ONLINE MIDTERMS #1

How many genetically different eggs could be formed by women with the following genotype?  Assume that all of these genes are assorted independently.

AaBbCcDdEEff

16

An organism has a diploid number of 2n = 22 chromosomes. At the end of meiotic anaphase II, what is the number of chromosomes present in each cell?

What you’re given

• 2n = 22

• That means:

  • n = 11

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The ONLY rule you need

👉 If something has its own centromere, it counts as a chromosome.

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What’s happening in anaphase II

• Sister chromatids split apart

• When they split:

  • each one becomes its own chromosome

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Count it like this

• Before splitting:
  👉 11 chromosomes, each made of 2 chromatids

• After splitting:
  👉 11 × 2 = 22 chromosomes

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So the answer is:

✅ 22 chromosomes per cell

An organism has a diploid number of 2n = 22 chromosomes. At the end of meiotic anaphase I, what is the number of chromosomes present in each cell?

Given

• 2n = 22

• n = 11

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What “end of meiotic anaphase I” REALLY means

At the end of anaphase I:

• Homologous chromosomes have already separated

• BUT

• The cell has NOT split yet

👉 There is still ONE cell, not two.

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Now count chromosomes in the cell

• Each homolog still counts as one chromosome (sister chromatids are together)

• All 22 chromosomes are still inside the same cell, just pulled toward opposite poles

So:
✅ 22 chromosomes are present in the cell

An organism has a diploid number of 2n = 22 chromosomes. At the end of mitotic anaphase, what is the number of chromosomes present in each cell?

Given

• 2n = 22

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What happens in mitotic anaphase

👉 Sister chromatids split apart

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The ONLY rule

🧠 Once chromatids split, each one counts as a chromosome

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Count it

• You start mitosis with:

  • 22 chromosomes

  • each has 2 chromatids

• In anaphase, chromatids split:

  • 22 × 2 = 44 chromosomes

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Why it’s 44

Because:

• The cell has not divided yet

• All 44 chromosomes are still inside the same cell

• They’re just being pulled to opposite sides

If your blood type is B, your sister is AB, your brother is A, and your mother is AB, “Your father’s genotype must be heterozygous — true or false?”

FALSE

Step 1: What does cream require?

From the chart:

👉 Cream = aa bb
AND
👉 Color must be allowed → C_ (CC or Cc)
(because cc = albino, no color)

So cream genotype = aa bb C_

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Step 2: Break the cross into pieces

Cross given:

AaBbCc × AabbCc

All genes assort independently → multiply probabilities.

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Step 3: Probability of aa

From Aa × Aa?
NO — careful:

• Parent 1: Aa

• Parent 2: Aa? → nope, AabbCc → Aa

So:

• Aa × Aa → aa = 1/4

✅ P(aa) = 1/4

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Step 4: Probability of bb

• Parent 1: Bb

• Parent 2: bb

Bb × bb →

• 1/2 Bb

• 1/2 bb

✅ P(bb) = 1/2

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Step 5: Probability of C_ (NOT albino)

Cc × Cc →

• 1/4 CC

• 1/2 Cc

• 1/4 cc ❌

So:

• C_ = 3/4

✅ P(C_) = 3/4

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Step 6: Multiply everything=3/32 (1/4×1/2×3/4)

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Step 7: Check the answer choices

Choices include:

• 1/64 ❌

• 1/32 ❌

• 1/16 ❌

• 1/8 ❌

• 3/16 ❌

• 1/4 ❌

• 1/2 ❌

Questions 7-10:

I^A, I^B, and i are three alleles encoding ABO blood group antigens.

A father with blood type AB and a mother with blood type B have a son with blood type B and a daughter with blood type A. 

Their daughter's genotype must be heterozygous.

T

Step 3: Is IA i heterozygous?

Yes:

• Two different alleles

• One dominant (IA), one recessive (i)

✅ Heterozygous

Questions 7-10:

I^A, I^B, and i are three alleles encoding ABO blood group antigens.

