This biology worksheet challenges students to complete Punnett squares for various monohybrid crosses involving guinea pigs, cattle, and humans.

Here are the solutions to the problems on your worksheet, broken down step-by-step.

16. Guinea Pig Hair Length

Key: Short hair ($S$) is dominant, Long hair ($s$) is recessive.

a. One guinea pig is $Ss$ and one is $ss$.
* The Cross: You put the alleles of one parent across the top ($S$, $s$) and the other down the side ($s$, $s$).
* The Squares:
* Top-Left: $S$ meets $s$ $\rightarrow$ $Ss$ (Short hair)
* Top-Right: $s$ meets $s$ $\rightarrow$ $ss$ (Long hair)
* Bottom-Left: $S$ meets $s$ $\rightarrow$ $Ss$ (Short hair)
* Bottom-Right: $s$ meets $s$ $\rightarrow$ $ss$ (Long hair)
* Results: You have two $Ss$ and two $ss$.

b. Both guinea pigs are heterozygous for short hair.
* "Heterozygous" means they have different alleles, so both parents are $Ss$.
* The Cross: Put $S, s$ across the top and $S, s$ down the side.
* The Squares:
* Top-Left: $S$ meets $S$ $\rightarrow$ $SS$ (Short hair)
* Top-Right: $s$ meets $S$ $\rightarrow$ $Ss$ (Short hair)
* Bottom-Left: $S$ meets $s$ $\rightarrow$ $Ss$ (Short hair)
* Bottom-Right: $s$ meets $s$ $\rightarrow$ $ss$ (Long hair)
* Results: Three squares result in Short hair ($SS$ or $Ss$), and one square results in Long hair ($ss$).

---

Part C: Monohybrid Cross Problems

17. Hornless vs. Horned Cattle
* Key: Hornless ($H$) is dominant, Horned ($h$) is recessive.
* Parents:
* Homozygous hornless bull: This means two dominant alleles $\rightarrow$ $HH$.
* Homozygous horned cow: This means two recessive alleles $\rightarrow$ $hh$.
* The Cross: Every offspring gets an $H$ from the dad and an $h$ from the mom.
* Result: All babies will be $Hh$. Because they have the big $H$, they will look hornless.

18. Tongue Rolling in Humans
* Key: Roller ($R$) is dominant, Non-roller ($r$) is recessive. *(Note: The problem uses a lowercase 't' for non-roller, but usually we match the letter. I will use $R$ and $r$ for clarity, treating 't' as the recessive allele).*
* Identify Parents:
* Father: He is a "non-roller." Since this is recessive, he must have two small letters. Genotype: $rr$ (or $tt$). Phenotype: Non-roller.
* Mother: She is "heterozygous." This means she has one of each. Genotype: $Rr$ (or $Rt$). Phenotype: Tongue roller (because she has the dominant $R$).
* The Punnett Square:
* Father ($rr$) goes on the side. Mother ($Rr$) goes on top.
* Box 1: $R$ + $r$ = $Rr$ (Roller)
* Box 2: $r$ + $r$ = $rr$ (Non-roller)
* Box 3: $R$ + $r$ = $Rr$ (Roller)
* Box 4: $r$ + $r$ = $rr$ (Non-roller)
* Probability: 2 out of the 4 boxes are Rollers ($Rr$). That is a 50% chance.

──────────────────────────────────────

Final Answer:

16a.
* Punnett Square contents: Top row: $Ss$, $ss$. Bottom row: $Ss$, $ss$.
* Expected number of offspring:
* Short hair ($SS$ or $Ss$): 2
* Long hair ($ss$): 2

16b.
* Punnett Square contents: Top row: $SS$, $Ss$. Bottom row: $Ss$, $ss$.
* Expected number of offspring:
* Short hair: 3
* Long hair: 1

17.
* Genotype of first generation: 100% $Hh$
* Phenotype of first generation: 100% Hornless

18.
* Father’s phenotype: Non-roller
* Father’s genotype: $rr$ (or $tt$)
* Mother’s phenotype: Tongue roller
* Mother’s genotype: $Rr$ (or $Rt$)
* Punnett Square contents: Two boxes are $Rr$ (or $Rt$), two boxes are $rr$ (or $tt$).
* Probability of having a child who is a tongue roller: 50% (or 1/2)