Chicks and hens flower

Sempervivum Picture Gallery A-D

Hens and Chicks; Have You Started Your Collection Yet?

These fascinating plants are members of the Crassulacaea, which also includes Echeveria, Sedum and Crassula. The many members of this tribe all have one trait in common; drought tolerance.

I can’t help myself – whenever I see one I don’t have yet, I have to have it!

The result is over 120 different ones, with more grown from seed.

They are versatile and tough garden plants, and also lend themselves to some really cool crafts where they adapt to challenging conditions, like being planted into a vertical wall mosaic, or covering a topiary.

What other plant can you think of that would not only survive this kind of treatment, but thrive on it?

Hens and chicks, stonecrop and many other hardy succulents are healthy and happy in the same kinds of conditions; bright light, well drained soil, and benign neglect.

This gallery of four pages shows these most intriguing plants listed alphabetically – click on the links at the bottom of the page for more.


Have a look at some other photographs of the Sempervivum on this page showing the dramatic colour changes that they undergo throughout the year: (Don’t miss the other pages in the Sempervivum Picture Gallery linked at the bottom of this page).

Hover your mouse over the photos to get a better view


Sempervivum List

Sempervivum Picture Gallery E-L

Sempervivum Picture Gallery M-P

Sempervivum Picture Gallery R-Z

Hens and Chicks Flowers: Do Hens And Chicks Plants Bloom

Hens and chicks have old-time charm and unbeatable hardiness. These little succulents are known for their sweet rosette form and numerous offsets or “chicks.” Do hens and chicks plants bloom? The answer is yes, but it spells demise for the flowering rosette in a life cycle that is unique among plants. Hens and chicks flowers are the plants way of producing seed and a new generation of beguiling succulents.

When Do Hens and Chicks Plants Bloom?

A rambunctious clump of hens and chicks have special allure to children and adults alike. The small plants are adaptable and resilient, producing flower-like clusters of varying sized rosettes. Gardeners new to the plants may say, “My hens and chicks are flowering,” and wonder if this is a natural occurrence. Blooms on hens and chicks plants are not only natural but an additional wonder with this fun, diminutive Sempervivum.

I love to walk the garden and see that my hens and chicks are flowering. This generally occurs in summer when the long warm days and bright light jar the plant’s instincts to form blooms. This signals the beginning or end of the

plant’s life cycle, depending upon whether you are a glass half empty or glass half full kind of gardener.

Hens will usually live for 3 years before they form flowers but, occasionally, stressed plants will bloom earlier. The tiny, starry flowers amp up the magic of these succulents, but it does mean the plant is forming seed and will die. Not to despair, though, because the lost plant will quickly fill in with a new rosette and the cycle will march on yet again.

About Hens and Chicks Flowers

A blooming hen on a hen and chicks plant is often referred to as a “rooster.” The individual rosettes will begin to elongate and lengthen vertically when it is time to produce flowers. The process lends an alien appearance to the normally low-growing plants, with flower stalks that can get from a few inches up to a foot in length.

Removing the budding stem can’t save the rosette. The blooms on hens and chicks plants are a part of a monocarpic process. That means they flower, seed and then die. There is nothing to be done about it so you might as well enjoy the pink, white or yellow flowers with bristling, erect stamen.

Their work will soon be done, but the plant should already have produced many smaller rosettes, the future of the line.

Hens and Chicks Flower Care

As with the entire plant, hens and chicks flower care consists of neglect. You can leave the bloom until it has finished and the stem and base rosette will dry out and die.

Clip off the stem rather than pulling it out of the living cluster or you may end up yanking some of the precious offsets. You may also choose to let nature take its course and leave the dying stem as proof of an interesting life cycle, which will eventually break off and compost in the area.

The young chicks will grow larger and fill in any gaps the parent plant made when bidding its fond farewell to this world. So enjoy the flowers and the guarantee of everlasting life this plant has in its offspring.