A father with blood type AB and a mother with blood type B have a son with blood type B and a daughter with blood type A. 

Their son's genotype must be homozygous.

F

Questions 7-10:

I^A, I^B, and i are three alleles encoding ABO blood group antigens.

A father with blood type AB and a mother with blood type B have a son with blood type B and a daughter with blood type A. 

 

If the couple is going to have a third child, what is the probability that this child will have AB blood type?

25%

Questions 7-10:

I^A, I^B, and i are three alleles encoding ABO blood group antigens.

A father with blood type AB and a mother with blood type B have a son with blood type B and a daughter with blood type A. 

If the couple is going to have a fourth child, it is impossible that this child will have blood type O.

T

Soft toenail is an extremely rare recessive disorder.  Mike whose aunt (his mother’s sister) has the condition marries Phoebe whose niece (her brother’s daughter) is affected by the condition.  No one else in either family has the condition.

Soft toenail is an X-linked recessive disorder.

 

F

What is the probability that Mike is carrier?

1/3

Soft toenail is an extremely rare recessive disorder.  Mike whose aunt (his mother’s sister) has the condition marries Phoebe whose niece (her brother’s daughter) is affected by the condition.  No one else in either family has the condition.

 What is the probability that Phoebe is a carrier?

1/2

Soft toenail is an extremely rare recessive disorder.  Mike whose aunt (his mother’s sister) has the condition marries Phoebe whose niece (her brother’s daughter) is affected by the condition.  No one else in either family has the condition.

 What is the probability Mike and Phoebe's first child will be affected?

Step 1: Mike’s carrier probability

• Mike’s aunt is affected → grandparents are Aa × Aa

• Mike’s mom is unaffected → carrier probability = (2/3)

• Dad is outsider → AA

• Mike gets “a” from mom with probability (1/2)

Mike carrier = (2/3) × (1/2) = (1/3)

Step 2: Phoebe’s carrier probability

• Phoebe’s niece is affected → brother must be Aa

• Brother’s spouse = AA

• Phoebe has (1/2) chance to be a carrier

Phoebe carrier = (1/2)

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Step 3: Probability both are carriers

(1/3) × (1/2) = (1/6)

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Step 4: Probability child is affected if both are carriers

• Aa × Aa → aa = (1/4)

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Step 5: Final probability

(1/6) × (1/4) = (1/24)

Soft toenail is an extremely rare recessive disorder.  Mike whose aunt (his mother’s sister) has the condition marries Phoebe whose niece (her brother’s daughter) is affected by the condition.  No one else in either family has the condition.

 If their first child has soft toenails, what is the probability that their second child will also be affected by the condition?

If their first child is affected by a recessive disorder, then:

👉 Both parents MUST be carriers (Aa)
No probabilities anymore — this is now 100% certain.

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So now the parents’ genotypes are fixed:

• Mike = Aa

• Phoebe = Aa

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Step 1: Probability second child is affected

For Aa × Aa:

• aa (affected) = (1/4)

Each child is an independent event, so the first child being affected does not change the odds for the second.

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✅ Final Answer:

(1/4)

Soft toenail is an extremely rare recessive disorder.  Mike whose aunt (his mother’s sister) has the condition marries Phoebe whose niece (her brother’s daughter) is affected by the condition.  No one else in either family has the condition.

 If they have four children, what is the probability that all four will have the same sex?

Key idea

• Each child has:

  • (1/2) chance male

  • (1/2) chance female

• Sex of each child is independent

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“All four same sex” means two possible cases

1. All boys

2. All girls

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Case 1: All boys(1/2)×(1/2)×(1/2)×(1/2)=(1/16)(1/2)×(1/2)×(1/2)×(1/2)=(1/16)Case 2: All girls(1/2)×(1/2)×(1/2)×(1/2)=(1/16)(1/2)×(1/2)×(1/2)×(1/2)=(1/16)

ADD THEM: 1/8

Soft toenail is an extremely rare recessive disorder.  Mike whose aunt (his mother’s sister) has the condition marries Phoebe whose niece (her brother’s daughter) is affected by the condition.  No one else in either family has the condition.

