# Test Cross help.

Online Science Education :: Subjects :: Biology :: Genetics :: 1 on 1 Tutoring

## Test Cross help.

My professor once again gave me problems to work out where in which I have no idea on how to start out on them...Can someone lead me in the right direction please?

a. Determine which genes are linked.

b. Draw a map that shows distances in map units.

(a) unlinked;

(b) completely linked (no crossing-over at all);

(c) 10 map units apart;

(d) 24 map units apart?

Any sort of direction or any suggestions on how to get started would be greatly appreciated.

Thanks in advance.

**1.**A three-point testcross was made in corn. The results and a partial recombination analysis are shown in the following display, which is typical of three-point testcrosses (p = purple leaves, + = green; v = virus- resistant seedlings, + = sensitive; b = brown midriff to seed, + = plain).a. Determine which genes are linked.

b. Draw a map that shows distances in map units.

**2.**If A/A • B/B is crossed to a/a • b/b, and the F1 is testcrossed, what percent of the testcross progeny will be a/a • b/b if the two genes are(a) unlinked;

(b) completely linked (no crossing-over at all);

(c) 10 map units apart;

(d) 24 map units apart?

Any sort of direction or any suggestions on how to get started would be greatly appreciated.

Thanks in advance.

**ph0bolus**- Student
- Posts : 4

Join date : 2008-06-28

## Re: Test Cross help.

First of all, three point test cross is performed to find out distance between genes on the same chromosome. Now let's figure out which gene is in the middle. To do this you need to know which are the double recombinant gametes. Double recombinants are rare, so you can see from the numbers that 7 and 8 are the smallest and, therefore, they are double recombinants. Your map for parental genotype is v - p - b.

Now you can work with numbers to figure out distances. To do that, you have to divide number of recombinants between 2 neighboring genes by the total number of samples. So, v-p is 2200/10000=0.22 or 22% or 22 m.u.

while p-b is 1500/10000=0.15 or 15% or 15 m.u.

You can verify the total distance between v-b by dividing the total number of recombinants v-b PLUS the 2x the number of double recombinants by the total samples: (3436+ (2)(72+60))/10000= 3700/10000 or 37 m.u. same as 22 m.u. + 15 m.u.

Since all distances are less than 50 m.u., all genes are linked.

For your second question:

The cross between two homozygous parents will result in all heterozygous F1 generation (all will be AaBb). The cross in question is AaBb x aabb (test cross is always to homozygous recessive).

a), c), and d) possible gametes are AB, Ab, aB, and ab. If you do the cross, you will find that when unlinked 4 out of 16 should be aabb or 25%. Map distance tells you the actual % or recombinant gametes. So, if A-B is 10 m.u. apart, then 10% progeny of test cross are recombinants. That means that 90% will have parental genotype. There are 2 parental genotypes: AB and ab, each of those will occur 45% of the time (90%/2). Thus, occurrence of aabb will be 45% in this case. You can work out 24 m.u. in the same way.

However, if genes are linked, then only parental genotypes are possible (AB or ab gametes) and test cross will result in 50% of each AaBb and aabb progeny.

Good luck!

Now you can work with numbers to figure out distances. To do that, you have to divide number of recombinants between 2 neighboring genes by the total number of samples. So, v-p is 2200/10000=0.22 or 22% or 22 m.u.

while p-b is 1500/10000=0.15 or 15% or 15 m.u.

You can verify the total distance between v-b by dividing the total number of recombinants v-b PLUS the 2x the number of double recombinants by the total samples: (3436+ (2)(72+60))/10000= 3700/10000 or 37 m.u. same as 22 m.u. + 15 m.u.

Since all distances are less than 50 m.u., all genes are linked.

For your second question:

The cross between two homozygous parents will result in all heterozygous F1 generation (all will be AaBb). The cross in question is AaBb x aabb (test cross is always to homozygous recessive).

a), c), and d) possible gametes are AB, Ab, aB, and ab. If you do the cross, you will find that when unlinked 4 out of 16 should be aabb or 25%. Map distance tells you the actual % or recombinant gametes. So, if A-B is 10 m.u. apart, then 10% progeny of test cross are recombinants. That means that 90% will have parental genotype. There are 2 parental genotypes: AB and ab, each of those will occur 45% of the time (90%/2). Thus, occurrence of aabb will be 45% in this case. You can work out 24 m.u. in the same way.

However, if genes are linked, then only parental genotypes are possible (AB or ab gametes) and test cross will result in 50% of each AaBb and aabb progeny.

Good luck!

**Alla**- Instructor
- Posts : 103

Join date : 2008-02-27

## Re: Test Cross help.

There was another one she gave us which was giving me problems...If you could help me yet again I would be greatful.

Which of the following linear asci in the Ascomycete fungus in Neurospora shows gene conversion at the arg2 locus?

Which of the following linear asci in the Ascomycete fungus in Neurospora shows gene conversion at the arg2 locus?

**ph0bolus**- Student
- Posts : 4

Join date : 2008-06-28

## Re: Test Cross help.

This one is really easy. Neurospora asci should have 4 dominant and 4 recessive spore types. So, 3 and 6 are not 'normal' and, therefore, conversion type.

P.S. Please, post additional problems as separate posts.

P.S. Please, post additional problems as separate posts.

**Alla**- Instructor
- Posts : 103

Join date : 2008-02-27

## Re: Test Cross help.

Hello

Thanks .... but the more information about this.

Thanks .... but the more information about this.

**tonydisalva**- Student
- Posts : 5

Join date : 2013-01-12

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