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

Answers

Answer 1
the recombination frequency would be between E and H meaning it will have
Answer 2

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 Bbee 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,

EHBehb

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

eHbEhB

Hence the simple recombinant gametes in region I are

EHbehB

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%

You can learn more about recombination frequency at

https://brainly.com/question/7299933

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A Geneticist Is Mapping The Chromosomes Of The Newly Captured Gremlin. Stripe Is Heterozygous For Three

Related Questions

In cocker spaniels, black coat color ( B) is dominant over red ( b), and solid color ( S) is dominant over spotted ( s). In the F 2 generation of a cross between BB ss with bb SS, what fraction of the offspring would be expected to be black and spotted

Answers

Answer:

The correct answer is -3/16

Explanation:

The cross between BBss and bbSS In cocker spaniels will form gametes that are - Bs and bS. As it is given that  black coat color ( B) is dominant over red ( b), and solid color ( S) is dominant over spotted ( s) then f1 gen would be :

      Bs      Bs

bS BbSs  BbSs

bS  BbSs  BbSs

Now F2 generation;

       BS        Bs        bS        bs

BS   BBSS   BBSs   BbSS   BbSs

Bs   BBSs    BBss   BbSs   Bbss

bS   BbSS   BbSs   bbSS    bbSs

bs   BbSs   Bbss   bbSs     bbss

Here black-spotted that is possible only in two cases, BBss or Bbss, are 3/16

The diagram below shows a cell placed in a solution.

A cell is shown placed inside a beaker. It is labeled Cell. The solution inside the beaker is labeled 2% salt solution and the solution inside the cell is labeled 4% salt solution.

What will most likely happen to the cell?

It will expand as water moves out of it.
It will shrink as water moves out of it.
It will expand as water moves into it.
It will shrink as water moves into it.

Answers

The answer is the second option, it will shrink as water moves out of it.


The cell will shrink in size, losing water due to osmosis.

The most likely to happen to the cell is ; ( C ) It will expand as water moves into it

The movement of water in and out of a cell is based on osmosis .The cell membrane of the cell will regulate the movement of water molecules in and out of the cell. based on the solute concentration inside the cell and outside the cell.

When a cell is placed in a hypotonic solution ( i.e. a solution with a lower solute concentration as seen in the 2% salt solution ) the water molecules from the hypotonic solution will move into the cell which will cause the cell to be Turgid/expand over time.  

Hence we can conclude that the most likely thing to happen to the cell is ;  It will expand as water moves into it

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Answers

Answer: i would say b!

Explanation:

hope i helped! :D

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