=pd.DataFrame([['Alex','m'],['Bob','m'],['Clarke','x']], columns=['Name', 'Gender'])

dfs = [A,B]

for k, v in enumerate(dfs):
    # The following line works as expected on A an B respectively, inplace is required to actually modify A,B as defined above
    dfs[k]=v.drop('Age',axis=1, inplace=True)
    # The following line doesn't do anything, I was expecting Anew (see above) 
    dfs[k] = pd.concat([v, C], axis=1)
    # The following line prints the expected result within the loop
    print(dfs[k])

# This just shows A, not Anew: To me tha tmeans A was never updated with dfs[k] as I thought it would. 
print(A)

Update

Try:

data = [['Alex',10],['Bob',12],['Clarke',13]]
data2 = ['m','m','x']
A = pd.DataFrame(data, columns=['Name','Age'])
B = pd.DataFrame(data, columns=['Name','Age'])
C = pd.DataFrame(data2, columns=['Gender'])
Anew = pd.DataFrame([['Alex','m'],['Bob','m'],['Clarke','x']], columns=['Name', 'Gender'])

dfs = [A, B]
for v in dfs:
    v.drop('Age', axis=1, inplace=True)
    v['Gender'] = C
print(A)
print(Anew)

Output:

>>> A
     Name Gender
0    Alex      m
1     Bob      m
2  Clarke      x

>>> Anew
     Name Gender
0    Alex      m
1     Bob      m
2  Clarke      x

If you use inplace=True, Pandas doesn't return a DataFrame so dfs is now None:

dfs[k]=v.drop('Age', axis=1, inplace=True)  # <- Remove inplace=True

Try:

dfs = [A, B]
for k, v in enumerate(dfs):
    dfs[k] = v.drop('Age', axis=1)
    dfs[k] = pd.concat([v, C], axis=1)
out = pd.concat([A, C], axis=1)

Output:

>>> out
     Name  Age Gender
0    Alex   10      m
1     Bob   12      m
2  Clarke   13      x