Tuberculosis is spread through the air when an infected person coughs or sneezes.
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Answer:
This is a trick question, and only because the question is vague
Explanation:
Use a Punnet Square to figure out the percentages.
If the parent with freckles (let's just call them Parent A) is homozygous dominant, their genotype would be FF. The parent without freckles (Parent B) has a genotype of ff.
In any situation, the probability of these two having a child that is homozygous dominant for freckles is 0%. (see punnet squares for more info).
Now here's where it gets a little more confusing because the question is kind of vague on this.
Assuming that Parent A is homozygous dominant (both genes for this are dominant (FF)), then the probability of having a heterozygous (non-identical genotype letters, meaning they carry the recessive trait) genotype is 100%. However, if Parent A is also heterozygous, then that percentage decreases to 50%
For Situation A (Parent A is homozygous dominant), the probability of a homozygous recessive genotype is 0%
For Situation B (Parent A is heterozygous dominant), the probability of a homozygous recessive child is 50%.
Please see the Punnet square I made for more information.
Key:
- Homozygous Dominant: FF
- Heterozygous Dominant: Ff
- Homozygous Recessive: ff
(heterozygous recessive does not exist, because that would mean one of the genes is dominant, which would overpower the recessive gene)
I really hope this helps, and if you're confused by this, please do not hesitate to let me know!