Problem 1
A rooster with gray feathers is mated with a hen of the same phenotype. Among their offspring, 15 chicks are gray, 6 are black, and 8 are white.

What is the simplest explanation for the inheritance of these colors in chickens?

Incomplete or codominance. Feather color is controlled by 2 genes B = black and b = white. The third phenotype is the result of a 50-50 mix of black and white to produce gray.

The 15 gray, 6 black, and 8 white birds represent a 2:1:1 ratio&emdash;the result of mating two heterozygous individuals: (Bb x Bb)

1 BB : 2 Bb : 1 bb

What offspring would you predict from the mating of a gray rooster and a black hen?

A gray rooster (Bb ) mated to a black hen (BB ) can be represented by the following Punnett square:

50% of the offspring should be gray (Bb ) and 50% black (BB )

Problem 2
In some plants, a true-breeding, red-flowered strain gives all pink flowers when crossed with a white-flowered strain: RR (red) x rr (white) —> Rr (pink).

If flower position (axial or terminal) is inherited as it is in peas what will be the ratios of genotypes and phenotypes of the generation resulting from the following cross: axial-red (true-breeding) x terminal-white?

Note: Axial (A ) is dominant over terminal (a ).

The genotypes of the parents are AARR and aarr. Therefore the gametes of the parents must be AR and ar so the genotype for all the offspring in the F1 generation will be AaRr, and their phenotype will be axial-pink.

What will be the ratios in the F2 generation?

The ratio of genotypes can be determined by examining the Punnett square below:

The ratio of phenotypes will be:

6 axial-pink 8 pink
3 axial-red 4 red
3 axial-white 4 white
2 terminal-pink 12 axial
1 terminal-white 4 terminal
1 terminal-red

Problem 3

Flower position, stem length, and seed shape were three characters that Mendel studied. Each is controlled by an independently assorting gene and has dominant and recessive expression as follows:

Character Dominant Recessive
Flower position Axial (A ) Terminal (a )
Stem length Tall (T ) Dwarf (t )
Seed shape Round (R ) Wrinkled (r)

If a plant that is heterozygous for all three characters were allowed to self-fertilize, what proportion of the offspring would be expected to be as follows: (Note – use the rules of probability (and show your work) instead of huge Punnett squares)

a) homozygous for the three dominant traits

AATTRR = 1/4 x 1/4 x 1/4 = 1/64

b) homozygous for the three recessive traits

aattrr = 1/4 x 1/4 x 1/4 = 1/64

c) heterozygous (assumed for each trait)

AaTtRr = 1/2 x 1/2 x 1/2 = 1/8

d) homozygous for axial and tall, heterozygous for seed shape

AATTRr = 1/4 x 1/4 x 1/2 = 1/32

Problem 4
A black guinea pig crossed with an albino guinea pig produced 12 black offspring. When the albino was crossed with a second one, 7 blacks and 5 albinos were obtained.

What is the best explanation for this genetic situation?

Black is dominant over white

Write genotypes for the parents, gametes, and offspring.

First cross:

Parent’s genotypes = BB (black) x bb (white)
gametes = B b
F1 offspring = all Bb

Second cross:

Parent’s genotypes = Bb (black) x bb (white)
gametes = B or b b
F1 offspring = Bb or bb

There should be 50% black to 50% white offspring in this cross.

