Introduction to Genetic Analysis
11th Edition
ISBN: 9781464109485
Author: Anthony J.F. Griffiths, Susan R. Wessler, Sean B. Carroll, John Doebley
Publisher: W. H. Freeman
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Chapter 4, Problem 36P
Summary Introduction
To determine: The genotypes of the spores in unordered tetrads that are parental ditypes, tetratypes and nonparental ditypes.
Introduction. The law of independent assortment was given in a dihybrid cross which is the cross between the alleles of two pairs of contrasting characters. The alleles produced by each character are independent to combine with the alleles of the other character.
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Students have asked these similar questions
From the fungal cross arg-6 ⋅ al-2 × arg-6+ ⋅ al-2+, whatwill the spore genotypes be in unordered tetrads thatare (a) parental ditypes? (b) tetratypes? (c) nonparental ditypes?
The following results are derived from crosses between Neurospora strain xy and strain ++:
Tetrad Class
3
4
ху
x+
x+
xy
++
++
++
+y
+y
ху
+y
25
++
3
124
4
(i)
Name the ascus type of each class from 1 to 4 as P, NP or T.
(ii)
Are genes x and y linked? Explain your answer.
(iii)
If they are linked, determine the map distance between the two genes. If they are
unlinked, provide all the information you can about why you draw this conclusion.
In tomato the mutant genes o (oblate=flattened fruit), p (peach=hairy fruit)
and s (compound inflorescence) were found to be in chromosome 2. The
test cross results are:
73
110
+
348
2
2
+
306
p
96
63
a) Construct the linkage map
b) What are the genotypes of the homozygous parents used in making the F1
heterozygote?
c) Compute for the coefficient of coincidence.
O o O O
+
Chapter 4 Solutions
Introduction to Genetic Analysis
Ch. 4 - Prob. 1PCh. 4 - Prob. 5PCh. 4 - Prob. 12PCh. 4 - Prob. 13PCh. 4 - Prob. 14PCh. 4 - Prob. 15PCh. 4 - Prob. 16PCh. 4 - Prob. 17PCh. 4 - Prob. 18PCh. 4 - Prob. 19P
Ch. 4 - Prob. 20PCh. 4 - Prob. 21PCh. 4 - Prob. 21.1PCh. 4 - Prob. 21.2PCh. 4 - Prob. 21.3PCh. 4 - Prob. 21.4PCh. 4 - Prob. 21.5PCh. 4 - Prob. 21.6PCh. 4 - Prob. 21.7PCh. 4 - Prob. 21.8PCh. 4 - Prob. 21.9PCh. 4 - Prob. 21.10PCh. 4 - Prob. 21.11PCh. 4 - Prob. 21.12PCh. 4 - Prob. 21.13PCh. 4 - Prob. 21.14PCh. 4 - Prob. 21.15PCh. 4 - Prob. 21.16PCh. 4 - Prob. 21.17PCh. 4 - Prob. 21.18PCh. 4 - Prob. 21.19PCh. 4 - Prob. 21.20PCh. 4 - Prob. 21.21PCh. 4 - Prob. 21.22PCh. 4 - Prob. 21.23PCh. 4 - Prob. 21.24PCh. 4 - Prob. 21.25PCh. 4 - Prob. 21.26PCh. 4 - Prob. 22PCh. 4 - Prob. 23PCh. 4 - Prob. 24PCh. 4 - Prob. 25PCh. 4 - Prob. 26PCh. 4 - Prob. 27PCh. 4 - Prob. 28PCh. 4 - Prob. 29PCh. 4 - Prob. 30PCh. 4 - Prob. 31PCh. 4 - Prob. 32PCh. 4 - Prob. 33PCh. 4 - Prob. 34PCh. 4 - Prob. 35PCh. 4 - Prob. 36PCh. 4 - Prob. 37PCh. 4 - Prob. 38PCh. 4 - Prob. 38.1PCh. 4 - Prob. 38.2PCh. 4 - Prob. 38.3PCh. 4 - Prob. 38.4PCh. 4 - Prob. 38.5PCh. 4 - Prob. 38.6PCh. 4 - Prob. 38.7PCh. 4 - Prob. 38.8PCh. 4 - Prob. 38.9PCh. 4 - Prob. 38.10PCh. 4 - Prob. 38.11PCh. 4 - Prob. 38.12PCh. 4 - Prob. 38.13PCh. 4 - Prob. 38.14PCh. 4 - Prob. 38.15PCh. 4 - Prob. 38.16PCh. 4 - Prob. 38.17PCh. 4 - Prob. 38.18PCh. 4 - Prob. 38.19PCh. 4 - Prob. 38.20PCh. 4 - Prob. 38.21PCh. 4 - Prob. 38.22PCh. 4 - Prob. 38.23PCh. 4 - Prob. 38.24PCh. 4 - Prob. 39PCh. 4 - Prob. 40PCh. 4 - Prob. 41PCh. 4 - Prob. 42PCh. 4 - Prob. 43PCh. 4 - Prob. 44PCh. 4 - Prob. 45PCh. 4 - Prob. 46PCh. 4 - Prob. 47PCh. 4 - Prob. 48PCh. 4 - Prob. 49PCh. 4 - Prob. 50PCh. 4 - Prob. 51PCh. 4 - Prob. 52PCh. 4 - Prob. 53PCh. 4 - Prob. 54PCh. 4 - Prob. 55PCh. 4 - Prob. 56PCh. 4 - Prob. 57PCh. 4 - Prob. 58PCh. 4 - Prob. 59PCh. 4 - Prob. 60PCh. 4 - Prob. 62PCh. 4 - Prob. 63PCh. 4 - Prob. 64PCh. 4 - Prob. 65PCh. 4 - Prob. 66PCh. 4 - Prob. 67PCh. 4 - Prob. 68PCh. 4 - Prob. 69P
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- The expected ratio of phenotypes among the progeny of a test cross is 1:1:1:1. Out of 200 total resulting progeny, 48 occur in one of the four phenotypic classes. Given this information, which of the following must also be true? a)At least one additional cell must also contain a count of 48. b)The progeny of this cross do not conform to a 1:1:1:1ratio. c)The value of observed - expected for this cell = -2. d)Since 48 is so close to the expected value, there is no need to calculate chi square before drawing a conclusion about the ratio.arrow_forwardIn a trihybrid cross of a tall, purple-flowered pea plant with round seeds (TtPpRr) with a short, white- flowered pea plant with round seeds (ttppRr), what is the probability: A) that a short, white-flowered plant with wrinkled seeds will be produced? B) that a tall, purple-flowered plant with round seeds will be produced? C) that a tall, white-flowered plant with round seeds will be produced?arrow_forwardYou cross two yeast strains one is an ade auxotroph the other is a pro auxotroph and allow the diploid to sporulate. When you score each spore in the ascus you find the following proportions: 518 PD, 8 NPD, and 225 T. a.) What are the genotypes of each spore in all three types of the tetrads. b) Are these genes linked why or why not? c.) If these genes are unlinked what would you expect the progeny numbers and ratios to be? d.) What is the formula to determine the most accurate distance between these genes? If linked what is the map distance?arrow_forward
