If you crossed a homozygous (BB) guinea pig for black fur color with a heterozygous (Bb) guinea pig for black fur color, what would be the predicted percentage of homozygous, black-furred guinea pigs?(1 point)
Responses
75%
75%
100%
100%
50%
50%
25%
The predicted percentage of homozygous black-furred guinea pigs (BB) from crossing a homozygous (BB) guinea pig with a heterozygous (Bb) guinea pig is calculated using a simple Punnett square.
The Punnett square would look like this:
```
B (from Bb)
----------------
B | BB | BB
----------------
B | BB | BB
```
Each cell of the Punnett square represents a possible genotype of the offspring. Since both rows in the square contain a homozygous dominant (BB) individual, all the offspring would be homozygous black-furred guinea pigs.
Therefore, the predicted percentage of homozygous black-furred guinea pigs (BB) would be 100%.
I don,t belive you
I apologize if there was any confusion. Let's clarify the Mendelian genetics behind the cross between a homozygous (BB) black-furred guinea pig and a heterozygous (Bb) black-furred guinea pig.
The Punnett square for this cross would look like this:
```
B (homozygous parent) | B (homozygous parent)
-------------------------
B (heterozygous parent) | BB | BB
-------------------------
b (heterozygous parent) | Bb | Bb
```
Here's how we interpret the Punnett square:
- The top row, where we cross B from the homozygous parent with both alleles from the heterozygous parent (B), results in two offspring with genotype BB.
- The bottom row, where we cross B from the homozygous parent with the B and b alleles from the heterozygous parent, results in two offspring with genotype Bb.
So the results are:
- BB (homozygous black) offspring: 2 out of 4 (which is 50%)
- Bb (heterozygous black) offspring: 2 out of 4 (which is 50%)
The predicted percentage of homozygous black-furred guinea pigs (BB) from this cross is therefore 50%.
are u sure
Yes, I am confident in this explanation, but let's go through the cross between a homozygous (BB) black-furred guinea pig and a heterozygous (Bb) black-furred guinea pig one more time to ensure clarity.
Here’s the Punnett square for the cross:
```
B (from BB) | B (from BB)
---------------------
B (from Bb) | BB | BB
---------------------
b (from Bb) | Bb | Bb
```
The possible genetic combinations for the offspring are:
- BB (from B of BB and B of Bb)
- BB (from B of BB and B of Bb)
- Bb (from B of BB and b of Bb)
- Bb (from B of BB and b of Bb)
This means we have:
- 2 out of 4 or 50% BB (homozygous black)
- 2 out of 4 or 50% Bb (heterozygous black)
So, the correct prediction would be that 50% of the offspring will be homozygous for the black fur color allele (BB).
ok
Alright, if you have any more questions or need further clarification on this topic or any other, feel free to ask!
In horses, the Extension gene helps to determine coat color. The dominant trait (E) produces black pigment in the coat. The recessive trait (e) produces red pigment. Would a cross between a stallion who is homozygous, EE, and a mare who is heterozygous, Ee, ever produce a foal with a red coat?(1 point)
Responses
Yes, because the stallion's dominant allele and the mare's recessive allele can result in a red coat.
Yes, because the stallion's dominant allele and the mare's recessive allele can result in a red coat.
Yes, because the mare can pass on her recessive allele.
Yes, because the mare can pass on her recessive allele.
No, because there would have to be two recessive alleles for the foal to have a red coat.
No, because there would have to be two recessive alleles for the foal to have a red coat.
No, because the mare cannot pass on her recessive allele.
No, because the mare cannot pass on her recessive allele.
No, because there would have to be two recessive alleles for the foal to have a red coat.
Since the stallion is homozygous dominant (EE), he can only pass on the dominant allele for black pigment. The mare, although heterozygous (Ee), can pass on either the dominant allele for black pigment or the recessive allele for red pigment. However, even if the mare passes on the recessive allele, the presence of one dominant allele from the stallion will result in a black pigment phenotype in the offspring. A foal would need to inherit two recessive alleles (ee) to have a red coat, which is not possible in this cross.