Problem 5
In sesame plants, the one-pod condition (P ) is dominant to the three-pod condition (p ), and normal leaf (L ) is dominant to wrinkled leaf (l) . Pod type and leaf type are inherited independently. Determinine the genotypes for the two parents for all possible matings producing the following offspring:

a. 318 one-pod normal, 98 one-pod wrinkled

Parental genotypes: PPLl x PPLl or PpLl x PPLl

b. 323 three-pod normal, 106 three-pod wrinkled

Parental genotypes: ppLl x ppLl

c. 401 one-pod normal

Parental genotypes: PPLL x PpLL or PPLl x PPLL or PPLL x PpLl etc (nine possible genotypes).

d. 150 one-pod normal, 147 one-pod wrinkled, 51 three-pod normal, 48 three-pod wrinkled. (a 3: 3: 1: 1 ratio)

Parental genotypes: PpLl x Ppll (see below for details)

3 One-pod normal (PPLl , PpLl , PpLl)

3 One-pod wrinkled (PPll , Ppll , Ppll)

1 Three-pod normal (ppLl)

1 Three-pod wrinked (ppll)

e. 223 one-pod normal, 72 one-pod wrinkled, 76 three-pod normal, 27 three-pod wrinkled (a 9: 3: 3: 1 ratio)

Parental genotypes: PpLl x PpLl

Problem 6
A man with group A blood marries a woman with group B blood. Their child has group O blood. What are the genotypes of these individuals?

Father = AO (or IAi)

Mother = BO (or IBi)

First Child = OO (or ii)

What other genotypes and in what frequencies, would you expect in offspring from this marriage?

Examine the Punnett square to determine the other genotypes possible.

The other genotypes for children are (according to Campbell’s system): 1/4 IAIB, 1/4 IAi, 1/4 IBi

Problem 7
Color pattern in a species of duck is determined by one gene with three alleles. Alleles H and I are codominant, and allele i is recessive to both. Note: this situation is similar to the ABO blood system.

How many phenotypes are possible in a flock of ducks that contains all the possible combinations of these three alleles?

As in the ABO blood system 4 phenotypes are possible in this case:

Genotype Phenotype
HH, Hi (H)
II, Ii (I)
ii (i)

Problem 8
Phenylketonuria (PKU) is an inherited disease caused by a recessive allele. If a woman and her husband are both carriers, what is the probability of each of the following?

Under these circumstances assume the following Punnett square to be true.

Where NN or Nn = normal conditions and nn = PKU

a. all three of their children will be of normal phenotype

3/4 x 3/4 x 3/4 = 27/64

b. one or more of the three children will have the disease (x)

1 – 27/64 = 37/64

All three have x 2 out of 3 has x 1 out of 3 has x
+ + =
x x o o o x
3 Combinations x o x o x o
o x x x o o
+ 3(3/4 x 1/4 x 1/4) + 3(3/4 x 3/4 x 1/4) =

Note: the probability of the disease (x) = 1/4 & the probability of being normal (o) = 3/4

c. all three children will have the disease

1/4 x 1/4 x 1/4 = 1/64

d. at least one child out of three will be phenotypically normal

(Note: Remember that the probabilities of all possible outcomes always add up to 1)

1 – 1/64 = 63/64

Problem 9
The genotype of F1 individuals in a tetrahybrid cross is AaBbCcDd. Assuming independent assortment of these four genes, what are the probabilities that F2 offspring would have the following genotypes?

a. aabbccdd = x x x = 1/256

b. AaBbCcDd = x x x = 1/16

c. AABBCCDD = x x x = 1/256

d. AaBBccDd = x x x = 1/64

e. AaBBCCdd = x x x = 1/128

Just remember that the probability of a heterozygote (Xx) = 2/4 or 1/2 and the probability of a homozygote XX or xx = 1/4

Problem 10
In 1981, a stray black cat with unusual rounded curled-back ears was adopted by a family in California. Hundreds of descendants of the cat have since been born, and cat fanciers hope to develop the “curl” cat into a show breed. Suppose you owned the first curl cat and wanted to develop a true breeding variety.

How would you determine whether the curl allele is dominant or recessive?

Mate the stray to a non-curl cat. If any offspring have the “curl” trait it is likely to be dominant. If the mutation is recessive, then on ly non-curl offspring will result.

How would you select for true-breeding cats?

You know that cats are true-breeding when curl crossed with curl matings produce only curl offspring.

How would you know they are true-breeding?

