Question 10 10 A test cross of AaBb x aabb yields the following babies: 35% babi

Question

Question 10 10 A test cross of AaBb x aabb yields the following babies: 35% babies with AB phenotype 35% babies with ab phenotype 15% babies with Ab phenotype 15% babies with aB phenotype Which of the following conclusions is correct? O The A and B genes are located on different chromosomes O The A and B genes are located 70 map units apart on the same chromosome O The A and B genes are located 35 map units apart on the same chromosome O The A and B genes are located 30 map units apart on the same chromosome O The A and B genes are located 15 map units apart on the same chromosome Quiz saved at 6:14pm Subs ch

Explanation / Answer

Answer:

The given cross can be represented in the form of a punnett square as below:

Parental gametes are: AB, Ab, aB, ab and ab, ab,ab, ab

So the four diferent genotypes are in same proportion:

AaBb:Aabb:aaBb:aabb = 1:1:1:1

So the expected percentage of AB, ab, Ab, aB is 100/4 = 25%, 25%, 25%, 25% each

The given percentage of different phenotypes is very different from the expected percentage. To check this a chi square test can be performed:

Degree of freedom = number of categories - 1 = 4 - 1 = 3

P-value corresponding to a Chi square value of 16 with 3 degree of freedom is P-Value = 0.001134. The result is significant at p < 0.05. Therefore, the Null hypothesis is rejected and based on the evidence, the observed percentage of four different phenotypes are different from the expected percentage. This indicates that the genes are linked.

Map distance can be calculated as below:

ab ab ab ab AB AaBb AaBb AaBb AaBb Ab Aabb Aabb Aabb Aabb aB aaBb aaBb aaBb aaBb ab aabb aabb aabb aabb

Related Questions

Navigate

• Browse (All)
• Browse Q
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.