Biology : asked on hayden5928
 23.04.2023

Determine what each letter represents in the Punnett square.

Letters A and B represent the genotypes of the

. 14

Step-by-step answer

17.02.2022, solved by verified expert
Unlock the full answer
15 students found this answer . helpful

Their are two Questions in my quiz instead of only one so here Are the answers. Determine what each letter represents in the Punnett square.

Letters A and B represent the genotypes of the parents.

Letters C, D, E, and F represent the genotypes of the Offspring.

Explanation:

Just did the quiz so ya

It is was helpful?

Faq

Biology
Step-by-step answer
P Answered by PhD
Letters A and B represent the genotypes of the Parents Letters C, D, E, and F represent the genotypes of the Offspring.

Explanation:

The punnet square is a square shaped figure or diagram used to anticipate the genotypes of offspring after a certain parental cross. It helps in determining the genotype as well as phenotype of the offspring if we know the genotype of parents.

The figure was devised for the very first time by Reginald C. Punnett and named after him. The square diagram can also be considered as a summary of parental and offspring alleles.It is important to understand the homozygosity and heterozygosity before using the punnet square in proper way.

Hope it help!

Biology
Step-by-step answer
P Answered by PhD
Letters A and B represent the genotypes of the Parents Letters C, D, E, and F represent the genotypes of the Offspring.

Explanation:

The punnet square is a square shaped figure or diagram used to anticipate the genotypes of offspring after a certain parental cross. It helps in determining the genotype as well as phenotype of the offspring if we know the genotype of parents.

The figure was devised for the very first time by Reginald C. Punnett and named after him. The square diagram can also be considered as a summary of parental and offspring alleles.It is important to understand the homozygosity and heterozygosity before using the punnet square in proper way.

Hope it help!

Biology
Step-by-step answer
P Answered by PhD

1. It involves multiple trials

2. 2 red: 2 white

3. Process 1 describes cloning and process 2 describes artificial selection

4. Furry Feet

5. Large scale insulin could be produced in relatively lesser time

6. A lowercase letter

7. 0%

8. rr

9. 50%

10. The allele for white flowers is recessive in the pea plant

Explanation:

1. Repetition, this means to do over and over. This would involve multiple trials and this is done to test the reliability of the results and determine if the hypothesis holds true or not.

2. If you put your cross in a Punnett square it would look like this:

      R    r                            

r     Rr   rr                          

r     Rr   rr

Based on your problem: R = red; r = white

If the genotype would have a dominant allele (R), then the dominant trait will come out, if it has only recessive alleles (r) then the recessive trait will come out. So as you can see there are 2 red and 2 white. So the ratio would be 2:2 or 1:1.

3. Process 1 is considered as cloning, and to be more specific, reproductive cloning. The aim of this is to produce an offspring that has the same genetic make up as the original or parent organism. Process 2 is an example of artificial selection. This is also known as selective breeding where breeders choose certain traits to pass on to the off springs by selecting the organisms that have the traits desired.

4. Furry feet of polar bears provide them extra traction when they walk on snow. The fur also provides extra warmth. They also have foot pads on the soles of their feet that help them as well. Together, polar bear have something that works like snow boots.

5. Large-scale insulin could be produced in relatively lesser time because bacteria reproduce through asexual means. They produce copies of themselves that are genetically identical. An advantage of asexual reproduction is the process occurs in a much faster rate. With that said, insulin supply would increase, which will help many diabetic patients.

6. Recessive alleles in a genotype are represented by a lowercase letter. On the other hand, dominant alleles are represented by a capital letter. Squares and circles are used in pedigree charts to represent the whole genotype of an organism for a specific trait.

7. The answer would be 0% because cystic fibrosis is a RECESSIVE gene disorder. For it to be expressed or observable in an individual , they should have two recessive alleles. As you can see in the end result, none of the offsprings have 2 recessive alleles. The genotype Aa, has a recessive allele, but it will not be expressed because the dominant trait will mask it.

8. When we talk about homozygous, this means the alleles are the same. And when you say heterozygous, this means that the organism would have a dominant and recessive allele. the recessive llele is represented by a lowercase letter. So if the organism is homozygous recessive, the genotype would have two lowercase letters. (rr)

9. The scenario shows the genotype is pp. This means that the pea plant is homozygous recessive. As explained before, recessive traits are represented by lowercase letters. So this would mean that the white color is a recessive trait. The answer would then be the allele for white flowers is recessive in the pea plant.

10. Mitosis and meiosis are both forms of cell division. Mitosis produces daughter cells that are exactly the same as the parent cell, while Meiosis produces daughter cells that are different. They produce what we call haploid cells, which has only half the chromosomes of the parent cell. Gametes or sex cells are produced through meiosis, while body cells are produced through mitosis.