A pure-bred “curl cat” is homozygous.

  1. If the trait is recessive any inividual with the “curl” condition is homozygous recessive.
  2. If the trait is dominant you can determine if the individual in question is true breeding (CC) or heterozygous (Cc) with a test cross (to a homozygous recessive individual).

Problem 11
What is the probability that each of the following pairs of parents will produce the indicated offspring (assume independent assortment of all gene pairs?

a. AABBCC x aabbcc —-> AaBbCc

(1)(1)(1) = 1

b. AABbCc x AaBbCc —–> AAbbCC

()()() = 1/32

c. AaBbCc x AaBbCc —–> AaBbCc

()()() = 1/8

d. aaBbCC x AABbcc —-> AaBbCc

(1)()(1) = 1/2
Problem 12
Karen and Steve each have a sibling with sickle-cell disease. Neither Karen, Steve, nor any of their parents has the disease, and none of them has been tested to reveal sickle-cell trait. Based on this incomplete information, calculate the probability that if this couple should have another child, the child will have sickle-cell anemia. In order for Karen and Steve to have siblings with sickle cell anemia their parents must be carriers (Nn). We also know that John and Carol are not homozygous recessive (nn) because they do not have the disease. Therefore the chance that Karen is a carrier is 2/3 (NN, Nn, nN) and the chance that Steve is a carrier is also 2/3. If they have a child and both Karen and Steve are carriers then the child has one chance in 4 of having sickle cell anemia. Since each event is independent of one another the overall probability of the child having sickle cell anemia is:

2/3 x 2/3 x 1/4 = 1/9.

Problem 13
Imagine that a newly discovered, recessively inherited disease is expressed only in individuals with type O blood, although the disease and blood group are independently inherited. A normal man with type A blood and a normal woman with type B blood have already had one child with the disease. The woman is now pregnant for a second time. What is the probability that the second child will also have the disease? Assume both parents are heterozygous for the “disease” gene.

Father AO
Mother BO

OO expresses the disease

The second child’s chance of having the disease is = x = 1/16

Problem 14
In tigers, a recessive allele causes an absence of fur pigmentation (a “white tiger”) and a cross-eyed condition. If two phenotypically normal tigers that are heterozygous at this locus are mated, what percentage of their offspring will be cross-eyed? What percentage will be white?

Using the Punnett square below where P = normal pigmenation and p = white

then 25% will be white (pp) and all of the white offspring will also be cross-eyed

Problem 15
In corn plants, a dominant allele I inhibits kernel color, while the recessive allele i permits color when homozygous. At a different locus, the dominant gene P causes purple kernel color, while the homozygous recessive genotype pp causes red kernels. If plants heterozygous at both loci are crossed, what will be the phenotypic ratio of the F1 generation? Phenotypic ratios: White (I _ _ _) = 12

Purple (i i P_ ) = 3

Red ( i i p p) = 1

The dominant allele I is epistatic to the p locus, and thus the F1 generation will be:

9 I_P_ : colorless

3 I_pp : colorless

3 i iP_ : purple

1 i i pp: red

Problem 16
The pedigree below traces the inheritance of alkaptonuria, a biochemical disorder. Affected individuals, indicated here by the filled-in circles and squares, are unable to break down a substance called alkapton, which colors the urine and stains body tissues. Does alkaptonuria appear to be caused by a dominant or recessive allele?


Fill in the genotypes of the individuals whose genotypes you know. What genotypes are possible for each of the other individuals?

If alkaptonuria is recessive George must be a carrier. See below.

If alkaptonuria is dominant Carla could not have the disease, as indicated in the pedigree chart, since the parents do not express the trait. See Below.

Problem 17
A man has six fingers on each hand and six toes on each foot. His wife and their daughter have the normal number of digits (5). Extra digits is a dominant trait. What fraction of this couple’s children would be expected to have extra digits?

