Cannabis Genetic Observations

[spacetraveller]

This is interesting to me, as i am also working on a project with what i believe is a semi dwarf plant.
Thankyou for supplying the link to the paper on this topic, i have been reading with much fascination.
This plant i am referring too, is a sativa dominant hybrid. Her
genetics are Rays choice by kiwi seeds x (rays choice x golden tiger original by Ace seeds).
I produced these seeds in 2016, and i have grown many in that time. 2 main phenos appear. The rarer is the malawi pheno with clumpy flower clusters, the other is a very narrow leaf variety with more spear shaped colas. Both types however can grow very large and can overwhelm any grower not prepared for the stretch and duration of growth.
Last year a short stature plant showed itself amongst a group i had started. I treated as per normal for these varieties, that is topping the main trunk after the 4th internode pair. Because of her shortness, instead of building a horizontal base for the plant to grow from, i let it grow upwards without further control techniques. She stretched for 9 weeks under 11/13 photoperiod, but only grew to 100 cm tall. Amazing!
In addition to the short stature, other phenotypic traits include reduced thickness of the trunk, branches and twigs. Also increased apparent density of the branches, as they are much more difficult to cut. Other physiological changes are also apparent , including a reduced tolerance for fertilizer, and differences in seed sprouting for the F1 compared to "normal" seeds.
To discover how these traits could be passed onto future generations, i crossed this plant with a pure sativa (malawi x panama f1) from Ace. Of the seeds i sprouted from the crossing, no short stature plants appeared leading me to conclude several possibilities.
1. The trait is an unrepeatable fluke.
2. The trait is controlled by a mutated recessive gene.
3. The trait is controlled by polygenic inheritance.
I selected a male F1 and backcrossed him to the "dwarf" plant anf the f2 seeds are currently developing nicely.
More experiments in a few months!

Pictured is the plant in full flower. 2nd plant is a clone in seed production. 3rd pic is the f1 male i used. After a 5 day veg period, he stretched for 7 weeks under an 11/13 photoperiod, and grew to 140cm in a 3 liter pot.

 

[acespicoli]

Like SamS told me one day, population genetics is the struggle we face as cannabis hobbiest.
He asked the question how many people do you know that can plant a field of 10,000 plants?

With small numbers we risk losing these special traits with every reproduction... short version

What we can do is keep cuttings and bx
To stack the dwarf sativa trait to better suit indoor growing

The whole indica/sativa class comes down to
A: effect thc/cbd chemo type
B: wild/domesticated
Leaf narrow or skinny.... etc
Of course if you growing thc there a main focus

Its adventageous for a wild plant to stretch in competition for light among other plants. Short domesticates leaves will widen and the seed structure will change due to selection.

To save 200+ seed
on the first run would be wise
A dwarf male... thats unusual in its self
A Single grain of pollen

I have seen some single pole males stack tight internodes without much stretch, with stalked glands... how much they pass to the female
GCA general combining ability remains to be seen.

In mandelian genetics you didnt get what you expected, id guess a male trait or a recessive since females are the doiminant traits selected for, and with sativa ... did you smoke sample the male? For cbd id try that as well if it shows again... things we do for breeding, smoking leaf...
Just selecting a line for a single trait is work?
Stem rub for terpene profile?
Best Luck hope to see more!
Thanks for sharing


Id take a dwarf indica looking plant that smelled and smoked like my favorite wild sativa any day right ?
Outdoors would it yield 5-15 lbs ?
Indoor Outdoor and Greenhouse

 

[gorilla ganja]

I'm interested in this topic as well. Thanks so much, @acespicoli, for posting the paper on Dwarf germplasm.
I'd like to see what you have been working on. It would be nice to see a list of varieties that might be suited to this type of breeding.

