Question
Sickle cell anemia is known to run in a family. A pedigree chart for this family is shown below.
The parents are shown at the top and from left to right appear as a half-shaded circle next to a half-shaded square. The two are connected by a straight line. Below the parent pair is a vertical straight line connecting to the next row of circles and squares which are the offspring. Starting from left an unshaded square, half-shaded circle, another half-shaded circle, and a completely shaded square are shown. All of these are also connected by a horizontal straight line.
What percentage of the offspring are carriers of the disease but do not have it?
Group of answer choices
100%
0%
25%
50%
The parents are shown at the top and from left to right appear as a half-shaded circle next to a half-shaded square. The two are connected by a straight line. Below the parent pair is a vertical straight line connecting to the next row of circles and squares which are the offspring. Starting from left an unshaded square, half-shaded circle, another half-shaded circle, and a completely shaded square are shown. All of these are also connected by a horizontal straight line.
What percentage of the offspring are carriers of the disease but do not have it?
Group of answer choices
100%
0%
25%
50%
Answers
GPT-4o mini
In sickle cell anemia, the condition is caused by a recessive allele (let’s denote it as "s"), while the normal allele is dominant (denote it as "S"). Individuals with the genotype "SS" are normal, those with "Ss" are carriers (asymptomatic), and individuals with the genotype "ss" have sickle cell anemia.
From your description of the pedigree:
- The parents are depicted as a half-shaded circle (female, carrier - genotype "Ss") next to a half-shaded square (male, carrier - genotype "Ss").
- The offspring consist of:
1. An unshaded square (normal, genotype "SS")
2. A half-shaded circle (carrier, genotype "Ss")
3. Another half-shaded circle (carrier, genotype "Ss")
4. A completely shaded square (affected, genotype "ss")
Now, let's analyze the potential offspring genotypes from two carriers (Ss x Ss):
To find the potential offspring genotypes, we can set up a Punnett square:
```
S | S
----------------
S | SS | SS
----------------
s | Ss | Ss
```
From this Punnett square, we see the following genotypic ratios:
- 25% (1 out of 4) of the offspring will be "SS" (normal)
- 50% (2 out of 4) will be "Ss" (carriers)
- 25% (1 out of 4) will be "ss" (affected)
Among the provided offspring:
1. SS (normal)
2. Ss (carrier)
3. Ss (carrier)
4. ss (affected)
Out of the four offspring, two are carriers (Ss).
To calculate the percentage of offspring that are carriers:
- Total offspring = 4
- Carrier offspring = 2
Percentage of carriers = (Number of carriers / Total offspring) × 100 = (2 / 4) × 100 = 50%
Thus, the percentage of offspring that are carriers of sickle cell anemia but do not have the disease is **50%**.
From your description of the pedigree:
- The parents are depicted as a half-shaded circle (female, carrier - genotype "Ss") next to a half-shaded square (male, carrier - genotype "Ss").
- The offspring consist of:
1. An unshaded square (normal, genotype "SS")
2. A half-shaded circle (carrier, genotype "Ss")
3. Another half-shaded circle (carrier, genotype "Ss")
4. A completely shaded square (affected, genotype "ss")
Now, let's analyze the potential offspring genotypes from two carriers (Ss x Ss):
To find the potential offspring genotypes, we can set up a Punnett square:
```
S | S
----------------
S | SS | SS
----------------
s | Ss | Ss
```
From this Punnett square, we see the following genotypic ratios:
- 25% (1 out of 4) of the offspring will be "SS" (normal)
- 50% (2 out of 4) will be "Ss" (carriers)
- 25% (1 out of 4) will be "ss" (affected)
Among the provided offspring:
1. SS (normal)
2. Ss (carrier)
3. Ss (carrier)
4. ss (affected)
Out of the four offspring, two are carriers (Ss).
To calculate the percentage of offspring that are carriers:
- Total offspring = 4
- Carrier offspring = 2
Percentage of carriers = (Number of carriers / Total offspring) × 100 = (2 / 4) × 100 = 50%
Thus, the percentage of offspring that are carriers of sickle cell anemia but do not have the disease is **50%**.