Question

20. SuperGreeneyes chromosome wa domate y ed woman has charen impossible to so 3 1/4 2/4 D o ne of the previous che C. C om D. 23. Assuming Ind what can you tell about the die Autosomal Dominant with C. X-Linked Dominant D st") E.Y.Linked a e ceive Recesse 21. Consider the following cross: (HhMm X HAMM). What fraction would be HOMM? 24. When an individual has an extra chromosome Ichromosome 21 in dow's syndrome's called? A. 0/16 D. 3/16 A. Monosom C. Trisomy E. None of the previous choices are correct B. Bar Bodies D. Diowy B. 1/16 E. 4/16 C. 2/16 25. Suppose a mutant form of DNA polymerase has a functional polymerizing activity, but not exonuclease activity proof reading function. Which of the following properties do you expect the mutant polymerase to have? -to-S A. It will polymerize DNA bi-directionally in both the 5'-to-3' direction and the 3'-to-S direction. B. It will polymerize DNA slower than normal C. It will polymerize RNA as well. D. It will generate more mismatched base pairs. E. It will generate less mismatched base pairs. following figure to answer questions 22-23: 26. For the following cross: DdEeFt X DdEeFt, assuming that each genes is on a different chromosome and each chromosome sorts independently, what ratio of the progeny woul DDEeFf? Hint: just asking you to do a trihybrid A. XXXX% DDEFf C.XXXX% DDEFfD.XX%X % DDEFE B.XXXX% DDEFE E. DdEeFfX DdEeFf cannot produce DDEFf off

Answer 1

Ans.20) Option C : 2/4

Explanation : As the eye color trait is associated with X chromosome green color is dominant over blue color. It means female parent is homozygous recessive for given trait. Both X chromosome in female contains recessive allele for blue color. Also male child always recieves its X chromosome from mother and Y chromosome from father. Therefore even father contain dominant green eye allele. No son will contain green eye color. As they will recieving X chromosome from mother. Both son out of four children will have blue colored eye.

Ans.21) Option E : 4/16

Explanation : This dihybrid cross can be divided into two monohybrid cross then result can be multiplied to obtain the result for dihybrid cross. So, from monohybrid cross between Hh and Hh we get 1/4 HH, 1/2 Hh, 1/4 hh. Similarly from monohybrid cross of MM and Mm we get 1/2 MM and 1/2Mm. In order to get a individual with genotype HhMM we have multiply respective ratios of genotype Hh i.e. 1/2 and MM i.e. 1/2. So we get a ratio of 1/4. Since we know that dihybrid cross gives 16 offsprings so net net ratio is 4/16.