Because the daughter is normal the man’s genotype must be heterozygous for the trait so:

if X = extra digits and x = normal (5) digits then:

50% of the offspring will be polydactylic

Problem 18
Imagine you are a genetic counselor, and a couple planning to start a family came to you for information. Charles was married once before, and he and his first wife had a child who has cystic fibrosis. The brother of his current wife Elaine died of cystic fibrosis. Cystic Fiborsis is a lethal recessive condition (a person with CF cannot have children).

What is the probability that Charles and Elaine will have a baby with cystic fibrosis? (Neither Charles nor Elaine has the disease)

The Probability that Elaine is a carrier is 2/3 (She does not have Cystic Fibrosis which eliminates one of the 4 possibilities. She does have 2 chances out of three of being a carrier.

The probability that the baby (?) has the disease (if Elaine is a carrier) is 1/4

The total probability is 2/3 x 1/4 or 1/6.

Problem 19
In mice, black color (B ) is dominant to white (b ). At a different locus, a dominant allele (A ) produces a band of yellow just below the tip of each hair in mice with black fur. This gives a frosted appearance known as agouti. Expression of the recessive allele (a ) results in a solid coat color. If mice that are heterozygous at both loci are crossed, what will be the expected phenotypic ratio of their offspring?

B = Black – dominant

A = Agouti – dominant

BbAa x BbAa =

Genotype Phenotype
BBAA agouti
BbAA agouti
BBAa agouti
BbAa agouti
BBaa black
Bbaa black
bbAA white
bbAa white
bbaa white

The phenotypic ratio is:

9 agouti: 4 white: 3 black

Problem 20
The pedigree below traces the inheritance of a vary rare biochemical disorder in humans. Affected individuals are indicated by filled-in circles and squares. Is the allele for this disorder dominant or recessive?

The allele is most likely dominant because the #2 individual (see below) with the trait marries a woman with the trait and 50% of their offspring are normal. If the trait were recessive one would expect the following:

100% of offspring would have the disease, which is not the case.

What genotypes are possible for the individuals marked 1, 2, and 3?

  1. Bb (heterozygous)
  2. Bb ( if 2’s genotype were bb he would not have the disease and if BB all his children would have the condition.)
  3. bb (all normal individuals are homozygous recessive)

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Sempervivum (Hens and Chicks) and Sempervivum heuffelii are resilient perennial succulents that make striking outdoor plantings. But just like an Agave century plant that send up a spike and goes to seed, a flowering semp rosette is at the end of its life. Below, we’ll show you the ways to manage blooming Sempervivum and keep your succulents alive longer.

Flowers of Sempervivum ‘Fame’

What happens when a Sempervivum blooms?

Semps can produce vegetative offsets (chicks) every year, so they do not need to flower to reproduce new seedlings annually. They are, however, a monocarpic plant, meaning that any rosette that does flower and go to seed will die. Think of it as the plant’s swan song at the end of its life as it makes way for a new generation.

Hens & Chicks are not typical grown for their flowers, but the blooms are small and come in shades of white, yellow, or pink. If your intention is to hybridize or grow Sempervivum from seed, then blooms are a welcome sight, but for many gardeners, flowering is cause to worry. Dead rosettes leave gaps in otherwise tidy Sempervivum clumps and a solitary, flowering mother hen with no chick offsets will be the last of her colony.

Lone, blooming Sempervivum ‘Kip’

What triggers a rosette to go to seed?

Flowering is an unpredictable process in Sempervivum. One year there’ll be none, and in another, a whole colony can bolt to produce seeds. It’s hard to know exactly what is signaling to the plants that the end is near and they need to propagate, because any form of stress can induce flowering. Drastic changes in light, temperature, and water can threaten a rosette’s survival and thus make it divert its resources into producing the next generation of seedlings.

All Sempervivum will eventually bloom after several years of growth, whether induced by stress or simply age. Signs that a rosette is preparing to flower include a closing in of the central leaves, stretching of the main stem, and tilting of the whole rosette.