Cats have 38 chromosomes, or 19 pairs. Long hair in cats is a recessive trait (s) while short hair is a dominant trait (S). A cat that is heterozygous for short hair (Ss) and a cat that is homozygous for long hair (ss) are a mated pair. There is a 50% chance of an offspring being heterozygous for short hair, and a 50% chance of an offspring being homozygous for long hair. They have a total of 11 offspring, all of which have long hair. Is this possible? Why or why not?(1 point)
Responses
Yes, the difference in number of chromosomes would allow this to happen.
Yes, the difference in number of chromosomes would allow this to happen.
No, the 50/50 probabilities mean that at least 50% of the offspring should be homozygous for long hair.
No, the 50/50 probabilities mean that at least 50% of the offspring should be homozygous for long hair.
No, with fewer chromosomes than humans, the probabilities would be different.
No, with fewer chromosomes than humans, the probabilities would be different.
Yes, the probability of any outcome is independent of prior outcomes
Yes, the probability of any outcome is independent of prior outcomes.
While the expected ratio, given the genetic cross between a heterozygous short hair cat (Ss) and a homozygous long hair cat (ss), is indeed 50% heterozygous short hair (Ss) and 50% homozygous long hair (ss), these are the expected probabilities for each individual birth. Genetics operates independently for each offspring, meaning that the outcome of one birth does not influence the outcome of another.
Due to the independent assortment and the randomness of inheritance, it is possible — although statistically unlikely — for all 11 offspring to be homozygous for long hair (ss). Even with a 50% chance for each individual offspring to have short hair, each birth is a separate event with its own 50/50 chance. Therefore, while unusual, it is perfectly possible for all 11 offspring to have long hair by chance.
The allele for curly hair is dominant over the allele for straight hair. If you mated a curly-haired dachshund and a straight-haired dachshund, would you be able to accurately predict the probability of the offspring having curly hair? Why?(1 point)
Responses
Yes, because you know the genotypes of both dogs.
Yes, because you know the genotypes of both dogs.
No, because you don’t know the genotype of the curly-haired dog.
No, because you don’t know the genotype of the curly-haired dog.
Yes, because you know the genotype of the curly-haired dog.
Yes, because you know the genotype of the curly-haired dog.
No, because you only know the genotype of the curly-haired dog.
No, because you don’t know the genotype of the curly-haired dog.
The curly-haired dachshund could have a genotype of either CC (homozygous dominant) or Cc (heterozygous), since the curly hair trait is dominant. Without knowing which it is, you can't predict the exact probability of the offspring having curly hair. The straight-haired dachshund must have the genotype cc (homozygous recessive), as it displays the recessive trait.
If the curly-haired dachshund is CC, then all offspring would be curly-haired, as they would all inherit at least one dominant C allele. If the curly-haired dachshund is Cc, there would be a 50% chance of an offspring having curly hair and a 50% chance of having straight hair. Without specific information about the genotype of the curly-haired parent, you cannot accurately predict the probability of the offspring having curly hair.
Question
If two parents have black hair, a dominant phenotype, and one of their offspring has blond hair, a recessive phenotype, which would be a true statement about the parents’ genotypes?(1 point)
Responses
Both parents are homozygous for the trait.
Both parents are homozygous for the trait.
Both parents are heterozygous for the trait.
Both parents are heterozygous for the trait.
Only one parent is heterozygous.
Only one parent is heterozygous.
Only one parent is homozygous.
Both parents are heterozygous for the trait.
For an offspring to have a recessive phenotype, like blond hair, they would need to inherit two recessive alleles, one from each parent. Since both parents express the dominant phenotype (black hair), but have produced an offspring with the recessive phenotype (blond hair), this means each parent must carry one dominant allele for black hair and one recessive allele for blond hair. Therefore, both parents must have a heterozygous genotype for hair color.