Chapter 6, Problem 1QSDC

In mice, a dominant allele that causes a short tail is located on chromosome 2. On chromosome 3, a recessive allele causing droopy ears is $\text{6 mu}$ away from another recessive allele that causes a flaky tail. A recessive allele that causes a jerker (uncoordinated) phenotype is located on chromosome 4. A jerker mouse with droopy ears and a short, flaky tail was crossed to a normal mouse. All the ${\text{F}}_{\text{1}}$ generation mice were phenotypically normal, except they had short tails. These ${\text{F}}_{\text{1}}$ mice were then testcrossed to jerker mice with droopy ears and long, flaky tails. If this testcross produces 400 offspring, what are the expected numbers of the 16 possible phenotypic categories?

Summary Introduction

To review:

The expected number of 16 phenotypic categories produced after the test cross of F₁ mice with jerker mice having droopy ears and long flaky tails.

Introduction:

Test crosses were proposed by Gregor Mendel for the first time.They are used to find the genotype of heterozygous individuals. In a test cross, the heterozygous individual is crossed with its corresponding homozygous recessive individual to determine its genotype.

Explanation of Solution

In mice, the dominant allele is a short tail which is located on chromosome 2. There is a recessive allele located on chromosome 3 which causes droopy ears. It is6mu (map unit) apart than another recessive allele which is for jerker phenotype that is located on chromosome 4. The cross is made between normal mouse and jerker mouse having droopy ears with a short flaky tail. The cross results in normal mouse with flaky tails. These F₁ mice were test crossed to jerker having droopy ears and long flaky tail mouse. It results in 400 offspring with 16 possible different phenotypic traits.

The possible number of phenotypic categories can be estimated by finding the percentage of recombination between various genes. Map distance is expressed by the map unit (mu), where 1 mu is equivalent to 1% of the recombination offspring in a testcross. The percentage of recombinants between normal/droopy ears and normal/flaky tail will be $6\%$ as it is present at a distance of 6 map units on the chromosome 3 and that of normal ears/flaky tail and droopy ears/non-flaky tail will be $3\%$. Half of the $94\%$ that are remaining will be normal ears/nonflaky tail and half will be droopy ears/flaky tail, that is, $\text{47\% droopy ears/flaky tail}$ and $\text{47\% normal ears/non-flaky tail}$.

The other two genes for the length of the tail and the gait of the mouse will be assorted independently of the two linked genes for ear and type of tail. There will be 0.5 or $\text{50\% normal gait}$ contrasted with $\text{50\% jerker}$ along with $\text{50\% long tail}$ as opposed to $\text{50\% short tail}$ for every phenotypic category. The relative number of each type of offspring will be calculated by multiplying these percentages together with the total number of offspring that were produced in the test cross, which is 400. It is calculated by using the formula of map distance.

$\text{Map distance = }\frac{\text{Number of recombinant offspring}}{\text{ Total number of offspring}}\text{ x 100 }$

Recombination frequency of offspring is dependent on the phenomena of crossing over. This crossing over in turn depends on the distance between the genes.

Percentage of recombinants of different phenotypic characters are as follows:

$\begin{array}{l}\text{Droopy ears/flaky tail = 0}\text{.47}\\ \text{Droopy ears/nonflaky tail = 0}\text{.03}\\ \text{Normal ears/flaky tail = 0}\text{.03}\\ \text{Normal ears/nonflaky tail = 0}\text{.47}\end{array}$

$\begin{array}{l}\text{Normal gait = 0}\text{.5}\\ \text{Jerker = 0}\text{.5}\\ \text{Long tail = 0}\text{.5}\\ \text{Short tail = 0}\text{.5}\end{array}$

On the basis of the above information, the number of offspring in the 16 phenotypic categories can be calculated by using map distance formula as shown below.

$\begin{array}{c}\text{Map distance }\text{X}{\text{P}}_{\text{non-recombinant}}{\text{X Total }}_{\text{offspring}}\text{= }\text{Number of individuals}\\ \text{0}\text{.47×0}\text{.5×0}\text{.5×400 = 47 }\end{array}$

Hence, each of the following phenotypes will produce 47 offspring each.

$\begin{array}{c}\text{47 droopy ears/flaky/long tail/normal gait}\\ \text{47 normal ears/nonflaky/long tail/normal gait}\\ \text{47 droopy ears/flaky/long tail/ jerker}\\ \text{47 normal ears/nonflaky/long tail/ jerker}\end{array}$

$\begin{array}{c}\text{47 droopy ears/flaky/short tail/jerker}\\ \text{47 droopy ears/flaky/short tail/normal gait}\\ \text{47 normal ears/nonflaky/short tail/jerker}\\ \text{47 normal ears/nonflaky/short tail/normal gait}\end{array}$

Similarly, by using the map distance formula, a number of offspring of the following traits can be determined:

$\begin{array}{c}\text{Map distance }\text{X}{\text{P}}_{\text{non-recombinant}}{\text{X Total }}_{\text{offspring}}\text{= }\text{Number of individuals }\\ 0.03\times 0.5\times 0.5\times 400=3\end{array}$

Hence, each of the following phenotypes will produce 3 offspring each.

$\begin{array}{c}\text{3 droopy ears/nonflaky/long tail/jerker}\\ \text{3 droopy ears/nonflaky/long tail/normal gait}\\ \text{3 normal ears/flaky/long tail/jerker}\\ \text{3 normal ears/flaky/long tail/normal gait}\end{array}$

$\begin{array}{c}\text{3 droopy ears/nonflaky/short tail/jerker}\\ \text{3 droopy ears/nonflaky/short tail/normal gait}\\ \text{3 normal ears/flaky/short tail/jerker}\\ \text{3 nomal ears/flaky/short tail/normal gait}\end{array}$

So, a test cross between normal F₁ mice with the recessive phenotypic traits will result in 400 offspring with 16 different phenotypic categories. The offspring are categorized into 8 phenotypic charactersthat produce 47 offspring and other 8 produce 3 offspring each with a different combination of phenotypes.

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Conclusion

Therefore, it can be concluded that the expected number of the 16 phenotypic categories are:

$\begin{array}{l}\text{Droopy ears/flaky/long tail/normal gait = 47}\hfill \\ \text{Droopy ears/flaky/long tail/jerker = 47}\hfill \\ \text{Normal ears/non-flaky/long tail/normal gait = 47}\hfill \\ \text{Normal ears/non-flaky/long tail/jerker = 47}\hfill \\ \text{Droopy ears/flaky/short tail/jerker = 47}\hfill \\ \text{Droopy ears/flaky/short tail/normal gait = 47}\hfill \\ \text{Normal ears/non-flaky/short tail/jerker = 47}\hfill \\ \text{Normal ears/non-flaky/short tail/normal gait = 47}\hfill \\ \text{Droopy ears/non-flaky/long tail/jerker = 3}\hfill \\ \text{Droopy ears/non-flaky/long tail/normal gait = 3,}\hfill \\ \text{Normal ears/flaky/long tail/jerker = 3,}\hfill \\ \text{Normal ears/flaky/long tail/normal gait = 3,}\hfill \\ \text{Droopy ears/non-flaky/short tail/jerker = 3,}\hfill \\ \text{Droopy ears/non-flaky/short tail/normal gait =3,}\hfill \\ \text{Normal ears/flaky/short tail/jerker = 3, and}\hfill \\ \text{Normal ears/flaky/short tail/normal gait = 3}\hfill \end{array}$