Peace GG

 

[spacetraveller]

What we can do is keep cuttings and bx
To stack the dwarf sativa trait to better suit indoor growing

Id take a dwarf indica looking plant that smelled and smoked like my favorite wild sativa any day right ?
Outdoors would it yield 5-15 lbs ?
Indoor Outdoor and Greenhouse

I'm guessing by stacking the dwarf traits you are implying the genetic basis is polygenic? I can't say one way or the other yet, need more evidence imo. I am hopeful it may be controlled by Mendelian genetics. Certainly, it is not a single gene dominant. My research with dwarf rice and other plants indicates these plants are often controlled by recessive inheritance. Time will tell.

Regarding this trait in indicas, I have already made seed with 2 phenos of gorilla glue, a kosher cookies and a Chiquita banana plant. I am near retirement and looking for a project.....

 

[Lugo]

Small plants ive grown and come across among Colombian landrace and my MBSkunk landrace hybrid.

I personally think most plants can be manipulated into a dwarf style plant with container size, light spectrum manipulation, elevation and nutrition but yes, if you grow enough of them your bound to run into one eventually.

And they are very cute too

 

[acespicoli]

So what was that scumbag doing all that time. Coulda swore they claimed they were doing work breeding Cannabis. Yet here we are. Everyone's weed smells like fucking shoes and we pretend it's skunk. Great leaps and bounds this ass fucker has taken the world.

question for sam the skunkman on the original haze

looks like a pic from the vintage marijuana pictures thread

www.icmag.com

If you wanna know what hes been doing.
You should ask him, dont want to answer for him, hes perfectly able to address your questions himself

Maybe youd like to share some of your breeding work here ?

 

[acespicoli]

PRINCIPLES OF BIOLOGY
Answers to Hardy-Weinberg practice questions​

Updated: 21 August 2000

---

POPULATION GENETICS AND THE HARDY-WEINBERG LAW
ANSWERS TO SAMPLE QUESTIONS​

Remember the basic formulas:
p2 + 2pq + q2 = 1 and p + q = 1
p = frequency of the dominant allele in the population
q = frequency of the recessive allele in the population
p2 = percentage of homozygous dominant individuals
q2 = percentage of homozygous recessive individuals
2pq = percentage of heterozygous individuals

1. PROBLEM #1.
   You have sampled a population in which you know that the percentage of the homozygous recessive genotype (aa) is 36%. Using that 36%, calculate the following:
   
   
   1. The frequency of the "aa" genotype. Answer: 36%, as given in the problem itself.
   2. The frequency of the "a" allele. Answer: The frequency of aa is 36%, which means that q2 = 0.36, by definition. If q2 = 0.36, then q = 0.6, again by definition. Since q equals the frequency of the a allele, then the frequency is 60%.
   3. The frequency of the "A" allele. Answer: Since q = 0.6, and p + q = 1, then p = 0.4; the frequency of A is by definition equal to p, so the answer is 40%.
   4. The frequencies of the genotypes "AA" and "Aa." Answer: The frequency of AA is equal to p2, and the frequency of Aa is equal to 2pq. So, using the information above, the frequency of AA is 16% (i.e. p2 is 0.4 x 0.4 = 0.16) and Aa is 48% (2pq = 2 x 0.4 x 0.6 = 0.48).
   5. The frequencies of the two possible phenotypes if "A" is completely dominant over "a." Answers: Because "A" is totally dominate over "a", the dominant phenotype will show if either the homozygous "AA" or heterozygous "Aa" genotypes occur. The recessive phenotype is controlled by the homozygous aa genotype. Therefore, the frequency of the dominant phenotype equals the sum of the frequencies of AA and Aa, and the recessive phenotype is simply the frequency of aa. Therefore, the dominant frequency is 64% and, in the first part of this question above, you have already shown that the recessive frequency is 36%.
      