Biology
Step-by-step answer
P Answered by PhD

1. It involves multiple trials

2. 2 red: 2 white

3. Process 1 describes cloning and process 2 describes artificial selection

4. Furry Feet

5. Large scale insulin could be produced in relatively lesser time

6. A lowercase letter

7. 0%

8. rr

9. 50%

10. The allele for white flowers is recessive in the pea plant

Explanation:

1. Repetition, this means to do over and over. This would involve multiple trials and this is done to test the reliability of the results and determine if the hypothesis holds true or not.

2. If you put your cross in a Punnett square it would look like this:

      R    r                            

r     Rr   rr                          

r     Rr   rr

Based on your problem: R = red; r = white

If the genotype would have a dominant allele (R), then the dominant trait will come out, if it has only recessive alleles (r) then the recessive trait will come out. So as you can see there are 2 red and 2 white. So the ratio would be 2:2 or 1:1.

3. Process 1 is considered as cloning, and to be more specific, reproductive cloning. The aim of this is to produce an offspring that has the same genetic make up as the original or parent organism. Process 2 is an example of artificial selection. This is also known as selective breeding where breeders choose certain traits to pass on to the off springs by selecting the organisms that have the traits desired.

4. Furry feet of polar bears provide them extra traction when they walk on snow. The fur also provides extra warmth. They also have foot pads on the soles of their feet that help them as well. Together, polar bear have something that works like snow boots.

5. Large-scale insulin could be produced in relatively lesser time because bacteria reproduce through asexual means. They produce copies of themselves that are genetically identical. An advantage of asexual reproduction is the process occurs in a much faster rate. With that said, insulin supply would increase, which will help many diabetic patients.

6. Recessive alleles in a genotype are represented by a lowercase letter. On the other hand, dominant alleles are represented by a capital letter. Squares and circles are used in pedigree charts to represent the whole genotype of an organism for a specific trait.

7. The answer would be 0% because cystic fibrosis is a RECESSIVE gene disorder. For it to be expressed or observable in an individual , they should have two recessive alleles. As you can see in the end result, none of the offsprings have 2 recessive alleles. The genotype Aa, has a recessive allele, but it will not be expressed because the dominant trait will mask it.

8. When we talk about homozygous, this means the alleles are the same. And when you say heterozygous, this means that the organism would have a dominant and recessive allele. the recessive llele is represented by a lowercase letter. So if the organism is homozygous recessive, the genotype would have two lowercase letters. (rr)

9. The scenario shows the genotype is pp. This means that the pea plant is homozygous recessive. As explained before, recessive traits are represented by lowercase letters. So this would mean that the white color is a recessive trait. The answer would then be the allele for white flowers is recessive in the pea plant.

10. Mitosis and meiosis are both forms of cell division. Mitosis produces daughter cells that are exactly the same as the parent cell, while Meiosis produces daughter cells that are different. They produce what we call haploid cells, which has only half the chromosomes of the parent cell. Gametes or sex cells are produced through meiosis, while body cells are produced through mitosis.

Biology
Step-by-step answer
P Answered by PhD

Answers and explanations:

1. For each genotype, indicate whether it is heterozygous (A) or homozygous (B).

Homozygous (A) and heterozygous (B) refers to the presence of two equal or different alleles for the same trait, respectively.

AA (A).Bb (B).Cc (B).Dd (B).Ee (B).ff (B).GG (A).HH (A).Jj (B).kk (A).Ll (B).Mm (B).nn (A).OO (A).Pp (B).2. For each of the genotypes below, determine the phenotype.

A) Purple flowers are dominant to white flowers

QQ Purple flowers.Qq Purple flowers.qq White flowers.

B) Brown eyes are dominant to blue eyes

BB Brown eyes.Bb Brown eyes.bb Blue eyes.

C) Round seeds are dominant to wrinkled

RR Round seeds.Rr Round seeds.rr Wrinled seeds.

D) Bobtails are recessive (long tails dominant)

TT Longtails.

Tt Longtails.

tt Bobtails.

3. For each phenotype, list the genotypes. (Use the letter of the dominant trait)

Straight hair is dominant to curly.

SS Straight.

Ss Straight.

ss Curly.

Pointed heads are dominant to round heads.

PP Pointed .

Pp Pointed .

pp Round.

4. Set up the Punnett square for each of the crosses listed below. The trait being studied is round seeds (dominant) and wrinkled seeds (recessive).

A) Rr x rr

Alelles     R       r

r             Rr      rr

r             Rr      rr

Results of the crossing

RR (Round seed homozygous) None.Rr (Round seed heterozygous) 50%.rr (Wrinkled seeds homozygous) 50%.

What percentage of the offspring will be round?