 What is the probability that their first child will be homozygous for the normal allele?

1. Mike’s Probability

Mike’s aunt (mother’s sister) has the condition (aa).

• Mike's Maternal Grandparents: Since they had an affected daughter (aa) but are not affected themselves, they must both be carriers (Aa).

• Mike's Mother: She is not affected, but her parents are Aa x Aa. The probability she is a carrier (Aa) is 2/3 (we exclude the aa possibility since she is healthy).

• Mike: If his mother is Aa, and we assume his father is AA (because the condition is "extremely rare"), Mike has a 1/2 chance of inheriting the a allele from her.

• Total probability Mike is Aa: 2/3×1/2=1/3.

• Probability Mike is AA: 1−1/3=2/3.

2. Phoebe’s Probability

Phoebe’s niece (her brother’s daughter) is affected (aa).

• Phoebe's Brother: For him to have an affected daughter, he must be a carrier (Aa).

• Phoebe's Parents: For her brother to be Aa (and assuming the niece's mother provided the other a), at least one parent must carry the allele. Since no one else is affected, we assume her parents are Aa x AA.

• Phoebe: In a cross of Aa x AA, Phoebe has a 1/2 chance of being Aa and a 1/2 chance of being AA.

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3. Calculating the Child's Genotype

To find the probability the child is AA, we look at the four possible parent combinations:

Scenario	Parent Genotypes	Probability of Scenario	Prob. Child is AA	Weighted Prob.
1	Mike AA, Phoebe AA	2/3×1/2=2/6	1	2/6
2	Mike AA, Phoebe Aa	2/3×1/2=2/6	1/2	1/6
3	Mike Aa, Phoebe AA	1/3×1/2=1/6	1/2	1/12
4	Mike Aa, Phoebe Aa	1/3×1/2=1/6	1/4	1/24

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Final Addition:

To get the total probability, we sum the weighted probabilities:

2/6+1/6+1/12+1/24=15/24

The gene for knuckle shape is found on the X chromosome. Straight is dominant to curled.  Assume recombination does not occur.  For each of the following families, identify the possible nondisjunction events (rare mistakes during meiosis) that could explain the phenotype of the offspring.

A man with curled knuckles and a woman with straight knuckles have a son with Klinefelter Syndrome (genotype XXY) and curled knuckles. Non-disjunction might have occurred in Meiosis I of the mother (the XX parent).

F

The gene for knuckle shape is found on the X chromosome. Straight is dominant to curled.  Assume recombination does not occur.  For each of the following families, identify the possible nondisjunction events (rare mistakes during meiosis) that could explain the phenotype of the offspring.

A man with curled knuckles and a woman with straight knuckles have a son with Klinefelter Syndrome (genotype XXY) and curled knuckles. Non-disjunction might have occurred in Meiosis I of the father (the XY parent).

T

The gene for knuckle shape is found on the X chromosome. Straight is dominant to curled.  Assume recombination does not occur.  For each of the following families, identify the possible nondisjunction events (rare mistakes during meiosis) that could explain the phenotype of the offspring.

A man with straight knuckles and a woman with curled knuckles have a son with Klinefelter Syndrome (genotype XXY) and curled knuckles. Non-disjunction might have occurred in Meiosis I of the mother (the XX parent).

T

The gene for knuckle shape is found on the X chromosome. Straight is dominant to curled.  Assume recombination does not occur.  For each of the following families, identify the possible nondisjunction events (rare mistakes during meiosis) that could explain the phenotype of the offspring.

A man with straight knuckles and a woman with curled knuckles have a son with Klinefelter Syndrome (genotype XXY) and curled knuckles. Non-disjunction might have occurred in Meiosis I of the father (the XY parent).

