💡 7th Grade Life Science: Genetics and Punnett Squares Practice Questions

Here's how to solve this genetics problem step-by-step:

• Step 1: Determine the genotypes of the parents.
  • A homozygous tall pea plant has two dominant alleles: \(TT\).
  • A homozygous short pea plant has two recessive alleles: \(tt\).
• Step 2: Set up the Punnett Square.
  We'll place the alleles from one parent along the top and the alleles from the other parent along the side. \[ \begin{array}{|c|c|c|} & T & T \\ t & Tt & Tt \\ t & Tt & Tt \\ \end{array} \]
• Step 3: Determine the genotypes of the offspring.
  • All four squares contain the genotype \(Tt\).
  • So, the genotype ratio is 100% \(Tt\).
• Step 4: Determine the phenotypes of the offspring.
  • Since tallness (T) is dominant over shortness (t), any plant with at least one 'T' allele will be tall.
  • All offspring have the \(Tt\) genotype, which means they will all be tall.
  • So, the phenotype ratio is 100% Tall.

✅ Answer:
The genotypes of the F1 offspring will all be \(Tt\).
The phenotypes of the F1 offspring will all be Tall.

Let's break down this cross:

• Step 1: Identify the genotypes of the parents.
  • Both parents are heterozygous tall, so their genotypes are \(Tt\) and \(Tt\).
• Step 2: Construct the Punnett Square.
  \[ \begin{array}{|c|c|c|} & T & t \\ T & TT & Tt \\ t & Tt & tt \\ \end{array} \]
• Step 3: Calculate the genotypic ratio.
  • One square has \(TT\).
  • Two squares have \(Tt\).
  • One square has \(tt\).
  • The genotypic ratio is \(1 TT : 2 Tt : 1 tt\).
• Step 4: Calculate the phenotypic ratio.
  • Plants with \(TT\) and \(Tt\) genotypes will be Tall (because T is dominant). There are 3 such plants.
  • Plants with the \(tt\) genotype will be Short. There is 1 such plant.
  • The phenotypic ratio is 3 Tall : 1 Short.

✅ Answer:
The predicted genotypic ratio is \(1 TT : 2 Tt : 1 tt\).
The predicted phenotypic ratio is 3 Tall : 1 Short.

Let's determine the fur color probabilities for the kittens:

• Step 1: Write down the genotypes of the parent cats.
  • Heterozygous black cat: \(Bb\)
  • White cat: \(bb\) (Since white is recessive, the cat must have two recessive alleles to show the trait).
• Step 2: Draw the Punnett Square.
  \[ \begin{array}{|c|c|c|} & B & b \\ b & Bb & bb \\ b & Bb & bb \\ \end{array} \]
• Step 3: Analyze the offspring genotypes.
  • Two squares show \(Bb\).
  • Two squares show \(bb\).
• Step 4: Determine the phenotypes and percentages.
  • Kittens with \(Bb\) genotype will have black fur (because B is dominant). This accounts for 2 out of 4 squares, or 50%.
  • Kittens with \(bb\) genotype will have white fur. This also accounts for 2 out of 4 squares, or 50%.

✅ Answer:
50% of the kittens are expected to have black fur.

Let's uncover the parents' genotypes:

• Step 1: Identify the trait and its alleles.
  • Bumpy skin (B) is dominant.
  • Smooth skin (b) is recessive.
• Step 2: Determine the genotype of the smooth-skinned offspring.
  • Since smooth skin is a recessive trait, the offspring must have two recessive alleles to show this trait. Therefore, the offspring's genotype is \(bb\).
• Step 3: Infer the parents' alleles.
  • For the offspring to have the genotype \(bb\), it must inherit one 'b' allele from each parent.
• Step 4: Consider the parents' phenotypes.
  • Both parents have bumpy skin, which means they must have at least one dominant 'B' allele.
• Step 5: Combine the information to find the parents' genotypes.
  • Each parent has bumpy skin (so they have B) AND each parent passed on a 'b' allele to their smooth-skinned offspring.
  • Therefore, both parents must carry both a dominant 'B' allele and a recessive 'b' allele.
  • This means both parents are heterozygous: \(Bb\).

✅ Answer:
Both bumpy-skinned parents must have the genotype \(Bb\) (heterozygous). This is because for them to have a smooth-skinned (\(bb\)) offspring, each parent must contribute a recessive 'b' allele. Since they themselves show the dominant bumpy skin trait, they must also possess a dominant 'B' allele.

Let's figure out the probability for their child:

• Step 1: Determine the man's genotype.
  • The man can roll his tongue, so he has at least one 'R' allele. His genotype is either \(RR\) or \(Rr\).
  • His mother cannot roll her tongue, meaning her genotype is \(rr\).
  • Since the man inherited one allele from his mother, he must have inherited a 'r' allele from her.
  • Therefore, the man's genotype must be \(Rr\).
• Step 2: Determine the woman's genotype.
  • The woman cannot roll her tongue. Since this is a recessive trait, her genotype must be \(rr\).
• Step 3: Set up the Punnett Square for the cross \(Rr \times rr\).
  \[ \begin{array}{|c|c|c|} & R & r \\ r & Rr & rr \\ r & Rr & rr \\ \end{array} \]
• Step 4: Calculate the probability for the child.
  • Out of the four possible outcomes in the Punnett Square:
  • Two are \(Rr\) (can roll tongue).
  • Two are \(rr\) (cannot roll tongue).
  • The probability of their child being able to roll their tongue (\(Rr\)) is 2 out of 4, or 50%.

