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A cross is made between a haploid strain of Neurospora of genotype nic + ad and

ID: 19054 • Letter: A

Question

A cross is made between a haploid strain of Neurospora of genotype nic+ ad and another haploid strain of nic ad+. From this cross, a total of 1000 linear asci are isolated and characterized as in the following table. Map the ad and nic loci in relation to centromeres and to each other.

nic+ ad

nic+ ad+

nic+ ad+

nic+ ad

nic+ ad

nic+ ad+

nic+ ad+

nic+ ad

nic+ ad+

nic+ ad+

nic+ ad

nic+ ad

nic+ ad+

nic+ ad+

nic+ ad

nic+ ad+

nic+ ad

nic ad

nic ad+

nic ad

nic ad

nic+ ad

nic+ ad+

nic+ ad

nic ad

nic ad+

nic ad

nic ad

nic ad+

nic ad

nic ad+

nic+ ad+

nic+ ad

nic+ ad+

nic+ ad

nic ad+

nic ad

nic ad+

nic+ ad+

nic+ ad

nic+ ad+

nic+ ad

nic ad+

nic ad

nic ad

nic ad+

nic ad+

nic ad

nic ad+

nic ad+

nic ad

nic ad

nic ad+

nic ad+

nic ad

nic ad+

808

1

90

5

90

1

5

nic+ ad

nic+ ad+

nic+ ad+

nic+ ad

nic+ ad

nic+ ad+

nic+ ad+

nic+ ad

nic+ ad+

nic+ ad+

nic+ ad

nic+ ad

nic+ ad+

nic+ ad+

nic+ ad

nic+ ad+

nic+ ad

nic ad

nic ad+

nic ad

nic ad

nic+ ad

nic+ ad+

nic+ ad

nic ad

nic ad+

nic ad

nic ad

nic ad+

nic ad

nic ad+

nic+ ad+

nic+ ad

nic+ ad+

nic+ ad

nic ad+

nic ad

nic ad+

nic+ ad+

nic+ ad

nic+ ad+

nic+ ad

nic ad+

nic ad

nic ad

nic ad+

nic ad+

nic ad

nic ad+

nic ad+

nic ad

nic ad

nic ad+

nic ad+

nic ad

nic ad+

808

1

90

5

90

1

5

Explanation / Answer

Please answer all of the questions in the space provided. Not all questions are of equal value, and the value of each question is one mark except where indicated. The total adds up to 20 marks. For rough work or extra space, you may use the backs of the pages.

Name:                                                               ID#

1. A tetrad analysis question from your textbook. A cross is made between a haploid strain of Neurospora of genotype nic+, ad- and another strain of genotype nic-, ad+ . From this cross, a total of 1000 linear asci were isolated and scored as to genotype into seven categories according to the data below. Each vertical column contains the genotypes of the eight ascospores in that category of ascus.

1                  2                  3                 4                 5                  6                 7

nic+ ad- nic+ ad+ nic+ ad+ nic+ad-                     nic+ad-       nic+ad+ nic+ad+

nic+ ad- x nic- ad+ nic+ ad- nic+ ad+ nic+ ad+ nic+ad-                        nic+ad-       nic+ad+ nic+ad+

nic+ ad-       nic+ ad+ nic+ ad-          nic-ad-        nic-ad+       nic-ad-        nic-ad

nic+ ad- nic+ ad+ nic+ ad-                nic-ad-        nic-ad+       nic-ad-        nic-ad

nic- ad+ mc- ad- mc- ad+ nic+ad+ nic+ad-                       nic+ad+ nic+ad

nic- ad+ nic- ad- nic- ad+ nic+ad+ nic+ad-                       nic+ad+ nic+ad

nic- ad+ nic- ad- nic- ad- nic- ad+ nic- ad+                             nic- ad-       nic-ad+

nic- ad+ nic- ad- nic- ad- nic- ad+ nic- ad+                             nic- ad-       nic-ad+

808              1                  90               5                  90                1                 5

PD               NPD            TT              TT               PD              NPD           TT

a)         (2 marks) Are the two genes linked or not? How can you tell without actually doing any calculations?

1 mark          The genes are linked.

1 mark          Look at the ratio of PD to NPD. Since PD>>NPD linkage is indicated.

b)      (3 marks) Map the two genes with respect to their distance from their respective centromere(s) and map the two genes with respect to each other, if applicable.

are asked for. Actual map not required. Just the calculations

nic                                             ad

44-5.05 -I. 44-.5.6

10.25______________________

•___________________________________________________             n

1 mark          nic - centromere = (5 + 90 + 1 + 5) 1/2 = 5.05 mu

1000

1 mark          ad - centromere      = 90 + 90 + 1 + 5 = 186/2 = 93/1000 = 9.3 mu.

1 mark          nic - ad                   = 50 (T + 6 NPD) / Total = 100 + 6 (2) = 112 (50) = 5600/1000 5.6 mu

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