- In rice, male sterility is controlled by maternal cytoplasmic elements. This phenotype renders the male part of rice plants (i.e. the stamen) unable to produce fertile pollen; the female parts, however, remain receptive to pollination by pollen from male fertile rice plants. However, the presence of a nuclear fertility restorer gene F restores fertility to male-sterile lines. Give the result(s) of the cross and explain the phenotype of the offspring.arrow_forwardIn a cross between a white-eyed female (ww) and a red-eyed male (w+Y), nearly all the progeny were either red-eyed females (w+w) or white-eyed males (wY). However, about 1 in every 2000 F1 flies had an "exceptional phenotype" and was either a white-eyed female or red-eyed male. How did Bridges explain this unexpected result? A) Crossing over B) Incomplete cytokinesis C) Incorrect synapsis D) Nondisjunction E) Pseudoautosomal regionarrow_forwardPlease label the tetrad type in the table as PD (parental ditype), NPD (non parental ditype) or T (tetratype) and answer the following questions a) Are the genes linked? Please explain SPECIFICALLY how you can distinguish between linked and unlinked genes in this instance. b) If the two genes are linked, calculate the % recombination between ser and thr. Show the formula used, as well as all of your calculations. c) Draw a single map illustrating the arrangement of the two genes on the chromosome with respect to each other and to the centromere of the chromosome. Make sure to map ALL three distancesarrow_forward
- Consider the following two nonhomologous wildtype chromosomes, where letters or numbers represent genes, the "-" represents the centromere of each chromosome, and chromosomes are shown on separate lines. ABCDE-FGHIJK 123-45678 Identify the type of rearrangement shown in each of the following (A-C) and then identify whether it is balanced or unbalanced. Assume that the individual is diploid and heterozygous for the rearrangement. A. ABCDE-FGHIJKGH 123-45678 Rearrangement: [Select] • Balanced or Unbalanced: [Select] B. ABCDGF-EHIJK 123-45678 Rearrangement: [Select] Balanced or Unbalanced: [Select]arrow_forwardAssume that the trihybrid cross AABBrr x aabbRR is made in a plant species. Assume that A and B are dominant alleles, but there is no dominance effect of alleles at the R locus. a) How many different gametes are possible in the F1generation? What are the genotypes of these gametes? b) What is the probability of the parental aabbRR genotype in the F2 progeny? c) What proportion of the F2 progeny would be expected to be homozygous for all three genes?arrow_forwardB4) Construct a Punnett square using the dihybrid cross TtGg × TtGg. (TT, Tt codes for tall plants, tt for dwarf plants;GG, Gg codes for green pods, gg for yellow pods.) Number the boxes from left to right, starting at the top(see below).B5) Describe the phenotypes of boxes 2, 8, 12, and 16.B6) Describe the genotypes of boxes 4, 5, 11, and 15.arrow_forward
- Alleles of genes A and B were analyzed in Neurospora according to the cross shown below. Ordered tetrads are summarized in each horizontal row with the number of tetrads in each category listed alongside. A) Analyze the data to determine the recombination frequency (RF) between A and B, along with any additional information that is available from this data. B) Draw a map of the chromosome or chromosomes with appropriate map distances. C) Use the Perkins formula to reanalyze any relationship between A and B.arrow_forwardImagine that you have done a cross between two strains of yeast, one of which has the genotype A B C and the other a b c, where the letters refer to three closely linked genes in the order given. You examine many tetrads resulting from this cross, and the two tetrads below are found that have atypical scoring patterns. In tetrad I, the spores are A B C, ABC, a B c, and a b c. In tetrad II, the spores are A B C, A b c, a b C, and a b c. Which tetrad or tetrads indicate(s) that gene conversion has occurred AND show(s) that recombination has occurred between genes A and C? O neither tetrads I and II O tetrad II O tetrad I O both tetrads I and IIarrow_forwardA Neurospora colony at the edge of a plate seemed to be sparse (low density) in comparison with the other colonies on the plate. This colony was thought to be a possible mutant, and so it was removed and crossed with a wild type of the opposite mating type. From this cross, 100 ascospore progeny were obtained. None of the colonies from these ascospores was sparse, all appearing to be normal. What is the simplest explanation of this result? How would you test your explanation? (Note: Neurospora is haploid.)arrow_forward
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