   
   
2. PROBLEM #2.
   Sickle-cell anemia is an interesting genetic disease. Normal homozygous individials (SS) have normal blood cells that are easily infected with the malarial parasite. Thus, many of these individuals become very ill from the parasite and many die. Individuals homozygous for the sickle-cell trait (ss) have red blood cells that readily collapse when deoxygenated. Although malaria cannot grow in these red blood cells, individuals often die because of the genetic defect. However, individuals with the heterozygous condition (Ss) have some sickling of red blood cells, but generally not enough to cause mortality. In addition, malaria cannot survive well within these "partially defective" red blood cells. Thus, heterozygotes tend to survive better than either of the homozygous conditions. If 9% of an African population is born with a severe form of sickle-cell anemia (ss), what percentage of the population will be more resistant to malaria because they are heterozygous (Ss) for the sickle-cell gene? Answer: 9% =.09 = ss = q2. To find q, simply take the square root of 0.09 to get 0.3. Since p = 1 - 0.3, then p must equal 0.7. 2pq = 2 (0.7 x 0.3) = 0.42 = 42% of the population are heterozygotes (carriers).
   
   
3. PROBLEM #3.
   There are 100 students in a class. Ninety-six did well in the course whereas four blew it totally and received a grade of F. Sorry. In the highly unlikely event that these traits are genetic rather than environmental, if these traits involve dominant and recessive alleles, and if the four (4%) represent the frequency of the homozygous recessive condition, please calculate the following:
   
   
   1. The frequency of the recessive allele. Answer: Since we believe that the homozygous recessive for this gene (q2) represents 4% (i.e. = 0.04), the square root (q) is 0.2 (20%).
   2. The frequency of the dominant allele. Answer: Since q = 0.2, and p + q = 1, then p = 0.8 (80%).
   3. The frequency of heterozygous individuals. Answer: The frequency of heterozygous individuals is equal to 2pq. In this case, 2pq equals 0.32, which means that the frequency of individuals heterozygous for this gene is equal to 32% (i.e. 2 (0.8)(0.2) = 0.32).
      
4. PROBLEM #4.
   Within a population of butterflies, the color brown (B) is dominant over the color white (b). And, 40% of all butterflies are white. Given this simple information, which is something that is very likely to be on an exam, calculate the following:
   
   
   1. The percentage of butterflies in the population that are heterozygous.
   2. The frequency of homozygous dominant individuals.
      Answers: The first thing you'll need to do is obtain p and q. So, since white is recessive (i.e. bb), and 40% of the butterflies are white, then bb = q2 = 0.4. To determine q, which is the frequency of the recessive allele in the population, simply take the square root of q2 which works out to be 0.632 (i.e. 0.632 x 0.632 = 0.4). So, q = 0.63. Since p + q = 1, then p must be 1 - 0.63 = 0.37. Now then, to answer our questions. First, what is the percentage of butterflies in the population that are heterozygous? Well, that would be 2pq so the answer is 2 (0.37) (0.63) = 0.47. Second, what is the frequency of homozygous dominant individuals? That would be p2 or (0.37)2 = 0.14.
      
5. PROBLEM #5.
   A rather large population of Biology instructors have 396 red-sided individuals and 557 tan-sided individuals. Assume that red is totally recessive. Please calculate the following:
   
   
   1. The allele frequencies of each allele. Answer: Well, before you start, note that the allelic frequencies are p and q, and be sure to note that we don't have nice round numbers and the total number of individuals counted is 396 + 557 = 953. So, the recessive individuals are all red (q2) and 396/953 = 0.416. Therefore, q (the square root of q2) is 0.645. Since p + q = 1, then p must equal 1 - 0.645 = 0.355.
   2. The expected genotype frequencies. Answer: Well, AA = p2 = (0.355)2 = 0.126; Aa = 2(p)(q) = 2(0.355)(0.645) = 0.458; and finally aa = q2 = (0.645)2 = 0.416 (you already knew this from part A above).
   3. The number of heterozygous individuals that you would predict to be in this population. Answer: That would be 0.458 x 953 = about 436.
   4. The expected phenotype frequencies. Answer: Well, the "A" phenotype = 0.126 + 0.458 = 0.584 and the "a" phenotype = 0.416 (you already knew this from part A above).
   5. Conditions happen to be really good this year for breeding and next year there are 1,245 young "potential" Biology instructors. Assuming that all of the Hardy-Weinberg conditions are met, how many of these would you expect to be red-sided and how many tan-sided? Answer: Simply put, The "A" phenotype = 0.584 x 1,245 = 727 tan-sided and the "a" phenotype = 0.416 x 1,245 = 518 red-sided ( or 1,245 - 727 = 518).
      