F

In Pleik-pleik flies, antenna shape is controlled by genes A and B.  They are two independently assorting autosomal genes.  Gene A has two alleles-dominant A and recessive a.  Gene B also has two alleles-dominant B and recessive b.  To form normal antenna, a fly needs both dominant alleles of genes A and B.  Flies homozygous for either recessive aa or bb will have abnormal antenna.  In addition, fly embryos simultaneously homozygous for both aa and bb are lethal. 

Two abnormal true-breeding flies mated.  All F₁ flies had normal antenna.  F₁ were allowed to intercross, and 960 F₂ were produced.

Among 960 F₂ flies, 384 of them would have abnormal antenna.

Step 1: Translate the rules into plain language

• Normal antenna → needs at least one A AND at least one B

  • genotype: A_ B_

• Abnormal antenna →

  • aa B_ OR A_ bb

• Lethal (dead) →

  • aa bb (never seen among F2 flies)

Step 2: What were the parents?

Two abnormal true-breeding flies mate
All F1 are normal

That tells us the parents must be:

• Parent 1: aa BB (abnormal because aa)

• Parent 2: AA bb (abnormal because bb)

These are true-breeding and complementary.

F1 Generation: aa BB * AAbb

F1: AaBb

They have:

• at least one A

• at least one B

✅ So all F1 are normal (matches the problem)

Step 4: F1 × F1 cross

Now cross: dihybrid cross (AaBb * AaBb)

A standard dihybrid cross gives 16 genotype combinations.

Step 5: Sort the 16 outcomes by phenotypeNormal antenna (A_ B_)

• 9/16

Abnormal antenna (aa B_ or A_ bb)

• 3/16 (aa B_)

• 3/16 (A_ bb)
  ➡ total 6/16

Lethal (aa bb)

• 1/16 ❌ (dies, not counted)

Step 6: Remove the lethal class

Only 15/16 of embryos survive.

Among survivors:

• Abnormal = 6

• Total survivors = 15

So:

Abnormal fraction=(6/15)Abnormal fraction=(6/15)

Step 7: Apply to 960 F2 flies 960×(6/15)

Simplify:

• 960 ÷ 15 = 64

• 64 × 6 = 384

In Pleik-pleik flies, antenna shape is controlled by genes A and B.  They are two independently assorting autosomal genes.  Gene A has two alleles-dominant A and recessive a.  Gene B also has two alleles-dominant B and recessive b.  To form normal antenna, a fly needs both dominant alleles of genes A and B.  Flies homozygous for either recessive aa or bb will have abnormal antenna.  In addition, fly embryos simultaneously homozygous for both aa and bb are lethal. 

Two abnormal true-breeding flies mated.  All F₁ flies had normal antenna.  F₁ were allowed to intercross, and 960 F₂ were produced.

The F₂ phenotypic ratio is 3 normal: 2 abnormal.

Step 1: Start from the F1 intercross

From earlier, we know:

• Parents: aa BB × AA bb

• F1: Aa Bb (all normal)

So F1 intercross: AaBb * AaBb

Step 2: Standard dihybrid outcomes (before lethality)

Out of 16 total:

• 9/16 → A_ B_ → normal

• 3/16 → aa B_ → abnormal

• 3/16 → A_ bb → abnormal

• 1/16 → aa bb → lethal ❌

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Step 3: Remove the lethal class

The aa bb (1/16) die, so they are not counted in the F2.

That leaves:

• 9 normal

• 6 abnormal

Total surviving = 15

Normal to Abnormal ratio 9:6 or 3:2

F2 phenotypic ratio = 3 normal : 2 abnormal

In Pleik-pleik flies, antenna shape is controlled by genes A and B.  They are two independently assorting autosomal genes.  Gene A has two alleles-dominant A and recessive a.  Gene B also has two alleles-dominant B and recessive b.  To form normal antenna, a fly needs both dominant alleles of genes A and B.  Flies homozygous for either recessive aa or bb will have abnormal antenna.  In addition, fly embryos simultaneously homozygous for both aa and bb are lethal. 