Sempervivum ‘Red Wings’ preparing to bloom

How do you prevent blooming?

You can minimize stress to reduce the incidence of blooming, but all rosettes will eventually bloom regardless. When this happens, you need to remove the bloom stalk as soon as it is noticeable. If you catch it too late, the stem will be tougher and the hormones that initiate flowering will already be at work. From our experience, the success of preventing flowering is highest when the bloom stalk is less than four inches tall.

Proceed confidently through the following surgical steps, knowing that an untreated rosette will certainly die and you’ve got nothing to lose.

  1. Separate the tighter, upturned leaves of the developing bloom from the low-lying layers of leaves at the base of the rosette.
  2. Using a sharp, clean knife, scoop the bloom stalk out from the rosette.
  3. Continue your normal schedule of watering and care while checking weekly for new growth.


This procedure is your last chance to get new chicks from a blooming Sempervivum rosette. It will not return you to the single, perfectly symmetrical mother hen you started with, but it can give you more offsets to transplant. Even if you follow this guide perfectly, some rosettes will keep trying to flower. You can continue to remove new bloom stalks, but be sure not to confuse them with emerging chicks. If you spot new offsets forming within the rosette, congratulations! You have successfully saved the rosette. If not, any non-flowering rosettes in the surrounding colony will survive and grow to fill the gap left by the one that went to seed and died.

From top to bottom: Sempevivum ‘Starshine’ preparing to bloom; recently cut rosette; one month after treatment with offsets forming; new offsets two months after treatment.


  • Sempervivum can take years to bloom, but a flowering rosette will die if it goes to seed.
  • Look for signs of elongation to catch a bloom stalk as soon as it begins to form.
  • Use a sharp, clean knife to cut out the tighter leaves on the developing bloom stalk.
  • Continue to cut out flowers and watch for offsets that you can cultivate or transplant.

We wish you all the best in your efforts to recover blooming Sempervivum rosettes. If all else fails, there’s alway the next batch of chicks!


Blooming semps happen to everyone

For more information on everything Sempervivum, from blooms to hybrids, check out Kevin Vaughn’s new book, Sempervivum: A Gardener’s Perspective of the Not-So-Humble Hens & Chicks.

Hen and Chicks … Blooming?

Last week, as I was walking into the office one morning, I suddenly noticed what appeared to be a strange swelling coming up out of the center of one of the succulents, commonly called “hen and chicks” (Sempervivium spp.) I decided to keep an eye on it as I passed daily since the plants were clustered at the sidewalk by the entrance.

After a couple of days, it began to appear to me that it was going to be a stalk of some kind. In rapt amazement, I watched as I saw flower buds form on that stalk, which was about 6” tall. In all of my plant gawking years, I have never seen such a sight! That hen and chick plant was going to bloom and I had no idea that it could do such a thing!

I then started receiving calls from other concerned citizens asking what was going on with their “hen and chicks” plants and they wondered if they needed to cut off the weird growth. When I explained what I thought was going to happen, every single person asked me if I was sure that something wasn’t wrong with the plant. Everyone informed me that these plants were at least 20 years old or more and they had never, ever flowered! So it was with a great sigh of relief that I finally observed my first “bloomin’ event” on the hen and chicks succulents. It was worth the wait!

Blooms of the succulent known as “Hen and Chicks” ( Sempervivium spp.)

My theory as to why this is happening this year and not in any other previous years… I think we have had the conditions which favor succulents! Namely, extremely dry to droughty conditions and very warm (upper 80’s & low 90’s) temperatures, accompanied by clear skies and lots of sunshine. Who could have guessed that such a beautiful result would come from such a parched, hot summer in Northeast Ohio!!!

