Chapter 5, Problem 44P

This problem leads you through the derivation of a corrected equation for RF in yeast tetrad analysis that takes into account double crossover (DCO) meioses.

A yeast strain that cannot grow in the absence of the amino acid histidine (his^-) is mated with a yeast strain that cannot grow in the absence of the amino acid lysine (lys^-). Among the 400 unordered tetrads resulting from this mating, 233 were PD, 11 were NPD, and 156 were T.

a.	What types of spores are in the PD, NPD, and T tetrads?
b.	Are the his and lys genes linked? How do you know?
c.	Using the simple equation RF = 100 × [NPD + (1/2) T]/total tetrads, calculate the distance in map units between the his and lys genes.
d.	If you think about all the kinds of meiotic events that could occur (refer to Fig. 5.24), you can see that the calculation you did in part (c) may substantially underestimate RF. What kinds of meioses (NCO, SCO, or DCO) generated each of the tetrad types in this cross?
e.	What incorrect assumptions does the simple RF equation you used in part (c) make about the meiotic events producing each type of tetrad? When could these assumptions nevertheless be correct?
f.	Use your answers to part (d) to determine the number of NCO, SCO, and DCO meioses that generated the 400tetrads.
g.	Use your answers to part (f) to write a general equation that relates the number of DCO meioses to the number of the various tetrad types. Then write another general equation that computes the number of SCO meioses as a function of the number of the various tetrad types.
h.	Based on your answer to part (f), calculate the average number of crossovers per meiosis (m) between his and lys.
i.	Use your answer to (h) to write an equation for m in terms of NCO, SCO, and DCO meioses.
j.	What is the relationship between RF and m?
k.	Use your answer to part (j) to write a corrected equation for RF in terms of SCO, DCO, and NCO meioses.
l.	Using your answer to part (g), rewrite the corrected RF equation from part (k) in terms of the numbers of the various tetrad types.
m.	The equation you just wrote in part (l) is a corrected equation for RF that takes into account double crossovers that would otherwise have been missed. Use this improved formula to calculate a more accurate distance between the his and lys genes than the one you calculated in part (c).

Summary Introduction

a.

To determine:

The forms of spores that are in PD, NPD, and T tetrads.

Introduction:

The cross between two yeast strains can be represented as:

$hi{s}^{-}ly{s}^{+}\times hi{s}^{+}ly{s}^{-}\to hi{s}^{-}hi{s}^{+}ly{s}^{+}ly{s}^{-}$

Explanation of Solution

The PD (parental ditype) asci would have $2hi{s}^{-}ly{s}^{+}:2hi{s}^{+}ly{s}^{-}$ spores. The NPD (nonparental ditype) would have $2hi{s}^{+}ly{s}^{+}:2hi{s}^{-}ly{s}^{-}$ spores. The T (tetratype) asci would have $1hi{s}^{-}ly{s}^{+}:1hi{s}^{+}ly{s}^{-}:1hi{s}^{+}ly{s}^{+}:1hi{s}^{-}ly{s}^{-}$ spores.

Summary Introduction

b.

To determine:

Whether the genes his and lys are linked.

Introduction:

The number of parental ditypes is equal to 233, and the number of nonparental ditypes is equal to 11.

Explanation of Solution

When the yeast strains that are dihybrid for linked genes undergo meiotic division, the number of parental ditypes (PDs) produced far exceeds the number of non-parental ditypes. Thus, the genes his and lys are linked.

Summary Introduction

c.

To determine:

The distance between the genes his and lys.

Introduction:

The distance between two linked genes can be calculated by the formula:

$\text{RF}=\frac{\left[\text{NPD}+\frac{1}{2}\left(\text{T}\right)\right]}{\text{Total tetrads}}\times 100$. Here, RF denotes recombination frequency

Explanation of Solution

The distance between the genes his and lys is:

$\begin{array}{rll}RF& =& \frac{\left[11+\frac{1}{2}\left(156\right)\right]}{233+11+156}\times 100\\ & =& 22.25\text{ map unit}\end{array}$

Thus, the distance between the genes his and lys is 22.25 map unit.

Summary Introduction

d.

To determine:

The types of meiotic divisions that produced each of the tetrad forms.

Introduction:

The different types of double crossover (DCO) meiosis are two-strand DCO, three-strand DCO, and four-strand DCO.

Explanation of Solution

The non-parental ditype asci are the result of four-strand double crossovers (DCOs) because the genes his and lys are linked. The three-strand double crossovers give tetratype asci, and two-strand double crossovers give parental ditype asci. Eleven tetrads underwent four-strand double crossovers, and 22 tetrads (tetratype asci) are the result of three-strand double crossovers. Another 11 asci that are parental ditypes underwent two-strand double crossovers. In total 44 $\left(11+22+11\right)$ asci underwent two crossovers and the remaining 134 $\left(156-22\right)$ tetratype asci underwent single crossover. The remaining 222 $\left(233-11\right)$ parental ditype asci underwent no (zero) recombination.

Summary Introduction

e.

To determine:

The incorrect assumption that is used in a simple RF equation and the time where these assumptions nevertheless are correct.

Introduction:

RF (recombination frequency) is used to calculate the physical distance that separates any two genes on the same chromosome.

