Chapter 7

Question 1

which statement best describes linkage?

Answer 1

it is a condition in which two or more genes do not show independent assortment

Question 2

in a testcross involving two heterozygous genes, equal numbers of recombinant and non recombinant progeny are produced.

from this result, what can be concluded ?

Answer 2

the genes are not linked

Question 3

a recombination frequency of 5% translates to what distance on a genetic map?

Answer 3

5 m.u

Question 4

three-point test crosses are often used to map genes

the two least frequency classes from such crosses usually represent which types of progeny?

Answer 4

double-crossover progeny

Question 5

when two genes are linked but quite far apart their estimated map distance, based on recombination frequencies, is often an underestimation of their true map distance.

what is the explanation fro this underestimation?

Answer 5

some double crossover events go undetected since they do not lead to recombinant progeny

Question 6

a genetic cross with two genes produces 400 offspring, and 20 of them have recombinant phenotypes.

what is the recombination frequency for this cross?

Answer 6

5%

20/ 400= 0.05–> 5%

Question 7

the recombination frequency between genes A and B is 23%b

what is the distance between A and B in map units?

Answer 7

23 m.u

Question 8

what information about recombination frequencies enables scientist to create linkage maps?

Answer 8

the recombination frequency is proportional to the distance between two genes

Question 9

which statement explains why the recombination frequency between two genes is always less than 50 %

Answer 9

genes with a recombination frequency near 50% are unlinked and have an equal likelihood of being inherited together or separately

Question 10

1. bbTT x true breed male & BBtt true breed female
2. BbTt x BbTt
   results:

dark-eyed, short-tailed (BBtt)= 22
dark-eyed, long-tailed (BbTt or BBTT)= 56
light-eyed, long tailed (bbTT)=23
light-eyed, short-tailed (bbtt)=4

what is the most likely explanation for why Diane sees light-eyed, short-tailed (bbtt) progeny in this cross?

Answer 10

recombination

Question 11

which statement is the definition of a map unit (centimorgan)?

Answer 11

it is the percent chance of a crossover between two locations on a chromosome

Question 12

F1 test cross progeny phenotype 
broad =621
fast, zigzag= 623
zigzag=211
fast, broad= 221
fast= 215
zigzag, broad= 207
fast, zigzag, broad= 8
wild type= 7 

select the order of the three genes

Answer 12

fbz

zbf

Question 13

test progeny phenotype 
curved= 277 
anomalous, buckled= 283 
wild type= 5
anomalous, buckled, curved=5
buckled, curved= 128
anomalous= 132
anomalous, curved= 86
buckled= 84 

calculate map distance between a and c

Answer 13

5+5+86+94 recombinants/ 1000 total = 0.18= 18 m.u.