Two abnormal true-breeding flies mated.  All F₁ flies had normal antenna.  F₁ were allowed to intercross, and 960 F₂ were produced.

What proportion of abnormal F₂ would be true breeding? 

Step 1: Recall the F1 cross

From earlier:

Parents:

aaBB * AAbb

F1: AaBb

F1 cross: AaBb * AaBb

Step 2: List the abnormal F2 genotypes

Abnormal antenna = aa B_ or A_ bb

From a dihybrid cross:

aa B_ (missing A)

• aa BB → true-breeding abnormal

• aa Bb → NOT true-breeding

Counts:

• aa BB = 1

• aa Bb = 2

A_ bb (missing B)

• AA bb → true-breeding abnormal

• Aa bb → NOT true-breeding

Counts:

• AA bb = 1

• Aa bb = 2

Step 3: Total abnormal genotypes (alive only)

Remember:

• aa bb is lethal → ignore it

So abnormal F2 (living):

Genotype	Count
aa BB	1
aa Bb	2
AA bb	1
Aa bb	2

Total abnormal = 6

Step 4: Which abnormal are true-breeding?

True-breeding means homozygous at both loci.

True-breeding abnormal genotypes:

• aa BB

• AA bb

That’s: 2/6 or 1/3

In Pleik-pleik flies, antenna shape is controlled by genes A and B.  They are two independently assorting autosomal genes.  Gene A has two alleles-dominant A and recessive a.  Gene B also has two alleles-dominant B and recessive b.  To form normal antenna, a fly needs both dominant alleles of genes A and B.  Flies homozygous for either recessive aa or bb will have abnormal antenna.  In addition, fly embryos simultaneously homozygous for both aa and bb are lethal. 

Two abnormal true-breeding flies mated.  All F₁ flies had normal antenna.  F₁ were allowed to intercross, and 960 F₂ were produced.

What proportion of normal F₂ would be true breeding? 

Step 1: Recall the F1 cross

From earlier:

Parents:

aaBB * AAbb

F1: AaBb

F1 cross: AaBb * AaBb

Step 2: Identify normal F2

Normal antenna requires A_ B_.

From a standard dihybrid cross:

• A_ B_ = 9/16 (before considering lethality)

The aa bb (1/16) are lethal, but note:
👉 aa bb are NOT normal anyway, so lethality does not change the makeup of the normal group.

So we still have 9 normal genotypes to consider.

Step 3: List the normal genotypes

Normal genotypes (A_ B_):

Genotype	Count
AA BB	1
AA Bb	2
Aa BB	2
Aa Bb	4

Total normal = 9

Step 4: Which normal ones are true breeding?

True breeding means homozygous at BOTH loci.

Among the normal genotypes, only:

• AA BB

That’s 1 genotype.

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Step 5: Form the proportion: true-breeding normal/total normal=1/9

A true-breeding black-flowered plant of tulip was cross-fertilized with a true-breeding yellow plant, and the F₁ progeny were all black.  When the F₁plants were allowed to self-fertilize.  Among F₂ plants, there are 252 black, 63 yellow, and 21 silver-flowered plants.

Flower colors of tulip are determined by two genes.

Step 1: Look at the F2 numbers

You’re given:

• 252 black

• 63 yellow

• 21 silver

First thing to do in any genetics problem like this 👉 check the ratio.

Divide everything by the smallest number (21):

• 252 ÷ 21 = 12

• 63 ÷ 21 = 3

• 21 ÷ 21 = 1

So the ratio is:

12 : 3 : 1

🚨 This is the key clue.

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Step 2: What does a 12 : 3 : 1 ratio mean?

A 12:3:1 phenotypic ratio is the textbook signature of dominant epistasis involving two genes.

