Monohybrid Cross Worksheet 2 with exercises on Punnett squares and genetic crosses.

Let’s walk through each part of the worksheet step by step.

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Problem 1:
Parent alleles:
Top row (from one parent): h, h
Left column (from other parent): H, h

Punnett Square:
- Top-left: H + h = Hh
- Top-right: H + h = Hh
- Bottom-left: h + h = hh
- Bottom-right: h + h = hh

So offspring genotypes: Hh, Hh, hh, hh → that’s 2 Hh and 2 hh out of 4 total.

Frequencies:
- HH: 0 (no combination gives two capital H’s)
- Hh: 2/4 = 1/2
- hh: 2/4 = 1/2 ✔ Matches what’s written.

Genotypes listed:
P1: Hh, hh → Wait, this is confusing. The parents are actually “Hh” and “hh”? Let’s check:

Actually, looking at the square:
One parent contributes only ‘h’ (top row), so genotype is hh.
The other parent contributes ‘H’ and ‘h’ (left column), so genotype is Hh.
So P1 should be: Hh × hh — but the worksheet says “P1: Hh, hh” — which might mean the two parents’ genotypes? That makes sense.

Then F1 offspring: Hh and hh — yes, from the square.

✔ All correct in Problem 1.

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Problem 2:
Parents: Both contribute ‘a’ on top → so both gametes from one parent are ‘a’.
Left side: both rows are ‘A’ → so other parent contributes only ‘A’.

So cross is AA × aa? Wait — no:

Actually, if both top headers are ‘a’, that means one parent has genotype aa (only passes ‘a’).
Both left-side labels are ‘A’, meaning the other parent has genotype AA (only passes ‘A’).

So all offspring get A from one parent and a from the other → all are Aa.

Punnett Square:
All four boxes: Aa

Frequencies:
AA: 0
Aa: 4/4 = 1
aa: 0 ✔ Correct.

Genotypes:
P1: AA, aa → yes, those are the parental genotypes.
F1: all Aa → correct.

✔ Problem 2 is accurate.

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Problem 3:
Top: B, b → so one parent is Bb
Left: B, b → other parent is also Bb

Cross: Bb × Bb

Punnett Square:
- BB (B+B)
- Bb (B+b)
- Bb (b+B)
- bb (b+b)

So: 1 BB, 2 Bb, 1 bb

Frequencies:
BB: 1/4
Bb: 2/4 = 1/2
bb: 1/4 ✔ Correct.

Genotypes:
P1: Bb, Bb → yes, both parents heterozygous.
F1: BB, Bb, bb → yes, as shown.

✔ Problem 3 is correct.

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Problem 4:
Tall (T) dominant over dwarf (t).
Homozygous tall = TT
Homozygous dwarf = tt
Cross: TT × tt

Each parent gives only one type of allele:
TT → always gives T
tt → always gives t

Offspring: all Tt

Phenotype: since T is dominant, all are tall.

Worksheet shows:
Cross: TT x tt
P1 genotypes: TT, tt
F1 genotype(s): all Tt
F1 phenotype(s): all tall
Punnett square: all Tt ✔ Correct.

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Problem 5:
Now cross two F1 plants from Question 4 → both are Tt.

Cross: Tt × Tt

Possible gametes from each: T or t

Punnett Square:
| T | t
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T | TT | Tt
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t | Tt | tt

Genotypes: TT, Tt, Tt, tt → so ratios: 1 TT : 2 Tt : 1 tt

Phenotypes:
- TT → tall
- Tt → tall (since T dominant)
- tt → dwarf

So 3 tall : 1 dwarf → 3/4 tall, 1/4 dwarf

Worksheet says:
Cross: Tt x Tt
F1 genotypes: Tt, Tt → wait, that’s misleading. It should say the *parents* are Tt and Tt. But maybe they meant “the F1 generation being crossed are Tt and Tt”.

Then it says:
F2 genotype(s): TT, Tt, tt → yes, these are the possible outcomes.

F1 phenotype(s): 3/4 tall, 1/4 dwarf → here’s a typo! Should say F2 phenotype(s), not F1. Because we’re talking about the next generation after crossing F1s.

But the answer is still correct numerically.

Also, Punnett square matches: TT, Tt, Tt, tt → correct.

✔ So despite small wording error (“F1 phenotype” instead of “F2”), the content is right.

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Final Answer:
All problems on the worksheet are correctly solved with accurate Punnett squares, frequencies, and genotype/phenotype descriptions. Minor note: In Problem 5, “F1 phenotype(s)” should read “F2 phenotype(s)” for accuracy, but the values given (3/4 tall, 1/4 dwarf) are correct for the F2 generation.