- 50%

B) Rr x Rr

Alelles     R       r

R            RR      Rr

r             Rr       rr

Results of the crossing

RR (Round seed homozygous) 25%.Rr (Round seed heterozygous) 50%.rr (Wrinkled seeds homozygous) 25%.

What percentage of the offspring will be round?

- 75 %

C) RR x Rr

Alelles     R       R

R            RR      RR

r             Rr       Rr

Results of the crossing

RR (Round seed homozygous) 50%.Rr (Round seed heterozygous) 50%.rr (Wrinkled seeds homozygous) None.

What percentage of the offspring will be round?

- 100%

5. Punnett squares.

A) A TT (tall) plant is crossed with a tt (short plant). What percentage of the  offspring will be tall?

The result of the cross between two pure lines (TT and tt) is always a 100% heterozygous offspring (Tt), with the phenotype of the dominant trait.

Alelles     T       T

t              Tt      Tt

t              Tt      Tt

100% of the offspring will be tall.

B) Tt X Tt

Alelles     T       t

T             TT      Tt

t              Tt      tt

Results of the crossing

TT (Tall plants homozygous) 25%.Tt (Tall plants heterozygous) 50%.tt (Short plants homozygous) 25%.

25% of the offspring will be short.

C) RR X Rr

Alelles     R       R

R            RR      RR

r             Rr       Rr

Results of the crossing

RR (Round seed homozygous) 50%.Rr (Round seed heterozygous) 50%.rr (Wrinkled seeds homozygous) None.

50% of the offspring will be homozygous (RR).

D) RR X rr

Parent Genotypes: Homozygous round seeded RR and homozygous wrinkled seed rr.

Alelles     R       R

r              Rr      Rr

r              Rr       Rr

There are no homozygous offspring at this crossing.

E) White/Yellow flowered (pea) plants.

Q for yellow flowers.

q for white flowers.

Parent Genotypes:  Pea plants white flowers  qq.

Alelles     q       q

q             qq      qq

q             qq      qq

100% of the offspring will be white flowered (qq).

F) White/Yellow flowered plants.

- Q for yellow flowers.

- q for white flowers.

Parent Genotypes:  White flowered plant  qq and yellow flowered plant (heterozygous) Qq.

Alelles     Q       q

q             Qq      qq

q             Qq      qq

Results of the crossing

QQ (Yellow flowered homozygous) noneQq (Yellow flowered heterozygous) 50%.qq (White flowered homozygous) 50%.

50% of the offspring will be yellow flowered (Qq).

G) Qq X Qq

Parent Genotypes:  Yellow flowered plant  Qq (both heterozygous).

Alelles     Q       q

Q             QQ      Qq

q              Qq      qq

Results of the crossing

QQ (Yellow flowered homozygous) 25%.Qq (Yellow flowered heterozygous) 50%.qq (White flowered homozygous) 25%.

75% of the offspring will have yellow flowers (QQ and Qq).

25% of the offspring will have white flowers (qq).

H) Guinea pigs, the allele for long hair is dominant.

C for longhaired guinea pig.

c for shorthaired guinea pig.

Heterozygous longhaired guinea pig have Cc genotype.Pure breeding longhaired guinea pig have CC genotype.Shorthaired guinea pig have cc genotype.

I) Guinea pigs CC X cc

Parent Genotypes:

Pure breeding longhaired guinea pig have CC genotype.Shorthaired guinea pig have cc genotype.

CC X cc

Alelles     C         C

c             Cc       Cc

c             Cc       Cc

100% of the offspring will have long hair (Cc)

J) Guinea pigs Cc X Cc

Alelles     C         c

C            CC       Cc

c             Cc       cc

Results of the crossing

CC (Long hair homozygous) 25%.Qq (Long hair heterozygous) 50%.qq (Short hair homozygous) 25%.
Biology
Step-by-step answer
P Answered by PhD

Answers and explanations:

1. For each genotype, indicate whether it is heterozygous (A) or homozygous (B).

Homozygous (A) and heterozygous (B) refers to the presence of two equal or different alleles for the same trait, respectively.

AA (A).Bb (B).Cc (B).Dd (B).Ee (B).ff (B).GG (A).HH (A).Jj (B).kk (A).Ll (B).Mm (B).nn (A).OO (A).Pp (B).2. For each of the genotypes below, determine the phenotype.

A) Purple flowers are dominant to white flowers

QQ Purple flowers.Qq Purple flowers.qq White flowers.

B) Brown eyes are dominant to blue eyes

BB Brown eyes.Bb Brown eyes.bb Blue eyes.

C) Round seeds are dominant to wrinkled

RR Round seeds.Rr Round seeds.rr Wrinled seeds.

D) Bobtails are recessive (long tails dominant)

TT Longtails.

Tt Longtails.

tt Bobtails.

3. For each phenotype, list the genotypes. (Use the letter of the dominant trait)

Straight hair is dominant to curly.