A true-breeding black-flowered plant of tulip was cross-fertilized with a true-breeding yellow plant, and the F₁ progeny were all black.  When the F₁plants were allowed to self-fertilize.  Among F₂ plants, there are 252 black, 63 yellow, and 21 silver-flowered plants.

All silver-flowered plants must be homozygous for color gene(s).

Step 1: Use the F2 numbers to identify the pattern

You already saw:

• 252 black

• 63 yellow

• 21 silver

Divide by 21:

12 : 3 : 1

That ratio tells us this is dominant epistasis involving two genes.

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Step 2: Set up the gene model (the only one that fits 12:3:1)

Let the two genes be A and B.

A standard dominant-epistasis setup that fits the data is:

• A_ _ → black

• aa B_ → yellow

• aa bb → silver

Key idea:
👉 Silver only appears when BOTH genes are recessive

Step 3: Why silver MUST be homozygous

Silver genotype = aa bb

That means:

• Homozygous recessive at gene A

• Homozygous recessive at gene B

There is no heterozygous genotype that produces silver:

• Aa bb → black (A masks everything)

• aa Bb → yellow

• Aa Bb → black

So silver shows up only when both loci are homozygous recessive.

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Step 4: Check this against the F2 ratio

In a dihybrid self-cross (AaBb × AaBb):

• aa bb occurs 1/16

• That exactly matches the “1” in the 12:3:1 ratio

So every silver plant:

• came from the 1/16 class

• must be aa bb

• therefore must be homozygous for all color genes

A true-breeding black-flowered plant of tulip was cross-fertilized with a true-breeding yellow plant, and the F₁ progeny were all black.  When the F₁plants were allowed to self-fertilize.  Among F₂ plants, there are 252 black, 63 yellow, and 21 silver-flowered plants.

It is impossible to produce a black offspring from a cross between a yellow and a silver parent.

Step 1: Lock in the gene model (from the 12:3:1 ratio)

From the F2 numbers
252 : 63 : 21 → 12 : 3 : 1

This forces a dominant-epistasis model with two genes:

• A_ _ → black (dominant epistasis)

• aa B_ → yellow

• aa bb → silver

There is no other model that gives 12:3:1.

Step 2: Write the genotypes of the parents in the statementYellow parent

Yellow = aa B_
So yellow can be:

• aa BB or

• aa Bb

Silver parent

Silver = aa bb
(always homozygous recessive at both genes)

Step 3: Cross yellow × silver

Let’s cross aa B_ × aa bb

Gene A

• Yellow: aa

• Silver: aa

So all offspring are aa
🚨 This is the key point.

There is no way to get an A allele.

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Gene B

• Yellow: B_

• Silver: bb

Offspring can be:

• aa Bb or

• aa bb

Step 4: Determine offspring phenotypes

Recall:

• Black requires A_ ❌

• Yellow = aa B_

• Silver = aa bb

Since all offspring are aa, black is impossible.

Possible offspring:

• aa Bb → yellow

• aa bb → silver

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✅ Final Answer (why the statement is true)

It is impossible to produce a black offspring from a yellow × silver cross because:

👉 Both parents are aa, so no offspring can receive a dominant A allele, and black requires A_.

A true-breeding black-flowered plant of tulip was cross-fertilized with a true-breeding yellow plant, and the F₁ progeny were all black.  When the F₁plants were allowed to self-fertilize.  Among F₂ plants, there are 252 black, 63 yellow, and 21 silver-flowered plants.

There are six possible different genotypes among those 252 F₂ black plant.

Step 1: Lock in the gene → phenotype rules (from earlier)

From the 12 : 3 : 1 ratio, we know the model is:

• Black → A_ _ (dominant A masks everything)

• Yellow → aa B_

• Silver → aa bb

So for a plant to be black, it must have at least one A.
Gene B can be anything.

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Step 2: List ALL genotypes that give black

We’ll list possibilities by gene A first.

