My prof says this is done wrong as the break point is between R and Y. Could you please help me out? With explanation? Thank you!
10. In a certain plant, a dominant allele for one gene, R, produces round fruit and the recessive allele, r, produces oblong fruit. A dominant allele of a second gene, Y, produces yellow flowers and the recessive allele, y, produces white flowers. The two genes are linked on chromosome IV. A plant (r y/r y) with normal chromosomes is crossed to a plant (R Y/R Y) homozygous for a reciprocal translocation between chromosomes IV and VIII. The presence of the translocation causes semi-sterility. The F1 progeny are phenotypically yellow flowered with round fruit and are semi-sterile. A backcross to the parent with normal chromosomes yields the 750 progeny shown in the table below.
Number of Progeny
Fertile
Semi-sterile
Yellow, round
3
283
White, oblong
295
2
Yellow, oblong
73
12
White, round
10
72
a. From these data, draw a map showing all relevant map distances (in centimorgans) for the genes and the translocation breakpoint. Show your work.
Answer:
Cross over gametes: Yellow, oblong & White, round
Total number of Progeny = 381
Frequency of cross over gametes=
83 / 381 = .22 Morgans
22 cM
85 / 369 = .23 Morgans
23 cM
The genes are located on the same side of the chromosome.
Break point R Y
. . 22cM .
No; the breakpoint is located between the two genes. Also you muat use both classes of dluble crossovers to calculate the two intervals. Please redo
b. Briefly explain the basis of the semi-sterility of translocation-bearing plants.
Answer: The semi-sterility in F1 plants are caused by reciprocal translocation. As the organism is heterozygous for reciprocal translocation meaning that offsprings are produced less than normal. This reciprocal translocation may have occurred at meiosis during chromosome segregation. No; the translocation already existed.
Please read your text and redo 3/10
My prof says this is done wrong as the break point is between R and Y. Could you please help me out? With explanation? Thank you!
10. In a certain plant, a dominant allele for one gene, R, produces round fruit and the recessive allele, r, produces oblong fruit. A dominant allele of a second gene, Y, produces yellow flowers and the recessive allele, y, produces white flowers. The two genes are linked on chromosome IV. A plant (r y/r y) with normal chromosomes is crossed to a plant (R Y/R Y) homozygous for a reciprocal translocation between chromosomes IV and VIII. The presence of the translocation causes semi-sterility. The F1 progeny are phenotypically yellow flowered with round fruit and are semi-sterile. A backcross to the parent with normal chromosomes yields the 750 progeny shown in the table below.
| Number of Progeny | ||
| Fertile | Semi-sterile | |
| Yellow, round | 3 | 283 |
| White, oblong | 295 | 2 |
| Yellow, oblong | 73 | 12 |
| White, round | 10 | 72 |
a. From these data, draw a map showing all relevant map distances (in centimorgans) for the genes and the translocation breakpoint. Show your work.
Answer:
Cross over gametes: Yellow, oblong & White, round
Total number of Progeny = 381
Frequency of cross over gametes=
83 / 381 = .22 Morgans
22 cM
85 / 369 = .23 Morgans
23 cM
The genes are located on the same side of the chromosome.
Break point R Y
. . 22cM .
No; the breakpoint is located between the two genes. Also you muat use both classes of dluble crossovers to calculate the two intervals. Please redo
b. Briefly explain the basis of the semi-sterility of translocation-bearing plants.
Answer: The semi-sterility in F1 plants are caused by reciprocal translocation. As the organism is heterozygous for reciprocal translocation meaning that offsprings are produced less than normal. This reciprocal translocation may have occurred at meiosis during chromosome segregation. No; the translocation already existed.
Please read your text and redo 3/10