6. PROBLEM #6.
   A very large population of randomly-mating laboratory mice contains 35% white mice. White coloring is caused by the double recessive genotype, "aa". Calculate allelic and genotypic frequencies for this population. Answer: 35% are white mice, which = 0.35 and represents the frequency of the aa genotype (or q2). The square root of 0.35 is 0.59, which equals q. Since p = 1 - q then 1 - 0.59 = 0.41. Now that we know the frequency of each allele, we can calculate the frequency of the remaining genotypes in the population (AA and Aa individuals). AA = p2 = 0.41 x 0.41 = 0.17; Aa = 2pq = 2 (0.59) (0.41) = 0.48; and as before aa = q2 = 0.59 x 0.59 = 0.35. If you add up all these genotype frequencies, they should equal 1.
   
   
7. PROBLEM #7.
   After graduation, you and 19 of your closest friends (lets say 10 males and 10 females) charter a plane to go on a round-the-world tour. Unfortunately, you all crash land (safely) on a deserted island. No one finds you and you start a new population totally isolated from the rest of the world. Two of your friends carry (i.e. are heterozygous for) the recessive cystic fibrosis allele (c). Assuming that the frequency of this allele does not change as the population grows, what will be the incidence of cystic fibrosis on your island? Answer: There are 40 total alleles in the 20 people of which 2 alleles are for cystic fibrous. So, 2/40 = .05 (5%) of the alleles are for cystic fibrosis. That represents p. Thus, cc or p2 = (.05)2 = 0.0025 or 0.25% of the F1 population will be born with cystic fibrosis.
   
   
8. PROBLEM #8.
   You sample 1,000 individuals from a large population for the MN blood group, which can easily be measured since co-dominance is involved (i.e., you can detect the heterozygotes). They are typed accordingly:
   
   

   N	NN	90	0.09
   MN	MN	420	0.42
   M	MM	490	0.49
   BLOOD TYPE	GENOTYPE	NUMBER OF INDIVIDUALS	RESULTING FREQUENCY

   
   Using the data provide above, calculate the following:
   
   
   1. The frequency of each allele in the population. Answer: Since MM = p2, MN = 2pq, and NN = q2, then p (the frequency of the M allele) must be the square root of 0.49, which is 0.7. Since q = 1 - p, then q must equal 0.3.
   2. Supposing the matings are random, the frequencies of the matings. Answer: This is a little harder to figure out. Try setting up a "Punnett square" type arrangement using the 3 genotypes and multiplying the numbers in a manner something like this:
      
      

      NN (0.09)	0.0441	0.0378	0.0081*
      MN (0.42)	0.2058	0.1764*	0.0378
      MM (0.49)	0.2401*	0.2058	0.0441
      	MM (0.49)	MN (0.42)	NN (0.09)

      Note that three of the six possible crosses are unique (*), but that the other three occur twice (i.e. the probabilities of matings occurring between these genotypes is TWICE that of the other three "unique" combinations. Thus, three of the possibilities must be doubled.
      MM x MM = 0.49 x 0.49 = 0.2401
      MM x MN = 0.49 x 0.42 = 0.2058 x 2 = 0.4116
      MM x NN = 0.49 x 0.09 = 0.0441 x 2 = 0.0882
      MN x MN = 0.42 x 0.42 = 0.1764
      MN x NN = 0.42 x 0.09 = 0.0378 x 2 = 0.0756
      NN x NN = 0.09 x 0.09 = 0.0081
      
