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Gene Segregation Problem Set

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A researcher studied six independently assorting genes in a plant. Each gene has a dominant and a recessive allele: R black stem, r red stem; D tall plant, d dwarf plant; C full pods, c con­stricted pods; O round fruit, o oval fruit; H hairless leaves, h hairy leaves; W purple flower, w white flower. From the cross
(P1) Rr Dd Oo cc Hh Ww X (P2) Rr dd Cc oo Hh ww,
(a) how many kinds of gametes can be formed by P1 ?
(b) How many genotypes are possible among the progeny of this cross?
(c) How many phenotypes are possible among the progeny?
(d) What is the probability of obtaining the Rr Dd cc Oo hh ww genotype in the progeny?
(e) What is the probability of obtaining a black, dwarf, constricted, oval, hairy, purple phenotype in the progeny?

2.B
Mendel testcrossed pea plants grown from yellow, round F1 seeds to plants grown from green, wrinkled seeds and obtained the following results: 31 yellow, round; 26 green, round; 27 yellow, wrinkled; and 26 green, wrinkled. Are these results consistent with the hypothesis that seed color and seed texture are controlled by independently assorting genes, each segregating two alleles?

5. GIVEN THAT AN ORGANISM HAS 4 METACENTRIC AND 8 TELOCENTRIC CHROMOSOMES BEFORE MEIOTIC DIVISION:
 HOW MANY CHROMOSOMES DOES THE ORGANISM HAVE IN PROPHASE I?

 HOW MANY CHROMATIDES DOES THE ORGANISM HAVE IN PROPHASE II?

6. GIVEN THAT AN ORGANISM HAS 10 METACENTRIC CHROMOSOMES BEFORE MEIOTIC DIVISION:
 HOW MANY CHROMOSOMES DOES THE ORGANISM HAVE IN PROPHASEII ?

 HOW MANY CHROMATIDS DOES THE ORGANISM HAVE IN PROPHASE I IF ALL CHROMOSOMES WERE METACENTRIC?

8. IF AN ORGANISM HAS A GENOTYPE OF AABbCCDdEe, HOW MANY DIFFERENT KINDS OF GAMETES CAN IT PRODUCE? NAME THEM.

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Solution Summary

Answers and explanations with Punnett squares of several gene segregation problems in plants.

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2A.
(a) (P1) Rr Dd Oo cc Hh Ww Because each of these alleles is unlinked each gamete will inherit one of the two alleles. The total possible combinations are as follows: RDOcHW, RDOcHw, RDOchW, RDOchw, RDocHW, RDocHw, RDochW, RDochw, RdOcHW, RdOcHw, RdOchW, RdOchw, RdocHW, RdocHw, RdochW, Rdochw, rDOcHW, rDOcHw, rDOchW, rDOchw, rDocHW, rDocHw, rDochW, rDochw, rdOcHW, rdOcHw, rdOchW, rdOchw, rdocHW, rdocHw, rdochW, rdochw. That's a total of 32 gamete combination. An easier way to calculate the number of gametes is to multiply the chances of inheriting each allele:
0.5 x 0.5 x 0.5 x 1 x 0.5 x 0.5= 1/32 Therefor there is a 1/32 chance of each allelic combination and 32 total different gamates possible.

(b) (P1) Rr Dd Oo cc Hh Ww X (P2) Rr dd Cc oo Hh ww
The simplest way to determine the genotypes possible from this cross is to create a Punnett square. See attached document for the details. To quickly determine the number of genotypes possible multiply the number of gametes possible for each parent (ie. 32 x 8= 256). There are 256 different genotypes possible.

(c) To determine the number of different ...

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