Explanation of Solution

The simple equation for RF between two genes only involves four-strand double crossovers that are non-parental ditypes and ignores three-strand and two-strand double crossovers. This equation assumes that tetratype asci are the result of single crossovers. This assumption is valid for the majority of tetratype asci but, a small proportion of tetratype asci are the result of three-strand double crossovers. Moreover, it is an oversimplification to think that all parental ditype asci are nonrecombinants because some of them are the result of two-strand double crossovers.

Summary Introduction

f.

To determine:

The number of different types of meioses that produced 400 tetrads.

Introduction:

The assemblage of four ascospores that produce from meiotic division in a single ascus is defined as a tetrad.

Explanation of Solution

The total of 11 tetrads underwent four-strand double crossovers (DCOs), and another 22 tetrads are the result of three-strand double crossovers. Also, another 11 asci underwent two-strand double crossovers. In total, 44 $\left(11+22+11\right)$ asci underwent double crossovers (DCOs), and the remaining 134 $\left(156-22\right)$ tetratype asci underwent single crossover (SCO). The remaining 222 $\left(233-11\right)$ parental ditype asci underwent no crossover (NCO).

Summary Introduction

g.

To determine:

The equation that relates the amount of DCO to the number of various tetrad forms and the equation that computes the amount of SCO as a function of a number of tetrad forms.

Introduction:

Double crossover (DCO) is a type of meiotic division in which two crossovers take place between a particular gene pair.

Explanation of Solution

DCO and SCO are the forms of meiotic division. The equation that relates DCO and SCO with various tetrad forms is as follows:

$\text{SCO}+\text{DCO}=\frac{1}{2}\left(\text{T}-2\text{NPD}\right)+4\left(\text{NPD}\right)$

Summary Introduction

h.

To determine:

The average number of crossovers per meiotic division (m) between the genes his and lys.

Introduction:

Meiosis refers to the process of two consecutive cell divisions that results in the formation of gametes.

Explanation of Solution

The mean number of crossovers per meiotic division (m) can be calculated by the following equation:

$\begin{array}{rll}\left(\text{T}-2\text{NPD}\right)+2\left(4\text{NPD}\right)& =& \text{T}+6\text{NPD}\\ & =& \mathrm{156+6}\left(11\right)\\ & =& 222\end{array}$

Summary Introduction

i.

To determine:

The equation for m in terms of NCO, SCO, and DCO meioses.

Introduction:

NCO (no crossover) is a type of meiotic division that takes place with no crossovers between a given gene pair.

Explanation of Solution

Double crossover is equal to 4 times of nonparental ditype, and the single crossover is equal to $\text{Tetratype}-2\text{NPD}$. No crossover is equal to $1-\left(\text{SCO}-\text{DCO}\right)$. The average number of crossovers per meiotic division (m) can be represented as $\text{m}=\text{T}+6\text{NPD}$

Here, $\text{NPD}=\frac{\text{DCO}}{4}$ and $T=SCO+\frac{DCO}{2}$.

Thus, $\text{m}=\text{SCO}+\frac{\text{DCO}}{2}+6\left(\frac{\text{DCO}}{4}\right)$

Summary Introduction

j.

To determine:

The relationship between RF and meiosis.

Introduction:

RF (recombination frequency) is equal to the number of recombinant spores divided by the total number of spores.

Explanation of Solution

The recombinant spores are the results of single crossovers (SCOs) and the double crossovers (DCOs). The total number of spores is equal to parents plus recombinants. Parental spores are the result of no crossovers (NCOs). NCOs, SCOs, and DCOs, are the types of meiotic division. Thus, the recombination frequency directly depends on the type of meiosis.

Summary Introduction

k.

To determine:

The corrected equation for RF in terms of SCO, DCO, and NCO meiotic divisions.

Introduction:

RF (recombination frequency) refers to the percentage of recombinant progeny and can be used to indicate the distance between two genes on the same chromosome.

Explanation of Solution

The corrected RF equation can be represented as follows:

$\text{RF}=\frac{\text{SCO}+\text{DCO}}{\text{Total asci}}\times 100$

Summary Introduction

l.

To determine:

The corrected RF equation in terms of the numbers of various tetrad forms.

Introduction:

Nonparental ditype is a form of fungal tetrad that contains about four recombinant ascospores.

Explanation of Solution

The corrected RF equation is as follows:

$\begin{array}{rll}\text{RF}& =& \frac{\frac{1}{2}\left(\text{T}-2\text{NPD}\right)+4\left(\text{NPD}\right)}{\text{Total asci}}\times 100\\ & =& \frac{\frac{1}{2}\left(156-2\left(11\right)\right)+4\left(11\right)}{400}\times 100\end{array}$

Summary Introduction

m.

To determine:

The more accurate distance between the genes his and lys based on the equation provides in part (l).

Introduction:

The map unit is defined as a unit of measure of RF. One map unit equals to 1% chance that one genetic locus is separated from a marker at second genetic locus due to crossover.

Explanation of Solution

The more accurate distance between the genes his and lys can be represented as:

$\begin{array}{rll}\text{RF}& =& \frac{\frac{1}{2}\left(156-2\left(11\right)\right)+4\left(11\right)}{400}\times 100\\ & =& \frac{\frac{1}{2}\left(156-22\right)+44}{400}\times 100\\ & =& 27.75\text{ map unit}\end{array}$