How to Grow Hens and Chicks

Hen & Chick Plant Propagation
Hens and Chicks produce numerous offspring, thus allowing them to “live forever”. The quantity and speed at which babies are produced depends on the variety. Sempervivums can be divided anytime during the spring/summer growing season. The baby chicks can be re-planted elsewhere or left to grow around the mother hen.

There are three common types of Hens and Chicks: Sempervivum, Jovibarba heuffelii and Jovibarba Rollers. They are all often called Sempervivum, but each type produces offspring in a different manner.

These grow babies on runners. Just pull off the chicks and plant elsewhere. It is best to remove the babies when the runner has begun to wither. Offsets root quickly and contact with soil is enough for them to start growing.
Jovibarba heuffelii
This species does not produce “chicks” on stolons. Instead the offspring of this plant are produced within the mother plant. To propagate it must be split with a knife.

Jovibarba Rollers
These types of Hens and Chicks produce lightly attached “chicks” that easily pop off and roll away from the mother plant.

Growing from the offsets preserves the characteristics of each cultivar. Seeds taken from the Sempervivum flowers generally produce plants that are untrue to type.

Sempervivum Life and Death Cycle

Once a hen plant produces a chick, that chick will begin producing its own babies after only 1 season. Sempervivum plants generally only live for 3 years, so the plants have 2 productive years before they die. After 3 years and having produced many baby plants a Sempervivum grows a tall center stalk that blooms before the plant dies. Cutting off the center stalk will not prevent the plant from dying.

It is extremely fun to grow Hens and Chicks and watch them mature and produce offsets. Their colors change drastically throughout the season due to maturity, temperatures, sunlight exposure, and other factors. Be sure to give your plants enough space to spread. Ideally they should have 4” for small plants and 6-8” for large varieties. Adequate space produces nicely formed rosettes.

If I’m about to tell you something you already know, excuse me. Maybe I just haven’t thought about hens and chicks (


) all that much. Which is weird since they are so common. But in the past several years (I can’t say exactly since I tend to underestimate passing time; perhaps that’s a defense mechanism), I’ve been trying to fill in a broken-concrete retaining wall that faces the burning-hot south sun. It’s been quite the process. Some of the cracks are filled with enough soil that plants like

creeping thyme, wire vine (


) and saxifrage have been able to get a good start. In other pockets, I’ve poured in soil mix and it just keep falling and falling. As I do this, I wonder where it’s going. If there are cracks that deep, won’t the wall collapse eventually? I try not to think about that.

Through trial and error, I’ve learned that if I stuff sphagnum moss into the hole around the roots and keep it in place with florist pins (or, bobby pins if I’m desperate), the plant

will have a much better chance of surviving. Of course, planting in spring or fall when it’s not hot is a good strategy, too, but that’s asking a little much of my impatient nature.

Hens and chicks have been one of my successes, though more at the top of the wall than in the actual cracks. They are spreading downward, but it’s a slow process. Which may be the reason I haven’t thought much about why they die out in the center. OK, I know you’re thinking that I’m dumb. And, about this, I’ll agree. But, in my defense, it doesn’t naturally follow that just because the mom has babies, she should die. If fact, that’s just wrong.

So, I took matters into my own hands. Since the mom has its chicks, blooms and then dies, I decided to behead it. Extreme, perhaps, but no more so than natural death in this case. Quicker, I’d say. As the flower stalks, which are pretty cool looking, began to arise from the hen, I snapped them off. I had some luck. My friends who have more botany knowledge than I tell me I’m goofy and that, if it hasn’t yet, the hen’s going to die soon. I don’t care. At least I tried.

Besides the whole idea of death at what I consider the peak of life, I don’t like plants that die in the center. So, my new solution is to plant a chick in the middle. I’ll let you know how it goes.

One more interesting fact, hens and chicks got their name, according to Yvonne Cave, author of “Succulents for the Contemporary Garden,” because of “the custom of planting the ridges of thatched roofs with these succulents to protect the houses from lightning (semper means forever or always and vivum means alive).” I don’t really get that, but whatever.

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