SS Straight.

Ss Straight.

ss Curly.

Pointed heads are dominant to round heads.

PP Pointed .

Pp Pointed .

pp Round.

4. Set up the Punnett square for each of the crosses listed below. The trait being studied is round seeds (dominant) and wrinkled seeds (recessive).

A) Rr x rr

Alelles     R       r

r             Rr      rr

r             Rr      rr

Results of the crossing

RR (Round seed homozygous) None.Rr (Round seed heterozygous) 50%.rr (Wrinkled seeds homozygous) 50%.

What percentage of the offspring will be round?

- 50%

B) Rr x Rr

Alelles     R       r

R            RR      Rr

r             Rr       rr

Results of the crossing

RR (Round seed homozygous) 25%.Rr (Round seed heterozygous) 50%.rr (Wrinkled seeds homozygous) 25%.

What percentage of the offspring will be round?

- 75 %

C) RR x Rr

Alelles     R       R

R            RR      RR

r             Rr       Rr

Results of the crossing

RR (Round seed homozygous) 50%.Rr (Round seed heterozygous) 50%.rr (Wrinkled seeds homozygous) None.

What percentage of the offspring will be round?

- 100%

5. Punnett squares.

A) A TT (tall) plant is crossed with a tt (short plant). What percentage of the  offspring will be tall?

The result of the cross between two pure lines (TT and tt) is always a 100% heterozygous offspring (Tt), with the phenotype of the dominant trait.

Alelles     T       T

t              Tt      Tt

t              Tt      Tt

100% of the offspring will be tall.

B) Tt X Tt

Alelles     T       t

T             TT      Tt

t              Tt      tt

Results of the crossing

TT (Tall plants homozygous) 25%.Tt (Tall plants heterozygous) 50%.tt (Short plants homozygous) 25%.

25% of the offspring will be short.

C) RR X Rr

Alelles     R       R

R            RR      RR

r             Rr       Rr

Results of the crossing

RR (Round seed homozygous) 50%.Rr (Round seed heterozygous) 50%.rr (Wrinkled seeds homozygous) None.

50% of the offspring will be homozygous (RR).

D) RR X rr

Parent Genotypes: Homozygous round seeded RR and homozygous wrinkled seed rr.

Alelles     R       R

r              Rr      Rr

r              Rr       Rr

There are no homozygous offspring at this crossing.

E) White/Yellow flowered (pea) plants.

Q for yellow flowers.

q for white flowers.

Parent Genotypes:  Pea plants white flowers  qq.

Alelles     q       q

q             qq      qq

q             qq      qq

100% of the offspring will be white flowered (qq).

F) White/Yellow flowered plants.

- Q for yellow flowers.

- q for white flowers.

Parent Genotypes:  White flowered plant  qq and yellow flowered plant (heterozygous) Qq.

Alelles     Q       q

q             Qq      qq

q             Qq      qq

Results of the crossing

QQ (Yellow flowered homozygous) noneQq (Yellow flowered heterozygous) 50%.qq (White flowered homozygous) 50%.

50% of the offspring will be yellow flowered (Qq).

G) Qq X Qq

Parent Genotypes:  Yellow flowered plant  Qq (both heterozygous).

Alelles     Q       q

Q             QQ      Qq

q              Qq      qq

Results of the crossing

QQ (Yellow flowered homozygous) 25%.Qq (Yellow flowered heterozygous) 50%.qq (White flowered homozygous) 25%.

75% of the offspring will have yellow flowers (QQ and Qq).

25% of the offspring will have white flowers (qq).

H) Guinea pigs, the allele for long hair is dominant.

C for longhaired guinea pig.

c for shorthaired guinea pig.

Heterozygous longhaired guinea pig have Cc genotype.Pure breeding longhaired guinea pig have CC genotype.Shorthaired guinea pig have cc genotype.

I) Guinea pigs CC X cc

Parent Genotypes:

Pure breeding longhaired guinea pig have CC genotype.Shorthaired guinea pig have cc genotype.

CC X cc

Alelles     C         C

c             Cc       Cc

c             Cc       Cc

100% of the offspring will have long hair (Cc)

J) Guinea pigs Cc X Cc

Alelles     C         c

C            CC       Cc

c             Cc       cc

Results of the crossing

CC (Long hair homozygous) 25%.Qq (Long hair heterozygous) 50%.qq (Short hair homozygous) 25%.
Biology
Step-by-step answer
P Answered by PhD
Tt. Tt. (All of the vices would be Tt)
Tt. Tt
Biology
Step-by-step answer
P Answered by PhD

Tt Tt Tt Tt

Explanation:

All answers inherit one dominant and one recessive trait.

Hope this is correct, I'm a tad rusty.

Try asking the Studen AI a question.

It will provide an instant answer!

FREE