      
   3. The probability of each genotype resulting from each potential cross. Answer: You may wish to do a simple Punnett's square monohybrid cross and, if you do, you'll come out with the following result:
      MM x MM = 1.0 MM
      MM x MN = 0.5 MM 0.5 MN
      MM x NN = 1.0 MN
      MN x MN = 0.25 MM 0.5 MN 0.25 NN
      MN x NN = 0.5 MN 0.5 NN
      NN x NN = 1.0 NN
      
9. PROBLEM #9.
   Cystic fibrosis is a recessive condition that affects about 1 in 2,500 babies in the Caucasian population of the United States. Please calculate the following.
   
   
   1. The frequency of the recessive allele in the population. Answer: We know from the above that q2 is 1/2,500 or 0.0004. Therefore, q is the square root, or 0.02. That is the answer to our first question: the frequency of the cystic fibrosis (recessive) allele in the population is 0.02 (or 2%).
   2. The frequency of the dominant allele in the population. Answer: The frequency of the dominant (normal) allele in the population (p) is simply 1 - 0.02 = 0.98 (or 98%).
   3. The percentage of heterozygous individuals (carriers) in the population. Answer: Since 2pq equals the frequency of heterozygotes or carriers, then the equation will be as follows: 2pq = (2)(.98)(.02) = 0.04 or 1 in 25 are carriers.
      
10. PROBLEM #10.
    In a given population, only the "A" and "B" alleles are present in the ABO system; there are no individuals with type "O" blood or with O alleles in this particular population. If 200 people have type A blood, 75 have type AB blood, and 25 have type B blood, what are the alleleic frequencies of this population (i.e., what are p and q)? Answer: To calculate the allele frequencies for A and B, we need to remember that the individuals with type A blood are homozygous AA, individuals with type AB blood are heterozygous AB, and individuals with type B blood are homozygous BB. The frequency of A equals the following: 2 x (number of AA) + (number of AB) divided by 2 x (total number of individuals). Thus 2 x (200) + (75) divided by 2 (200 + 75 + 25). This is 475/600 = 0.792 = p. Since q is simply 1 - p, then q = 1 - 0.792 or 0.208.
    
    
11. PROBLEM #11.
    The ability to taste PTC is due to a single dominate allele "T". You sampled 215 individuals in biology, and determined that 150 could detect the bitter taste of PTC and 65 could not. Calculate all of the potential frequencies. Answer: First, lets go after the recessives (tt) or q2. That is easy since q2 = 65/215 = 0.302. Taking the square root of q2, you get 0.55, which is q. To get p, simple subtract q from 1 so that 1 - 0.55 = 0.45 = p. Now then, you want to find out what TT, Tt, and tt represent. You already know that q2 = 0.302, which is tt. TT = p2 = 0.45 x 0.45 = 0.2025. Tt is 2pq = 2 x 0.45 x 0.55 = 0.495. To check your own work, add 0.302, 0.2025, and 0.495 and these should equal 1.0 or very close to it. This type of problem may be on the exam.
    
    
12. PROBLEM #12. (You will not have this type of problem on the exam)
    What allelic frequency will generate twice as many recessive homozygotes as heterozygotes? Answer: We need to solve for the following equation: q2 (aa) = 2 x the frequency of Aa. Thus, q2 (aa) = 2(2pq). Or another way of writing it is q2 = 4 x p x q. We only want q, so lets trash p. Since p = 1 - q, we can substitute 1 - q for p and, thus, q2 = 4 (1 - q) q. Then, if we multiply everything on the right by that lone q, we get q2 = 4q - 4q2. We then divide both sides through by q and get q = 4 - 4q. Subtracting 4 from both sides, and then q (i.e. -4q minus q = -5q) also from both sides, we get -4 = -5q. We then divide through by -5 to get -4/-5 = q, or anotherwards the answer which is 0.8 =q. I cannot imagine you getting this type of problem in this general biology course although if you take algebra good luck.
    
    

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