Gene A options for black

• AA

• Aa

Now combine each with all possible B genotypes.

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Case 1: AA __

Possible B genotypes:

• BB

• Bb

• bb

So we get:

1. AA BB

2. AA Bb

3. AA bb

(All are black because A is present.)

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Case 2: Aa __

Possible B genotypes:

• BB

• Bb

• bb

So we get:
4. Aa BB
5. Aa Bb
6. Aa bb

(All are black because A is present.)

There are six possible different genotypes among the 252 black F2 plants because any genotype with at least one A allele produces black, and gene B can be BB, Bb, or bb.

A true-breeding black-flowered plant of tulip was cross-fertilized with a true-breeding yellow plant, and the F₁ progeny were all black.  When the F₁plants were allowed to self-fertilize.  Among F₂ plants, there are 252 black, 63 yellow, and 21 silver-flowered plants.

Among 63 F₂ yellow plants, how many are true breeding?

Step 1: Recall the gene → color rules (from the 12 : 3 : 1 ratio)

From earlier, the only model that fits is:

• Black → A_ _

• Yellow → aa B_

• Silver → aa bb

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Step 2: What genotypes can be yellow?

Yellow requires:

• aa (homozygous recessive at gene A)

• at least one B

So yellow genotypes are:

1. aa BB → yellow (true breeding)

2. aa Bb → yellow (not true breeding)

Only aa BB is true breeding.

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Step 3: Ratio of yellow genotypes in F2

From the F1 self-cross (AaBb × AaBb):

Among the yellow class (aa B_), the breakdown is:

• aa BB = 1

• aa Bb = 2

So the ratio within yellow is:

1 true breeding : 2 not true breeding

That means:

• 1/3 of yellow plants are true breeding

• 2/3 are not

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Step 4: Apply this to the 63 yellow plants: 63 × 1/3=21

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✅ Final Answer

21 of the 63 yellow F2 plants are true breeding

A true-breeding black-flowered plant of tulip was cross-fertilized with a true-breeding yellow plant, and the F₁ progeny were all black.  When the F₁plants were allowed to self-fertilize.  Among F₂ plants, there are 252 black, 63 yellow, and 21 silver-flowered plants.

Among 252 F₂ black plants, how many are true breeding?

Step 1: Lock in the color rules (from the 12 : 3 : 1 ratio)

We already know the only model that fits is:

• Black → A_ _

• Yellow → aa B_

• Silver → aa bb

So any genotype with at least one A is black.

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Step 2: List ALL black genotypes in the F2

From the F1 self-cross (AaBb × AaBb), the black genotypes are:

Genotype	Count
AA BB	1
AA Bb	2
AA bb	1
Aa BB	2
Aa Bb	4
Aa bb	2

Total black = 12 (this is the “12” in the 12:3:1 ratio)

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Step 3: Which black genotypes are true breeding?

True breeding = homozygous at BOTH genes.

Among the black genotypes, that means:

• AA BB ✅

• AA bb ✅

These are the only ones that will always produce black offspring when selfed.

So:

• True-breeding black = 2

• Total black = 12

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Step 4: Convert to a proportion 2/12:1/6

So 1/6 of black F2 plants are true breeding.

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Step 5: Apply to the 252 black plants: 252×1/6=42

What is the genotype of I-2?

I^Bi

What is the genotype of I-4?

I^Ai

What is the genotype of III-4?

I^Ai

What is the genotype of II-5?

I^AI^B

What is the genotype of II-3?

I^Bi

What is the genotype of II-2?

I^Bi

What is the genotype of III-1?

or more

more

What is the probability that IV-1's phenotype is O.

• Mom has IB and i. Probability she gives i is 50% (1/2).

• Dad has IA (or IB) and i. Probability he gives i is 50% (1/2).

• For the baby to be ii, both "coin flips" must land on i.

• 1/2×1/2=1/4.

What is the probability that IV-1's blood type is AB?

1/8