1) A White eye (w) female Drosophila is crossed with diminutive (dm) males. The F1 wild type females are then crossed to the double mutant, white eye, diminutive males and produce the following:
white eye, normal size 487
normal eye, diminutive 475
normal eye, normal size 19
white eye, diminutive
12
Could you show me the crosses using the proper symbols?
What is the map distance between white and diminutive?
2) You have a line of Drosophila that has both yellow body and vermillion (v) eyes. Can you show the crosses, using the proper symbols to construct a line of flies that has only yellow bodies. I believe they are both X linked.
If you could help me with that would be great
Sure, let's tackle each question one by one:
1) To represent the crosses using proper symbols, we use the following conventions:
- A capital letter represents a dominant allele, and a lowercase letter represents a recessive allele.
- The "+" symbol represents the wild-type allele, meaning the normal or most common form of a gene.
- A slash (/) between two alleles indicates a heterozygote, meaning an individual with two different alleles for a particular gene.
- A comma (,) separates genes that are on different chromosomes.
Based on the information provided, we can depict the crosses as follows:
Parental cross:
w/w (female) x dm/Y (male)
F1 cross:
w/+ (female) x w/Y (male)
F2 cross:
w/w (female) x dm/Y (male)
The results of the F2 cross are given as follows:
white eye, normal size = 487
normal eye, diminutive = 475
normal eye, normal size = 19
white eye, diminutive = 12
To determine the map distance between the white and diminutive genes, we can calculate the recombination frequency. The recombination frequency is the percentage of recombinant offspring compared to the total number of offspring. In this case, the recombinant offspring are the white eye, diminutive flies.
Recombination frequency = (recombinant offspring / total offspring) * 100
= (12 / (487 + 475 + 19 + 12)) * 100
≈ (12 / 993) * 100
≈ 1.21%
Therefore, the map distance between the white and diminutive genes is approximately 1.21 cM (centimorgans).
2) To construct a line of flies with only yellow bodies, we need to perform appropriate crosses. Since both the yellow body (y) and vermillion eye (v) genes are X-linked (on the X chromosome), we need to ensure that the desired traits are carried on the X chromosome.
To achieve this, we can perform the following crosses:
Parental cross:
XvXv (female with vermillion eyes) x Xy (male with yellow body)
F1 cross:
XvXy (female with vermillion eyes and yellow body) x Xy (male with yellow body)
F2 cross:
XvXy (female with vermillion eyes and yellow body) x Xy (male with yellow body)
By continuously crossing the F2 flies from the previous cross with flies carrying the desired traits (yellow body), we can eventually obtain a line of flies with only yellow bodies.
1) Crosses using the proper symbols:
Parental cross: w (white eye) female x dm (diminutive) male
F1 cross: F1 females (wild type) x w dm (white eye, diminutive) males
F1 females (wild type) x w dm (white eye, diminutive) males:
487 white eye, normal size
475 normal eye, diminutive
19 normal eye, normal size
12 white eye, diminutive
To determine the map distance between white and diminutive, we need to use the formula:
Map Distance = (Number of recombination events / Total number of offspring) x 100
From the given data, we can determine the number of recombination events between white and diminutive by adding the number of normal eye, diminutive and the number of white eye, normal size:
Number of recombination events = 475 (normal eye, diminutive) + 12 (white eye, normal size) = 487
Total number of offspring from the F1 cross = 487 (white eye, normal size) + 475 (normal eye, diminutive) + 19 (normal eye, normal size) + 12 (white eye, diminutive) = 993
Map Distance = (487 / 993) x 100 = 49.04%
Therefore, the map distance between white and diminutive is approximately 49.04%.
2) Crosses to construct a line of flies with only yellow bodies:
Parental cross: yellow body (yb) male x vermillion eyes (v) female
F1 cross: F1 males (yellow body, vermillion eyes) x F1 females (yellow body, vermillion eyes)
F1 males (yellow body, vermillion eyes) x F1 females (yellow body, vermillion eyes):
All offspring will have both yellow bodies and vermillion eyes.
To create a line of flies with only yellow bodies, you need to perform a cross between F1 males (yellow body, vermillion eyes) and F1 females (yellow body, vermillion eyes) and then select the offspring that have yellow bodies only. Continuously selecting and breeding those offspring will eventually yield a line of flies with only yellow bodies.