Let's address the first question regarding the genetic distance between genes Aa and Bb.
Genes Aa and Bb are in coupling (cis) arrangement on chromosome 14. After a testcross, they are in trans arrangement in 20 out of 100 progeny. The genetic distance between two genes is calculated based on the frequency of recombination events.
The recombination frequency (RF) is given by the formula:
\[ \text{RF} = \frac{\text{Number of recombinant progeny}}{\text{Total number of progeny}} \]
In this case:
\[ \text{RF} = \frac{20}{100} = 0.20 \]
Genetic distance in map units (centiMorgans, cM) is equivalent to the recombination frequency multiplied by 100:
\[ \text{Genetic distance} = 0.20 \times 100 = 20 \text{ map units} \]
So, the answer is:
b. 20 map units
- a. 2 map units: Incorrect, as the recombination frequency is 20%, not 2%.
- b. 20 map units: Correct, as the recombination frequency is 20%, which translates to 20 map units.
- 0,2 map units: Incorrect, as this seems to be a typographical error and does not match the calculated genetic distance.
- d. not enough information to make a determination: Incorrect, as sufficient information is provided to calculate the genetic distance.
Two parents both have the genotype Zz. We need to determine the probability that they will have a child with the genotype Zz.
Using a Punnett square, we can determine the possible genotypes of the offspring:
| Parent 1 \ Parent 2 | Z | z |
|---------------------|----|----|
| Z | ZZ | Zz |
| z | Zz | zz |
The possible genotypes of the offspring are:
There are 4 possible outcomes, and 2 of them are Zz.
The probability of having a child with the genotype Zz is:
\[ \frac{2}{4} = \frac{1}{2} \]
So, the answer is:
C. \( \frac{1}{2} \)
- A \( \frac{1}{4} \): Incorrect, as this is the probability for either ZZ or zz, not Zz.
- B: No value provided, so cannot be considered.
- C \( \frac{1}{2} \): Correct, as the probability of Zz is 50%.
- \( \frac{1}{8} \): Incorrect, as this is not a possible outcome for this scenario.
- The genetic distance between genes Aa and Bb is 20 map units.
- The probability that two Zz parents will have a child with the genotype Zz is \( \frac{1}{2} \).
Answered
