A geneticist is mapping the chromosomes of the newly captured gremlin. Stripe is heterozygous for three linked genes with alleles Ee, Hh, and Bb, that determine if gremlins are evil (E), have hair (H), and biting teeth (B). In order to determine if the three genes are linked, a standard testcross was done, and the 1000 offspring had the following genotypes: 48 ee Hh bb 36 ee hh Bb 400 ee Hh Bb 4 Ee Hh Bb 426 Ee hh bb 46 Ee hh Bb 38 Ee Hh bb 2 ee hh bb What is the recombination frequency between genes E and H

Recombination frequency determines if two genes are linked or if they segregate independently from each other. The recombination frequency between E and H is 8%

What are linked genes?

Linked genes are those that are too close to each other in the same chromosome and they do not segregate independently.

These are the linked genes that do not exhibit an independent distribution, and they are inherited together more frequently.

What is recombination frequency?

Recombination frequency can be defined as the crossing over events frequency occurring between two genes.

The value of recombination frequency indicates weather genes are linked or not. The maximum recombination frequency is always 50%.

When RF is 50% or more, we can assume that genes are not linked.

In the exposed example, we can recognize the parental gametes in the descendants because their phenotypes are the most frequent, while the double recombinants are the less frequent.

Parental)

400 ee Hh Bb

426 Ee hh bb

Double recombinant)

4 Ee Hh Bb

2 ee hh bb

Simple recombinant)

48 ee Hh bb

36 ee hh Bb

46 Ee hh Bb

38 Ee Hh bb

We will call Region I to the area between genes E and H and Region II to the area between genes H and B.

------------E------------H--------------B-----------

/---RI------//------RII-----/

To calculate the recombination frequency between E and H we just need to consider region I. We will call P1 to the recombination frequency in this region.

P1 = (R + DR) / N

Where:

R is the number of simple recombinants in this region,
DR is the number of double recombinants in each region, and
N is the total number of individuals.

Now, let us analyze the information.

We know that individuals carrying double recombinant gametes are

Ee Hh Bb
ee hh bb

We know that the recessive parent from the testcross could only provide the following gametes ⇒ ehb

This leads us to assume that the followings are the double recombinant gametes from the trihybrid parent,

EHB
ehb

Since these gametes are the product of a double recombination event in region I and II, we can assume that the parental gametes of the trihybrid individual are

eHb
EhB

Hence the simple recombinant gametes in region I are

EHb

ehB

Now, according to this analysis, we know that

Simple recombinants in region I

36 ee hh Bb

38 Ee Hh bb

Double recombinants

4 Ee Hh Bb

2 ee hh bb

Total number of individuals

N = 1000

Recombination frequency in Region I

P1 = (R + DR) / N

P1 = (38 + 36 + 4 + 2) / 1000

P1 = 80 / 1000

P1 = 0.08 = 8%

The recombination frequency between genes E and H is P1 = 8%

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