✅ Answer:
There is a 50% probability that their first child will be able to roll their tongue.

Let's deduce the parent genotypes based on the offspring:

• Step 1: Analyze the offspring phenotypes.
  • The appearance of smooth-furred offspring (recessive trait) is the key clue.
  • For an offspring to have smooth fur, its genotype must be \(rr\).
• Step 2: Determine what this means for the parents.
  • If an offspring is \(rr\), it must have received one 'r' allele from each parent.
• Step 3: Consider the parents' phenotypes.
  • Both parents have rough fur, which is a dominant trait. This means each parent must have at least one 'R' allele.
• Step 4: Combine the information.
  • Each parent has rough fur (so they have R) AND each parent passed on a 'r' allele to their smooth-furred offspring.
  • Therefore, both parents must be heterozygous: \(Rr\).
• Step 5: Verify with a Punnett Square (\(Rr \times Rr\)).
  \[ \begin{array}{|c|c|c|} & R & r \\ R & RR & Rr \\ r & Rr & rr \\ \end{array} \] This cross predicts a phenotypic ratio of 3 Rough : 1 Smooth. The observed ratio of 7 Rough : 3 Smooth (approximately 70% Rough, 30% Smooth) is very close to the expected 75% Rough, 25% Smooth, confirming that both parents were heterozygous.

✅ Answer:
Both parent guinea pigs must have been heterozygous, with the genotype \(Rr\).

Let's deduce the genotype of the blue-winged parent:

• Step 1: Identify the genotypes of the parents based on their phenotypes.
  • Yellow-winged butterfly: Since yellow is recessive, its genotype must be \(bb\).
  • Blue-winged butterfly: Since blue is dominant, its genotype is either \(BB\) or \(Bb\).
• Step 2: Analyze the offspring.
  • All offspring have blue wings. This means all offspring have at least one 'B' allele.
• Step 3: Consider possible crosses.
  • Option 1: If the blue-winged parent was heterozygous (\(Bb\)).
    • Cross: \(Bb \times bb\)
    • Punnett Square: \[ \begin{array}{|c|c|c|} & B & b \\ b & Bb & bb \\ b & Bb & bb \\ \end{array} \]
    • This cross would produce 50% blue-winged (\(Bb\)) and 50% yellow-winged (\(bb\)) offspring. This contradicts the observation that all offspring have blue wings.
  • Option 2: If the blue-winged parent was homozygous dominant (\(BB\)).
    • Cross: \(BB \times bb\)
    • Punnett Square: \[ \begin{array}{|c|c|c|} & B & B \\ b & Bb & Bb \\ b & Bb & Bb \\ \end{array} \]
    • This cross would produce 100% blue-winged (\(Bb\)) offspring. This matches the observation!

✅ Answer:
The most likely genotype of the blue-winged parent is \(BB\) (homozygous dominant). If the blue-winged parent were heterozygous (\(Bb\)), then a cross with a yellow-winged parent (\(bb\)) would result in 50% blue-winged and 50% yellow-winged offspring. Since all offspring were blue-winged, the blue-winged parent must have passed on a 'B' allele to every single offspring, which only happens if it is \(BB\).

Let's determine the parents' genotypes based on their puppies:

• Step 1: Identify the genotypes of the parents based on their phenotypes.
  • Female dog has normal ears. Since normal ears (f) are recessive, her genotype must be \(ff\).
  • Male dog has floppy ears. Since floppy ears (F) are dominant, his genotype is either \(FF\) or \(Ff\).
• Step 2: Analyze the offspring.
  • Half the puppies have floppy ears and half have normal ears.
  • For a puppy to have normal ears, its genotype must be \(ff\).
• Step 3: Deduce the male parent's genotype.
  • The female parent is \(ff\), so she can only pass on an 'f' allele to her offspring.
  • For a puppy to be \(ff\) (normal ears), it must receive one 'f' from the mother and one 'f' from the father.
  • Since some puppies are \(ff\), the floppy-eared male dog must have contributed an 'f' allele.
  • Because the male dog also shows the dominant floppy ear trait, his genotype must be \(Ff\).
• Step 4: Verify with a Punnett Square (\(Ff \times ff\)).
  \[ \begin{array}{|c|c|c|} & F & f \\ f & Ff & ff \\ f & Ff & ff \\ \end{array} \] This cross predicts a 1:1 ratio of floppy ears (\(Ff\)) to normal ears (\(ff\)), which perfectly matches the observed litter (half floppy, half normal).

✅ Answer:
The male dog's genotype is \(Ff\) (heterozygous floppy ears).
The female dog's genotype is \(ff\) (homozygous normal ears).